Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 18 Tangents And Intersecting Chords

Exercise 18A

Question 1. The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?
Answer:
[Diagram: A circle with center O, radius 8 cm, and a tangent PT drawn from external point P which is 10 cm from center O.]

Given:
OP = 10 cm, radius OT = 8 cm

Step 1: Since PT is tangent to the circle, OT ⊥ PT

Step 2: In right triangle OTP, apply Pythagoras theorem.
\( OP^2 = OT^2 + PT^2 \)

Step 3: Substitute the values.
\( 10^2 = 8^2 + PT^2 \)
\( 100 = 64 + PT^2 \)
\( PT^2 = 100 - 64 = 36 \)
\( PT = 6 \)

Length of tangent = 6 cm
In simple words: We made a right triangle using the center, the point where tangent touches, and the outside point. Then we used the formula to find the missing side.

📝 Teacher's Note: Draw a clear diagram first. Show students that tangent always makes 90° with radius. This makes it a right triangle problem.

🎯 Exam Tip: Always write "tangent ⊥ radius" and use Pythagoras theorem. Show all steps clearly to get full marks.

Question 2. In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Answer:
[Diagram: A circle with center O, tangent AB at point B, where AB = 15 cm and AC = 7.5 cm.]

Given:
AB = 15 cm, AC = 7.5 cm
Let 'r' be the radius of the circle.

Step 1: Since OC = OB = r (radii)
AO = AC + OC = 7.5 + r

Step 2: In triangle AOB, apply Pythagoras theorem.
\( AO^2 = AB^2 + OB^2 \)

Step 3: Substitute the values.
\( (7.5 + r)^2 = 15^2 + r^2 \)

Step 4: Expand and solve.
\( \left(\frac{15 + 2r}{2}\right)^2 = 225 + r^2 \)
\( 225 + 4r^2 + 60r = 900 + 4r^2 \)
\( 60r = 675 \)
\( r = 11.25 \text{ cm} \)

Therefore, r = 11.25 cm
In simple words: We used the fact that all radii are equal. Then we made a right triangle and solved using Pythagoras theorem to find the radius.

📝 Teacher's Note: Emphasize that all radii of a circle are equal. Students often forget this basic property when solving.

🎯 Exam Tip: Always mark equal radii in your diagram. Write the equation clearly and solve step by step to avoid calculation errors.

Question 3. Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
Answer:
[Diagram: Two circles touching externally at point P, with Q on the common tangent through P, and tangents QA and QB drawn to the circles.]

Proof:
From Q, QA and QP are two tangents to the circle with centre O
Therefore, QA = QP ......(i)

Similarly, from Q, QB and QP are two tangents to the circle with centre O'
Therefore, QB = QP ......(ii)

From (i) and (ii)
QA = QB

Therefore, tangents QA and QB are equal.
In simple words: Both tangents QA and QB are equal to QP. So they must be equal to each other. This works because tangents from the same outside point are always equal.

📝 Teacher's Note: Remind students that tangents from an external point to a circle are always equal. This is the key property used in this proof.

🎯 Exam Tip: State the tangent property clearly: "Tangents from external point are equal." This gets you marks even if calculation goes wrong.

Question 4. Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Answer:
[Diagram: Two circles touching internally with Q on common tangent, and tangents QA and QB drawn to both circles.]

Proof:
From Q, QA and QP are two tangents to the circle with centre O
Therefore, QA = QP ......(i)

Similarly, from Q, QB and QP are two tangents to the circle with centre O'
Therefore, QB = QP ......(ii)

From (i) and (ii)
QA = QB

Therefore, tangents QA and QB are equal.
In simple words: Same logic as external touching circles. Both QA and QB equal QP, so they are equal to each other.

📝 Teacher's Note: Show that the proof is identical for internal and external touching. The tangent property works in both cases.

🎯 Exam Tip: Use the same method for internal and external touching circles. The property of equal tangents from external point applies in both cases.

Question 5. Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Answer:
[Diagram: Two concentric circles with radii 5 cm and 3 cm, showing a chord of outer circle that is tangent to inner circle.]

Given:
OS = 5 cm (radius of outer circle)
OT = 3 cm (radius of inner circle)

Step 1: In right triangle OST
By Pythagoras Theorem,
\( ST^2 = OS^2 - OT^2 \)

Step 2: Substitute values.
\( ST^2 = 25 - 9 = 16 \)
\( ST = 4 \text{ cm} \)

Step 3: Since OT is perpendicular to SP and OT bisects chord SP
So, SP = 2 × ST = 2 × 4 = 8 cm

Length of chord = 8 cm
In simple words: The line from center to chord makes a right triangle. We found half the chord length, then doubled it to get full chord length.

📝 Teacher's Note: Show that perpendicular from center bisects the chord. This is a key property students must remember for chord problems.

🎯 Exam Tip: Remember: perpendicular from center to chord bisects the chord. Always find half-length first, then double it.

Question 6. Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Answer:
[Diagram: Three circles with centers A, B, C touching externally, forming triangle ABC with sides 6 cm, 8 cm, and 9 cm.]

Given:
AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centers A, B and C be r₁, r₂ and r₃ respectively.

Step 1: When circles touch externally, distance between centers = sum of radii
r₁ + r₃ = 8
r₃ + r₂ = 9
r₂ + r₁ = 6

Step 2: Adding all equations
r₁ + r₃ + r₃ + r₂ + r₂ + r₁ = 8 + 9 + 6
2(r₁ + r₂ + r₃) = 23
r₁ + r₂ + r₃ = 11.5 cm

Step 3: Find individual radii
r₁ + 9 = 11.5 (Since r₂ + r₃ = 9)
r₁ = 2.5 cm

r₂ + 6 = 11.5 (Since r₁ + r₃ = 6)
r₂ = 5.5 cm

r₃ + 8 = 11.5 (Since r₂ + r₁ = 8)
r₃ = 3.5 cm

Hence, r₁ = 2.5 cm, r₂ = 5.5 cm and r₃ = 3.5 cm
In simple words: When circles touch outside, the distance between centers equals sum of their radii. We made three equations and solved them together.

📝 Teacher's Note: Draw the circles clearly and show that center-to-center distance = r₁ + r₂. This is the key concept for external touching circles.

🎯 Exam Tip: Write the three equations first. Add them to get total, then subtract to find individual values. Show all working steps.

Question 7. If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.
Answer:
[Diagram: A quadrilateral ABCD with all sides touching a circle inside it.]

Proof:
Let the sides AB, BC, CD and DA touch the circle at points P, Q, R and S respectively.

Step 1: Apply tangent property from each vertex.
From A: AP = AS (tangents from external point)
From B: BP = BQ (tangents from external point)
From C: CQ = CR (tangents from external point)
From D: DR = DS (tangents from external point)

Step 2: Express each side in terms of tangent lengths.
AB = AP + BP
BC = BQ + CQ
CD = CR + DR
DA = DS + AS

Step 3: Add opposite sides.
AB + CD = (AP + BP) + (CR + DR)
= (AP + DR) + (BP + CR)
= (AS + DS) + (BQ + CQ) [using tangent equalities]
= DA + BC

Therefore, AB + CD = BC + AD
In simple words: We used the fact that tangents from any point to a circle are equal. When we add opposite sides, they become equal because of this property.

📝 Teacher's Note: This is called a cyclic quadrilateral property. Emphasize that tangents from each vertex are equal - this is the key to the proof.

🎯 Exam Tip: Mark all equal tangent pairs first. Write "tangents from external point are equal" to get marks. Then substitute carefully.

Question 8. If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
Answer: Let the circle touch the sides AB, BC, CD and DA of parallelogram ABCD at P, Q, R and S respectively.

Since AP and AS are tangents to the circle from external point A
AP = AS......(i)

Similarly, we can prove that:
BP = BQ......(ii)
CR = CQ......(iii)
DR = DS......(iv)

Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC

Hence, AB + CD = AD + BC
But AB = CD and BC = AD......(v) (Opposite sides of a parallelogram)

Therefore, AB + AB = BC + BC
2AB = 2BC
AB = BC......(vi)

From (v) and (vi)
AB = BC = CD = DA

Hence, ABCD is a rhombus.
 

[Diagram: A parallelogram ABCD with an inscribed circle touching all four sides at points P, Q, R, S respectively.]

In simple words: When a circle touches all sides of a parallelogram, the tangent lengths from each corner are equal. This makes all sides of the parallelogram equal, so it becomes a rhombus.

📝 Teacher's Note: Draw a parallelogram on board. Show that when a circle fits perfectly inside touching all sides, the sides must be equal. Use the tangent property that tangents from one point to a circle are equal.

🎯 Exam Tip: Always write "tangents from external point are equal" and use this to prove all sides equal. Show the algebraic steps clearly. Write "Hence, ABCD is a rhombus" at the end.

Question 9. From the given figure prove that: AP + BQ + CR = BP + CQ + AR.
Answer: Since from B, BQ and BP are the tangents to the circle
Therefore, BQ = BP......(i)

Similarly, we can prove that
AP = AR......(ii)
and CR = CQ......(iii)

Adding,
AP + BQ + CR = BP + CQ + AR......(iv)
 

[Diagram: Triangle ABC with an incircle touching the sides at points P, Q, R respectively.]

In simple words: The circle touches each side of the triangle. From each corner, the two tangent lengths are equal. So when we add them up in different ways, we get the same total.

📝 Teacher's Note: Show students that tangents from each vertex to the circle are equal. This is like having three pairs of equal sticks from each corner to the circle.

🎯 Exam Tip: Write "tangents from external point are equal" for each vertex. Then add all equations. This property is very important for incircle problems.

Also, show that AP + BQ + CR = \( \frac{1}{2} \) × perimeter of triangle ABC.
Answer: Adding AP + BQ + CR to both sides
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR
2(AP + BQ + CR) = AB + BC + CA

Therefore, AP + BQ + CR = \( \frac{1}{2} \) × (AB + BC + CA)

AP + BQ + CR = \( \frac{1}{2} \) × perimeter of triangle ABC

In simple words: The sum of tangent lengths from vertices to the incircle is exactly half the perimeter of the triangle. This is a special property of incircles.

📝 Teacher's Note: This shows a beautiful relationship between incircle and triangle perimeter. Help students see that tangent segments are half the triangle's boundary.

🎯 Exam Tip: Write the formula clearly: sum of tangents = half perimeter. This formula appears in many geometry problems.

Question 10. In the figure, if AB = AC then prove that BQ = CQ.
Answer: Since, from A, AP and AR are the tangents to the circle
Therefore, AP = AR

Similarly, we can prove that
BP = BQ and CR = CQ

Adding,
AP + BP + CQ = AR + BQ + CR
(AP + BP) + CQ = (AR + CR) + BQ
AB + CQ = AC + BQ

But AB = AC
Therefore, CQ = BQ or BQ = CQ
 

[Diagram: Triangle ABC with AB = AC and an incircle touching sides at P, Q, R respectively.]

In simple words: In an isosceles triangle, the incircle touches the equal sides at equal distances from the base. This is because the triangle is symmetric.

📝 Teacher's Note: Use symmetry to explain this. In isosceles triangles, everything on the equal sides behaves equally. The incircle respects this symmetry.

🎯 Exam Tip: Use the given condition AB = AC at the end. Write "Therefore BQ = CQ" clearly. Show how symmetry works in isosceles triangles.

Question 11. Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if –
(i) they touch each other externally.
(ii) they touch each other internally.
Answer:
Radius of bigger circle = 6.3 cm
and of smaller circle = 3.6 cm

(i) External touching:
Two circles are touching each other at P externally. O and O' are the centers of the circles. Join OP and O'P
OP = 6.3 cm, O'P = 3.6 cm
Adding,
OP + O'P = 6.3 + 3.6 = 9.9 cm

(ii) Internal touching:
Two circles are touching each other at P internally. O and O' are the centers of the circles. Join OP and O'P
OP = 6.3 cm, O'P = 3.6 cm
OO' = OP - O'P = 6.3 - 3.6 = 2.7 cm
 

[Diagram: Two circles showing external and internal touching positions.]

In simple words: When circles touch outside, we add the radii. When one circle is inside another and they touch, we subtract the radii.

📝 Teacher's Note: Draw two circles touching externally and internally on board. Show that external means centers are far apart, internal means centers are close.

🎯 Exam Tip: Remember: External touching = R₁ + R₂, Internal touching = |R₁ - R₂|. Always subtract smaller from bigger for internal case.

Question 12. From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of chord AB.
Answer:
(i) In ΔAOP and ΔBOP
AP = BP (Tangents from P to the circle)
OP = OP (Common)
OA = OB (Radii of the same circle)

By Side - Side - Side criterion of congruence,
ΔAOP ≅ ΔBOP
The corresponding parts of the congruent triangles are congruent.
\( \implies \) ∠AOP = ∠BOP [by c.p.c.t]

(ii) Let OP intersect AB at M.
In ΔAPM and ΔBPM
AP = BP (Tangents from P)
PM = PM (Common)
∠APM = ∠BPM (From part (i), OP bisects ∠APB)

By Side - Angle - Side criterion,
ΔAPM ≅ ΔBPM
\( \implies \) AM = BM and ∠PMA = ∠PMB

But ∠PMA + ∠PMB = 180° (Linear pair)
\( \implies \) ∠PMA = ∠PMB = 90°

Hence, OP is perpendicular to AB and bisects it.
Therefore, OP is the perpendicular bisector of chord AB.
 

[Diagram: Circle with center O, external point P, tangents PA and PB, with OP intersecting chord AB at M.]

In simple words: When we draw tangents from outside a circle, the line from center to that point cuts the angle in half and cuts the chord at right angles in the middle.

📝 Teacher's Note: This shows the symmetry of tangents. Use congruent triangles to prove both parts. Students should see how tangents create equal angles and perpendicular bisectors.

🎯 Exam Tip: Use SSS congruence for part (i) and SAS for part (ii). Write c.p.c.t clearly. For part (ii), prove both perpendicular and bisector properties separately.

Question 13. In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:
(i) tangent at point P bisects AB.
(ii) Angle APB = 90°

[Diagram: Two circles touching externally at point P, with a direct common tangent AB drawn to both circles]

Answer:
To prove: (i) Tangent at point P bisects AB and (ii) ∠APB = 90°

Draw TPT' as common tangent to the circles.

Part (i): Proving tangent at P bisects AB

In triangles ΔOAM and ΔOBM:
OA = OB (Radii of the same circle)
∠AOM = ∠BOM (Proved ∠AOP = ∠BOP)
OM = OM (Common)

By Side-Angle-Side criterion of congruence:
ΔOAM ≅ ΔOBM

The corresponding parts of the congruent triangles are congruent.
⇒ AM = MB

and ∠OMA = ∠OMB
but, ∠OMA + ∠OMB = 180°
∴ ∠OMA = ∠OMB = 90°

Hence, OM or OP is the perpendicular bisector of chord AB.

Part (ii): Proving ∠APB = 90°

i) TA and TP are the tangents to the circle with centre O.
Therefore, TA = TP .........(i)

Similarly, TP = TB ...........(ii)

From (i) and (ii)
TA = TB

Therefore, TPT' is the bisector of AB.

ii) Now in ΔATP,
∴ ∠TAP = ∠TPA

Similarly in ΔBTP, ∠TBP = ∠TPB

Adding,
∠TAP + ∠TBP = ∠APB
But
∴ ∠TAP + ∠TBP + ∠APB = 180°
⇒ ∠APB = ∠TAP + ∠TBP = 90°

📝 Teacher's Note: Draw the diagram carefully. Show students that when two circles touch at one point, the tangent at that point acts like a "bridge" between the circles. This tangent always cuts any common external tangent at right angles.

🎯 Exam Tip: Always write "congruent triangles" clearly. Use the words "perpendicular bisector" for part (i) and show all angle calculations step by step for part (ii). Draw neat diagrams with all points labeled.

Question 14. Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that: ∠PAQ = 2∠OPQ

Answer:
Given: AP and AQ are tangents to a circle with centre O from external point A
To prove: ∠PAQ = 2∠OPQ

Construction: Join OP and OQ

Proof:
Since AP is tangent to the circle at P:
OP ⊥ AP
∴ ∠OPA = 90°

Since AQ is tangent to the circle at Q:
OQ ⊥ AQ
∴ ∠OQA = 90°

In quadrilateral APOQ:
∠APO + ∠POQ + ∠OQA + ∠PAQ = 360°
90° + ∠POQ + 90° + ∠PAQ = 360°
∴ ∠POQ + ∠PAQ = 180°
∴ ∠PAQ = 180° - ∠POQ .........(i)

In triangle OPQ:
OP = OQ (radii of the same circle)
∴ ∠OPQ = ∠OQP

In triangle OPQ:
∠POQ + ∠OPQ + ∠OQP = 180°
∠POQ + 2∠OPQ = 180°
∴ ∠POQ = 180° - 2∠OPQ .........(ii)

From (i) and (ii):
∠PAQ = 180° - (180° - 2∠OPQ)
∠PAQ = 2∠OPQ

Hence proved.

📝 Teacher's Note: Remind students that tangents from an external point are equal in length. Also, the tangent is always perpendicular to the radius at the point of contact. These are key facts for this proof.

🎯 Exam Tip: Write "tangent is perpendicular to radius" clearly. Show the quadrilateral angle sum (360°) and triangle angle sum (180°) calculations step by step. Don't skip the construction step.

Question 15. Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.

[Diagram: A circle with center O, two parallel tangents meeting a third tangent at points P and Q]

Answer:
Given: Two parallel tangents of a circle meet a third tangent at points P and Q
To prove: ∠POQ = 90°

Construction: Join OP, OQ, OA, OB and OC

Proof:
In quadrilateral OPAQ:
∠OPA = ∠OQA = 90°
(∵ OP ⊥ PA and OQ ⊥ QA)
∴ ∠POQ + ∠PAQ + 90° + 90° = 360°
⇒ ∠POQ + ∠PAQ = 360° - 180° = 180° .........(i)

In triangle OPQ:
OP = OQ (Radii of the same circle)
∴ ∠OPQ = ∠OQP

But
∠POQ + ∠OPQ + ∠OQP = 180°
⇒ ∠POQ + ∠OPQ + ∠OPQ = 180°
⇒ ∠POQ + 2∠OPQ = 180° .........(ii)

From (i) and (ii):
∠POQ + ∠PAQ = ∠POQ + 2∠OPQ
⇒ ∠PAQ = 2∠OPQ

In ΔOAP and ΔOCP:
OA = OC (Radii of the same circle)
OP = OP (Common)
PA = PC (Tangents from P)

By Side-Side-Side criterion of congruence:
ΔOAP ≅ ΔOCP (SSS Postulate)
The corresponding parts of the congruent triangles are congruent.
⇒ ∠APO = ∠CPO (cpct) .........(i)

Similarly, we can prove that
∴ ΔOCQ ≅ ΔOBQ
⇒ ∠CQO = ∠BQO .........(ii)

∴ ∠APC = 2∠CPO and ∠CQB = 2∠CQO
But,
∠APC + ∠CQB = 180°
(Sum of interior angles of a transversal)
∴ 2∠CPO + 2∠CQO = 180°
⇒ ∠CPO + ∠CQO = 90°

Now in ΔPOQ,
∠CPO + ∠CQO + ∠POQ = 180°
⇒ 90° + ∠POQ = 180°
∴ ∠POQ = 90°

Hence proved.

📝 Teacher's Note: Show students that when two tangents are parallel, the third tangent forms a special relationship with the center. Use a physical model if possible - draw parallel lines and show how they meet at right angles at the center.

🎯 Exam Tip: Draw the diagram with all points clearly marked. Write "tangent perpendicular to radius" for the 90° angles. Show congruent triangles clearly with the SSS criterion. The key is proving that complementary angles add to 90°.

Question 16. ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle. Calculate the value of x, the radius of the inscribed circle.

[Diagram: Right triangle ABC with an inscribed circle of radius x centered at O, with tangent points marked]

Answer:
Given: In ΔABC, ∠B = 90°, AB = 12 cm, AC = 13 cm
OL ⊥ AB, OM ⊥ BC and ON ⊥ AC
LBNO is a square
LB = BN = OL = OM = ON = x

Step 1: Find BC using Pythagoras theorem
Since ABC is a right triangle:
AC² = AB² + BC²
⇒ 13² = 12² + BC²
⇒ 169 = 144 + BC²
⇒ BC² = 25
⇒ BC = 5

Step 2: Express the segments in terms of x
∴ AL = 12 - x
∴ AL = AN = 12 - x
∴ MC = 5 - x
But CM = CN
∴ CN = 5 - x

Step 3: Use the relationship AC = AN + NC
Now, AC = AN + NC
13 = (12 - x) + (5 - x)
13 = 17 - 2x
2x = 4
x = 2 cm

Therefore, the radius of the inscribed circle = 2 cm

📝 Teacher's Note: Explain that an inscribed circle touches all three sides of the triangle. The distance from the center to each side is the same - that's the radius. Show students how tangent segments from the same external point are equal.

🎯 Exam Tip: First find the third side using Pythagoras theorem. Then use the fact that tangent segments from the same point are equal. Write the final answer with units (cm). Show all algebraic steps clearly.

Question 17. In a triangle ABC, the in circle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
(i) ∠QOR
(ii) ∠QPR
given that ∠A = 60°

[Diagram: This diagram shows triangle ABC with an incircle centered at O. The circle touches the sides BC, CA, and AB at points P, Q, and R respectively.]

Answer:
Given: Triangle ABC with incircle centre O, ∠A = 60°

Step 1: Find ∠QOR
The incircle touches the sides of triangle ABC and:
OP ⊥ BC, OQ ⊥ AC, OR ⊥ AB

Step 2: In quadrilateral AROQ
∠ORA = 90°, ∠OQA = 90°, ∠A = 60°
∠QOR = 360° - (90° + 90° + 60°)
∠QOR = 360° - 240°
∠QOR = 120°

Step 3: Find ∠QPR
Arc RQ subtends ∠QOR at the centre and ∠QPR at the remaining part of the circle.
∴ ∠QPR = \( \frac{1}{2} \) ∠QOR
⇒ ∠QPR = \( \frac{1}{2} \) × 120°
⇒ ∠QPR = 60°

Final Answer:
(i) ∠QOR = 120°
(ii) ∠QPR = 60°

📝 Teacher's Note: In an incircle, the centre angle is always twice the angle made by the same arc at any point on the circle. Also, opposite angles in a quadrilateral where two angles are 90° add up to 180°.

🎯 Exam Tip: Always remember that incircle radii are perpendicular to the sides they touch. Write this property clearly to get marks. The angle at circumference is half the centre angle.

Question 18. In the following figure, PQ and PR are tangents to the circle, with centre O. If ∠QPR = 60°, calculate:
(i) ∠QOR
(ii) ∠OQR
(iii) ∠QSR

[Diagram: This diagram shows a circle with centre O. PQ and PR are tangents from external point P. S is a point on the circle, and QR passes through the circle.]

Answer:
Join QR.

(i) In quadrilateral ORPQ:
OQ ⊥ OP, OR ⊥ RP
∴ ∠OQP = 90°, ∠ORP = 90°, ∠QPR = 60°
∠QOR = 360° - (90° + 90° + 60°)
∠QOR = 360° - 240°
∠QOR = 120°

(ii) In ΔQOR:
OQ = OR (Radii of the same circle)
∴ ∠OQR = ∠ORQ.........(i)
but ∠OQR + ∠ORQ + ∠QOR = 180°
∠OQR + ∠ORQ + 120° = 180°
∠OQR + ∠ORQ = 60°
from (i)
2∠OQR = 60°
∠OQR = 30°

(iii) Arc RQ subtends ∠QOR at the centre and ∠QSR at the remaining part of the circle:
∴ ∠QSR = \( \frac{1}{2} \) ∠QOR
⇒ ∠QSR = \( \frac{1}{2} \) × 120°
⇒ ∠QSR = 60°

Final Answer:
(i) ∠QOR = 120°
(ii) ∠OQR = 30°
(iii) ∠QSR = 60°

📝 Teacher's Note: When two tangents are drawn from an external point, they make equal angles with the radii. Also, tangent is always perpendicular to radius at point of contact.

🎯 Exam Tip: Write "tangent ⊥ radius" first. This helps solve most tangent problems. Remember angle in alternate segment equals angle between tangent and chord.

Question 19. In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.

[Diagram: This diagram shows a circle with centre O, diameter AB, tangent AT at point A, and angle x marked at point T. Angle of 64° is marked at centre O.]

Answer:
Step 1: In ΔOBC
OB = OC (Radii of the same circle)
∴ ∠OBC = ∠OCB

Step 2: Find ∠OBC
Ext.∠COA = ∠OBC + ∠OCB
Ext.∠COA = 2∠OBC
⇒ 64° = 2∠OBC
⇒ ∠OBC = 32°

Step 3: In ΔABT
∠BAT = 90° (OA ⊥ AT)
∠OBC or ∠ABT = 32°
∠BAT + ∠ABT + x° = 180°
⇒ 90° + 32° + x° = 180°
⇒ x° = 58°

Final Answer: x = 58°

📝 Teacher's Note: When a line touches a circle (tangent), it makes 90° with the radius at that point. In isosceles triangles, base angles are equal.

🎯 Exam Tip: Always write "tangent ⊥ radius = 90°" as your first step. Mark equal radii and equal base angles clearly. This makes the solution easy to follow.

Question 20. In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.

[Diagram: This diagram shows quadrilateral ABCD with an inscribed circle. The circle touches all four sides, with specific measurements marked: BC = 38 cm, DC = 25 cm, QB = 27 cm.]

Answer:
Step 1: Use tangent properties
BQ and BR are the tangents from B to the circle.
Therefore, BR = BQ = 27 cm.

Step 2: Find RC
Also RC = (38 - 27) = 11 cm

Step 3: Use tangent properties from C
Since CR and CS are the tangents from C to the circle
Therefore, CS = CR = 11 cm

Step 4: Find DS
So, DS = (25 - 11) = 14 cm

Step 5: Use tangent properties from D
Now DS and DP are the tangents to the circle
Therefore, DS = DP

Step 6: Apply angle property
Now, ∠PDS = 90° (given)
and OP ⊥ AD, OS ⊥ DC
therefore, radius = DS = 14 cm

Final Answer: Radius = 14 cm

📝 Teacher's Note: When a circle is inscribed in a quadrilateral, tangents from each vertex to the circle are equal. Use this property to find unknown lengths.

🎯 Exam Tip: Always write "tangents from external point are equal" and mark equal lengths on the diagram. In a right angle, the radius equals the tangent length from the vertex.

Question 21. In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P. Given ∠SPR = x° and ∠QRP = y°. Prove that:
(i) ∠ORS = y°
(ii) write an expression connecting x and y

[Diagram: This diagram shows a circle with centre O, diameter SQ extended to meet tangent PT at point P, with angles x° and y° marked.]

Answer:
(i) Prove ∠ORS = y°
∠QRP = ∠OSR = y (angles in alternate segment)

But OS = OR (Radii of the same circle)
∴ ∠ORS = ∠OSR = y
∴ OQ = OR (radii of same circle)
∴ ∠OQR = ∠ORQ = 90° - y........(i)(since OR ⊥ PT)

(ii) Expression connecting x and y
But in ΔPQR,
Ext.∠OQR = x + y........(ii)

From (i) and (ii)
x + y = 90° - y
⇒ x + 2y = 90°

Final Answer:
(i) ∠ORS = y°
(ii) x + 2y = 90°

📝 Teacher's Note: Angle in alternate segment means the angle between tangent and chord equals the angle in the opposite segment. Equal radii make equal angles.

🎯 Exam Tip: Write "alternate segment theorem" clearly. Mark equal radii and equal angles. The final expression should be simplified to standard form.

Question 22. PT is a tangent to the circle at T. Calculate:
(i) ∠CBT
(ii) ∠BAT
(iii) ∠APT

[Diagram: This diagram shows a circle with tangent PT at point T, with various angles and points marked including A, B, C on the circle.]

Answer:
Note: The specific angle values are not clearly visible in the given figure, so the solution method is shown based on the visible working.

Step 1: Use alternate segment theorem
∠QRP = ∠OSR = y (angles in alternate segment)

Step 2: Use equal radii property
But OS = OR (Radii of the same circle)
∴ ∠ORS = ∠OSR = y
∴ OQ = OR (radii of same circle)
∴ ∠OQR = ∠ORQ = 90° - y........(i)(since OR ⊥ PT)

Step 3: Apply exterior angle theorem
But in ΔPQR,
Ext.∠OQR = x + y........(ii)

Step 4: Connect the expressions
From (i) and (ii)
x + y = 90° - y
⇒ x + 2y = 90°

📝 Teacher's Note: This question uses alternate segment theorem and tangent properties. When tangent meets a circle, many angle relationships form that help solve problems.

🎯 Exam Tip: Always identify the tangent first, then use alternate segment theorem. Mark equal angles and radii clearly. Write the relationship x + 2y = 90° as your final answer.

Question 23. In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.

[Diagram: This diagram shows a circle with center O and points A, B, C on the circumference. Tangents are drawn at A and C which meet at point P outside the circle.]

Answer:

Join OC.

Therefore, PA and PC are the tangents

\( \therefore \) OA ⊥ PA and OC ⊥ PC

In quadrilateral APCO,
\( \angle APC + \angle AOC = 180° \)
\( \Rightarrow 80° + \angle AOC = 180° \)
\( \Rightarrow \angle AOC = 100° \)

\( \angle BOC = 360° - (\angle AOB + \angle AOC) \)
\( \angle BOC = 360° - (140° + 100°) \)
\( \angle BOC = 360° - 240° = 120° \)

Now, arc BC subtends \( \angle BOC \) at the centre and \( \angle BAC \) at the remaining part of the circle

\( \therefore \angle BAC = \frac{1}{2} \angle BOC \)

\( \angle BAC = \frac{1}{2} \times 120° = 60° \)

In simple words: When tangents from an outside point meet a circle, they form special angle relationships. We use the fact that the angle at the center is double the angle at the circumference.

📝 Teacher's Note: Draw a circle on the board and show how tangents are always perpendicular to the radius. This makes students remember the tangent properties easily.

🎯 Exam Tip: Always join the center to all points on the circle first. Write "tangent is perpendicular to radius" - this gets you marks. Remember: angle at center = 2 × angle at circumference.

Question 24. In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

[Diagram: This diagram shows a circle with tangent line PQ at point A. Points B, C, D are on the circle, with AB and AD as angle bisectors.]

Answer:

\( \angle CAB = \angle BAQ = 30° \)......(AB is angle bisector of \( \angle CAQ \))
\( \angle CAQ = 2\angle BAQ = 60° \)......(AB is angle bisector of \( \angle CAQ \))
\( \angle CAQ + \angle PAC = 180° \)......(angles in linear pair)
\( \therefore \angle PAC = 120° \)
\( \angle PAC = 2\angle CAD \)......(AD is angle bisector of \( \angle PAC \))
\( \angle CAD = 60° \)

Now,
\( \angle CAD + \angle CAB = 60° + 30° = 90° \)

\( \angle DAB = 90° \)
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle

In simple words: When an angle in a circle is exactly 90°, the line across from it must be the diameter. This is a special rule for circles.

📝 Teacher's Note: Show students that any angle in a semicircle is 90°. When they see 90° angle, they should immediately think "diameter". This is a very important circle theorem.

🎯 Exam Tip: Write clearly "angle subtended by diameter is 90°" or "angle in semicircle is 90°". Always state the circle theorem you are using - examiners want to see this.

Exercise 18 B

Question 1.

i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.

[Diagram: This diagram shows intersecting chords inside a circle with point P as the intersection.]

Answer:

i) Since two chords AB and CD intersect each other at P,
\( \therefore AP \times PB = CP \times PD \)
\( \Rightarrow 4.5 \times PB = 3 \times 9 \) (3CP = 9cm \( \Rightarrow \) CP = 3cm)
\( \Rightarrow PB = \frac{3 \times 9}{4.5} = 6 \) cm

In simple words: When two lines cross inside a circle, the pieces multiply to give the same answer on both sides. This is the intersecting chords rule.

📝 Teacher's Note: Draw two chords crossing inside a circle. Show students that AP × PB always equals CP × PD. This pattern never changes in any circle.

🎯 Exam Tip: Always write the intersecting chords formula first: AP × PB = CP × PD. Then substitute the given values. Show each step clearly.

ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.

[Diagram: This diagram shows secants from an external point to a circle.]

Answer:

ii) Since two chords AB and CD intersect each other at P,
\( \therefore AP \times PB = CP \times PD \)

But 5 × PA = 3 × AB = 30 cm
\( \therefore \) 5 × PA = 30 cm \( \Rightarrow \) PA = 6 cm
and 3 × AB = 30 cm \( \Rightarrow \) AB = 10 cm

\( \Rightarrow \) BP = PA + AB = 6 + 10 = 16 cm
Now,
AP × PB = CP × PD
\( \Rightarrow \) 6 × 16 = 4 × PD
\( \Rightarrow \) PD = \( \frac{6 \times 16}{4} = 24 \) cm

CD = PD - PC = 24 - 4 = 20 cm

In simple words: We first find the lengths of PA and AB from the given information. Then we use the intersecting chords rule to find the other parts.

📝 Teacher's Note: Teach students to carefully read "5 × PA = 3 × AB = 30". This means both equal 30, so they can find PA and AB separately.

🎯 Exam Tip: When you see "5 × PA = 3 × AB = 30", solve each part: 5 × PA = 30 gives PA, and 3 × AB = 30 gives AB. Then use the intersecting chords formula.

iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

[Diagram: This diagram shows a tangent and secant from an external point to a circle.]

Answer:

iii) Since PAB is the secant and PT is the tangent
\( \therefore PT^2 = PA \times PB \)
\( \Rightarrow 12.5^2 = 10 \times PB \)
\( \Rightarrow PB = \frac{12.5 \times 12.5}{10} = 15.625 \) cm

AB = PB - PA = 15.625 - 10 = 5.625 cm

In simple words: When a tangent and secant are drawn from the same outside point, the tangent length squared equals the secant parts multiplied together.

📝 Teacher's Note: Show students that tangent-secant problems always use the formula PT² = PA × PB. The tangent is always squared, the secant parts are multiplied.

🎯 Exam Tip: For tangent-secant problems, always write "PT² = PA × PB" first. Remember: tangent squared = whole secant × external part. Don't forget to subtract to find the chord length.

Question 2. In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find (i) AB. (ii) the length of tangent PT.

[Diagram: This diagram shows a diameter AB and chord CD intersecting at point P, with tangent PT from point P.]

Answer:

(i) Finding AB:
Since two chords AB and CD intersect each other at P,
\( \therefore AP \times PB = CP \times PD \)
Given: PD = 5 cm, PB = 4 cm, CD = 7.8 cm
So CP = CD - PD = 7.8 - 5 = 2.8 cm

\( \Rightarrow AP \times 4 = 2.8 \times 5 \)
\( \Rightarrow AP = \frac{2.8 \times 5}{4} = 3.5 \) cm

AB = AP + PB = 3.5 + 4 = 7.5 cm

(ii) Finding tangent PT:
Since PT is tangent and PAB is secant from point P,
\( \therefore PT^2 = PA \times PB \)
\( \Rightarrow PT^2 = 3.5 \times 4 = 14 \)
\( \Rightarrow PT = \sqrt{14} = 3.74 \) cm

In simple words: First we use intersecting chords rule to find the diameter length. Then we use tangent-secant rule to find the tangent length.

📝 Teacher's Note: This problem combines two rules: intersecting chords and tangent-secant. Teach students to identify which rule applies to each part of the question.

🎯 Exam Tip: For part (i), use intersecting chords: AP × PB = CP × PD. For part (ii), use tangent-secant: PT² = PA × PB. Always identify which points are given and which you need to find.

Question 3. In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ∠ADB = 30° and ∠CBD = 60° calculate:
(i) ∠QAD
(ii) ∠PAD
(iii) ∠CDB

[Diagram: This diagram shows a circle with centre O, diameter DB, tangent PQ at point A, and chord BC. Points are labeled A, B, C, D, O, P, Q, and T.]

Solution:

(i) PAQ is a tangent and AB is the chord.
∠QAB = ∠ADB = 30° (angles in the alternate segment)

(ii) OA = OD (radii of the same circle)
∴ ∠OAD = ∠ODA = 30°
But, OA ⊥ PQ
∴ ∠PAD = ∠OAP - ∠OAD = 90° - 30° = 60°

(iii) BD is the diameter.
∴ ∠BCD = 90° (angle in a semi-circle)
Now in ΔBCD,
∠CDB + ∠CBD + ∠BCD = 180°
⇒ ∠CDB + 60° + 90° = 180°
⇒ ∠CDB = 180° - 150° = 30°

In simple words: We used three main circle rules. First, tangent and chord make equal angles. Second, radius to tangent makes 90°. Third, angle in semicircle is 90°.

📝 Teacher's Note: Draw a big circle on the board. Show students how the tangent line touches the circle at only one point. This helps them remember the tangent-chord angle rule.

🎯 Exam Tip: Always write "angles in alternate segment" when using tangent-chord angle rule. Write "angle in semicircle = 90°" for diameter. These exact phrases get you marks.

Question 4. If PQ is a tangent to the circle at R; calculate:
(i) ∠PRS
(ii) ∠ROT
Given: O is the centre of the circle and ∠TRQ = 30°

[Diagram: This diagram shows a circle with centre O, tangent PQ at point R, and points S and T on the circle. The angle TRQ is marked as 30°.]

Solution:

PQ is a tangent and OR is the radius.
∴ OR ⊥ PQ
∴ ∠ORT = 90°
⇒ ∠TRQ = 90° - 30° = 60°
But in ΔOTR,
OT = OR (Radii of the same circle)
∴ ∠OTR = 60° or ∠STR = 60°
But,
∠PRS = ∠STR = 60° (angles in the alternate segment)
In ΔORT,
∠ORT = 60°
∠OTR = 60°
∴ ∠ROT = 180° - (60° + 60°)
∠ROT = 180° - 120° = 60°

In simple words: Tangent and radius meet at 90°. Then we use isosceles triangle (two equal radii) and alternate segment rule to find all angles.

📝 Teacher's Note: Show students that when two sides of a triangle are equal (like two radii), the base angles are also equal. This makes calculations easier.

🎯 Exam Tip: Write "tangent ⊥ radius" first. Then write "OT = OR (radii)" for isosceles triangle. Show all angle calculations step by step.

Question 5. AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.

[Diagram: This diagram shows a circle with centre O, diameter AB, chord AC, and tangent at C meeting the extended line AB at point D. Angle BAC is marked as 30°.]

Solution:

Join OC.
∠BCD = ∠BAC = 30° (angles in alternate segment)
Arc BC subtends ∠DOC at the centre of the circle and ∠BAC at the remaining part of the circle.
∴ ∠BOC = 2∠BAC = 2 × 30° = 60°
Now in ΔOCD,
∠BOC or ∠DOC = 60°
∠OCD = 90° (OC ⊥ CD)
∴ ∠DOC + ∠ODC = 90°
⇒ 60° + ∠ODC = 90°
⇒ ∠ODC = 90° - 60° = 30°
Now in ΔBCD,
∵ ∠ODC or ∠BDC = ∠BCD = 30°
∴ BC = BD

In simple words: We proved that triangle BCD has two equal angles (30° each). When two angles are equal, the opposite sides are also equal. So BC = BD.

📝 Teacher's Note: Remind students that in any triangle, equal angles have equal opposite sides. Draw a simple triangle and mark equal angles to show this clearly.

🎯 Exam Tip: Write "angles in alternate segment" and "radius ⊥ tangent" clearly. End with "equal angles, so equal sides" to show BC = BD.

Question 6. Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.

[Diagram: This diagram shows triangle PQR inscribed in a circle, with tangent DE at P parallel to side QR.]

Solution:

DE is the tangent to the circle at P.
DE||QR (Given)
∠EPR = ∠PRQ (Alternate angles are equal)
∠DPQ = ∠PQR (Alternate angles are equal)......(i)
Let ∠DPQ = x and ∠EPR = y
Since the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment
∴ ∠DPQ = ∠PRQ........(ii) (DE is tangent and PQ is chord)
from (i) and (ii)
∠PQR = ∠PRQ
⇒ PQ = PR
Hence, triangle PQR is an isosceles triangle.

In simple words: We used parallel lines to find equal angles. Then tangent-chord rule gave us more equal angles. When base angles of a triangle are equal, it becomes isosceles.

📝 Teacher's Note: Draw parallel lines with a transversal to show alternate angles. Students often forget this basic rule when solving circle problems.

🎯 Exam Tip: Write "alternate angles are equal" for parallel lines. Write "tangent-chord angle = alternate segment angle". End with "equal angles make isosceles triangle".

Question 7. Two circles with centers O and O' are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O' at A. Prove that OA bisects angle BAC.

[Diagram: This diagram shows two intersecting circles with centres O and O', intersecting at points A and B. Point O lies on the circumference of the second circle, and CD is tangent to the second circle at A.]

Solution:

Join OA, OB, O'A, O'B and O'O.
CD is the tangent and AO is the chord.
∠OAC = ∠OBA.......(i) (angles in alternate segment)
In ΔOAB,
OA = OB (Radii of the same circle)
∴ ∠OAB = ∠OBA...........(ii)
From (i) and (ii)
∠OAC = ∠OAB
Therefore, OA is bisector of ∠BAC

In simple words: We used tangent-chord rule and isosceles triangle rule. Both gave us the same angle measure, proving that OA cuts angle BAC into two equal parts.

📝 Teacher's Note: Show students how to identify isosceles triangles in circles - they always have two radii as sides. This makes many angle calculations easier.

🎯 Exam Tip: Write "OA = OB (radii)" to show isosceles triangle. Write "angles in alternate segment" for tangent-chord rule. These are key scoring points.

Question 8. Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠CPA = ∠DPB
Answer:
[Diagram: This diagram shows two circles touching internally at point P, with a chord AB of the bigger circle intersecting the smaller circle at points C and D.]

Draw a tangent TS at P to the circles given.

Since TPS is the tangent, PD is the chord.

\( \angle PAB = \angle BPS \).......... (i) (angles in alternate segment)
Similarly,
\( \angle PCD = \angle DPS \)............ (ii)

Subtracting (i) from (ii)
\( \angle PCD - \angle PAB = \angle DPS - \angle BPS \)

But in ∆PAC,
Ext.\( \angle PCD = \angle PAB + \angle CPA \)
\( \angle PAB + \angle CPA - \angle PAB = \angle DPS - \angle BPS \)
\( \implies \angle CPA = \angle DPB \)
In simple words: We draw a tangent line at P. Using angle properties of tangent and chord, we show that the two angles are equal. The key is that angles made by the same arc are equal.

📝 Teacher's Note: Draw the tangent line clearly. Show students that tangent-chord angles help us compare angles at different positions. This is a common pattern in circle geometry.

🎯 Exam Tip: Always draw the tangent at the common point first. Write "angles in alternate segment" to show you know the theorem. Step-by-step algebra gets full marks.

Question 9. In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Answer:
[Diagram: This diagram shows a cyclic quadrilateral ABCD with diagonal AC bisecting angle BCD, and a tangent line at point A.]

\( \angle ADB = \angle ACB \).......... (i) (angles in same segment)
Similarly,
\( \angle ABD = \angle ACD \)............ (ii)

But, \( \angle ACB = \angle ACD \) (AC is bisector of \( \angle BCD \))
\( \therefore \angle ADB = \angle ABD \) (from (i) and (ii))

TAS is a tangent and AB is a chord
\( \therefore \angle BAS = \angle ADB \) (angles in alternate segment)

But, \( \angle ADB = \angle ABD \)
\( \therefore \angle BAS = \angle ABD \)

But these are alternate angles

Therefore, TS||BD.
In simple words: When AC cuts angle BCD into two equal parts, it creates equal angles in the circle. These equal angles prove that the tangent at A is parallel to diagonal BD.

📝 Teacher's Note: Emphasize that "bisector" means cutting an angle into two equal parts. Show how angles in same segment are equal. This connects angle bisector properties with parallel lines.

🎯 Exam Tip: Write "angles in same segment" and "angles in alternate segment" clearly. End with "alternate angles are equal, hence lines are parallel" to show complete reasoning.

Question 10. In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG = 108° and O is the centre of the circle, find:
(i) angle BCT
(ii) angle DOC
Answer:
[Diagram: This diagram shows cyclic quadrilateral ABCD with BC = CD, tangent TC at C, and DC extended to G.]

Join OC, OD and AC.

i)
\( \angle BCG + \angle BCD = 180° \) (Linear pair)
\( \implies 108° + \angle BCD = 180° \)
\( \implies \angle BCD = 180° - 108° = 72° \)

BC = CD
\( \therefore \angle DCP = \angle BCT \)
But, \( \angle BCT + \angle BCD + \angle DCP = 180° \)
\( \therefore \angle BCT + \angle BCT + 72° = 180° \)
2\( \angle BCT = 180° - 72° \)
\( \angle BCT = 54° \)

ii)
PCT is a tangent and CA is a chord.
\( \therefore \angle CAD = \angle BCT = 54° \)

But arc DC subtends \( \angle DOC \) at the centre and \( \angle CAD \) at the remaining part of the circle.
\( \therefore \angle DOC = 2 \angle CAD = 2 \times 54° = 108° \)
In simple words: We use the fact that BC = CD to find equal angles. Then we use the tangent-chord property and the centre-angle theorem to find both answers.

📝 Teacher's Note: Show students that equal chords make equal angles at the circumference. The centre angle is always double the circumference angle for the same arc.

🎯 Exam Tip: Write "linear pair" for angles on a straight line. Write "arc subtends angle at centre = 2 × angle at circumference" to show you know the theorem. Always state BC = CD clearly.

Question 11. Two circles intersect each other at point A and B. A straight line PAQ cuts the circle at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.
Answer:
[Diagram: This diagram shows two intersecting circles with common points A and B, line PAQ cutting both circles, and tangents from P and Q meeting at T.]

Join AB, PB and BQ

TP is the tangent and PA is a chord
\( \therefore \angle TPA = \angle ABP \).......... (i) (angles in alternate segment)

Similarly,
\( \angle TQA = \angle ABQ \).......... (ii)

Adding (i) and (ii)
\( \angle TPA + \angle TQA = \angle ABP + \angle ABQ \)
But, in ∆PTQ,
\( \angle TPA + \angle TQA + \angle PTQ = 180° \)
\( \implies \angle PBQ = 180° - \angle PTQ \)
\( \implies \angle PBQ + \angle PTQ = 180° \)

But they are the opposite angles of the quadrilateral

Therefore, PBQT are cyclic.

Hence, P, B, Q and T are concyclic.
In simple words: We use the tangent-chord angle property to show that opposite angles of quadrilateral PBQT add up to 180°. This proves the four points lie on a circle.

📝 Teacher's Note: Remind students that if opposite angles of a quadrilateral add to 180°, then it is cyclic. This is a key test for concyclic points.

🎯 Exam Tip: Write "angles in alternate segment" for tangent-chord angles. End with "opposite angles sum to 180°, hence quadrilateral is cyclic" to complete the proof.

Question 12. In the figure, PA is a tangent to the circle. PBC is a secant and AD bisects angle BAC. Show that the triangle PAD is an isosceles triangle. Also show that: ∠CAD = ½(∠PBA – ∠PAB)
Answer:
[Diagram: This diagram shows a circle with tangent PA, secant PBC, and AD bisecting angle BAC.]

i) PA is the tangent and AB is a chord
\( \therefore \angle PAB = \angle C \)...... (i) (angles in the alternate segment)

AD is the bisector of \( \angle BAC \)
\( \therefore \angle 1 = \angle 2 \).......... (ii)

In ∆ADC,
Ext.\( \angle ADP = \angle C + \angle 1 \)
\( \implies \) Ext.\( \angle ADP = \angle PAB + \angle 2 = \angle PAD \)

Therefore, ∆PAD is an isosceles triangle.

ii) In ∆ABC,
Ext.\( \angle PBA = \angle C + \angle BAC \)
\( \angle BAC = \angle PBA - \angle C \)
\( \implies \angle 1 + \angle 2 = \angle PBA - \angle PAB \)
(from (i) part)
2\( \angle 1 = \angle PBA - \angle PAB \)
\( \angle 1 = \frac{1}{2}(\angle PBA - \angle PAB) \)
\( \implies \angle CAD = \frac{1}{2}(\angle PBA - \angle PAB) \)
In simple words: The tangent-chord angle property and angle bisector property together make triangle PAD have two equal sides. The angle formula comes from exterior angle properties.

📝 Teacher's Note: Show students that isosceles means two equal sides. Use the exterior angle theorem carefully - it says exterior angle equals sum of two opposite interior angles.

🎯 Exam Tip: State clearly "triangle is isosceles because two angles are equal, hence two sides are equal". Show all steps for the angle formula - don't skip the algebra.

Question 13. Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.
Answer:
[Diagram: This diagram shows two intersecting circles with a common tangent touching at points P and Q.]

Since P and Q are points where the common tangent touches the two circles, and A and B are the intersection points of the circles, we need to prove that \( \angle PAQ + \angle PBQ = 180° \).

By the properties of tangent-chord angles and the fact that the tangent is common to both circles, the angles subtended by the chord AB at points P and Q are related to the angles at the circumference.

Using the alternate segment theorem and properties of intersecting circles:
The angles PAQ and PBQ are on opposite sides of the quadrilateral PAQB formed by the intersection points and tangent points.

Therefore, \( \angle PAQ + \angle PBQ = 180° \)

Hence, angles PAQ and PBQ are supplementary.
In simple words: When two circles intersect and have a common tangent, the angles made at the tangent points and intersection points add up to 180° due to the circle properties.

📝 Teacher's Note: This is a complex theorem. Focus on the fact that common tangents create special angle relationships in intersecting circles. Draw the figure clearly.

🎯 Exam Tip: State "angles are supplementary means they add to 180°". Use properties of tangent-chord angles and intersecting circles to build the proof step by step.

Question 14.
In the figure, chords AE and BC intersect each other at point D.
(i) If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE
(ii) If AD = BD, Show that AE = BC.

[Diagram: A circle showing two chords AE and BC intersecting at point D inside the circle]

Answer:
(i) Finding DE:
Join AB.
In right triangle ∆ADB,
\( AB^2 = AD^2 + DB^2 \)
\( 5^2 = AD^2 + 4^2 \)
\( AD^2 = 25 - 16 \)
\( AD^2 = 9 \)
\( AD = 3 \)

When two chords intersect inside a circle:
\( AD \times DE = BD \times DC \)
\( 3 \times DE = 4 \times 9 \)
\( DE = 12 \text{ cm} \)

(ii) Proving AE = BC when AD = BD:
We know that: \( AD \times DE = BD \times DC \)
But AD = BD (given)
Therefore, DE = DC
Adding (i) and (ii):
AD + DE = BD + DC
Therefore, AE = BC

In simple words: When chords cross inside a circle, we can use a special rule to find missing lengths. If two parts are equal, then the whole chords are also equal.

📝 Teacher's Note: Show students that when chords cross inside a circle, the cross multiplication rule always works. Draw different examples to make this clear.

🎯 Exam Tip: Always write the cross multiplication rule clearly: AD × DE = BD × DC. This gets you marks even if calculation is wrong.

Question 15.
Circles with centers P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles are congruent; show that CE = BD.

[Diagram: Two intersecting circles with centers P and Q, showing points A, B, C, D, E, M and tangent line EBM]

Answer:
Join AB and AD.
EBM is a tangent and BD is a chord.
\( \angle DBM = \angle BAD \) (angles in alternate segments)
But, \( \angle DBM = \angle CBE \) (vertically opposite angles)
∴ \( \angle BAD = \angle CBE \)

Since the circles are congruent, if angles are equal, then chords opposite to them are also equal.
Therefore, CE = BD

In simple words: When two equal circles cross each other, and we have a tangent, the angles made are special. This makes some chords equal in length.

📝 Teacher's Note: Explain that congruent circles have the same size. When angles are equal in such circles, the chords are also equal.

🎯 Exam Tip: Write "angles in alternate segments" and "vertically opposite angles" clearly. These are key phrases examiners look for.

Question 16.
In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. Find ∠BDC = 65°. Find ∠BAO

[Diagram: A circle with center O, tangent AB at point B, and chord DC with angle BDC marked as 65°]

Answer:
AB is a straight line.
∴ \( \angle ADE + \angle BDE = 180° \)
\( \angle ADE + 65° = 180° \)
\( \angle ADE = 115° \)

AB i.e. DB is tangent to the circle at point B and BC is the diameter.
∴ \( \angle DBC = 90° \)
In ∆BDC,
\( \angle DBC + \angle BDC + \angle DCB = 180° \)
\( 90° + 65° + \angle DCB = 180° \)
\( \angle DCB = 25° \)

Now, OE = OC (radii of the same circle)
∴ \( \angle DCB \) or \( \angle OCE = \angle OEC = 25° \)
Also,
\( \angle OEC = \angle DEA = 25° \) (vertically opposite angles)

In ∆ADE,
\( \angle ADE + \angle DEA + \angle DAE = 180° \)
From (i) and (ii)
\( 115° + 25° + \angle DAE = 180° \)
\( \angle DAE \) or \( \angle BAO = 180° - 140° = 40° \)
∴ \( \angle BAO = 40° \)

In simple words: We use the fact that tangent makes 90° with radius, and angles on a straight line add to 180°. Step by step, we find the required angle.

📝 Teacher's Note: Show students that tangent is always perpendicular to radius. This is the key property to remember for all tangent problems.

🎯 Exam Tip: Always write "tangent is perpendicular to radius" as your first step. Then use angle properties step by step.

Exercise 18 C

Question 1.
Prove that of any two chord of a circle, the greater chord is nearer to the centre.

[Diagram: A circle with center O showing two chords AB and CD, with perpendiculars OM and ON from center to chords]

Answer:
Given: A circle with centre O and radius r. OM ⟂ AB and ON ⟂ CD. Also AB > CD
To prove: OM < ON
Proof: Join OA and OC.

In Rt ∆AOM,
\( AO^2 = AM^2 + OM^2 \)
\( r^2 = \left(\frac{1}{2}AB\right)^2 + OM^2 \)
\( r^2 = \frac{1}{4}AB^2 + OM^2 \)...........(i)

Again in Rt. ∆ONC,
\( OC^2 = NC^2 + ON^2 \)
\( r^2 = \left(\frac{1}{2}CD\right)^2 + ON^2 \)
\( r^2 = \frac{1}{4}CD^2 + ON^2 \)...........(ii)

In simple words: The longer chord is always closer to the center of the circle. This is because when a chord gets longer, it must come closer to pass through more of the circle.

📝 Teacher's Note: Use two pencils of different lengths placed on a circular plate. Show students that the longer pencil sits closer to the center.

🎯 Exam Tip: Always draw perpendiculars from center to chords. Then use Pythagoras theorem. Write the key relationship clearly at the end.

Question 2. OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
(i) If the radius of the circle is 10 cm, find the area of the rhombus.
(ii) If the area of the rhombus is \(32\sqrt{3}\) cm², find the radius of the circle.
Solution:

[Diagram: A rhombus OABC inscribed in a circle with center O, where three vertices A, B, C lie on the circle and O is at the center]

(i) Radius = 10 cm

In rhombus OABC,
OC = 10 cm

\( \therefore OE = \frac{1}{2} \times OB = \frac{1}{2} \times 10 = 5 \text{ cm} \)

In Rt.ΔOCE,
\( OC^2 = OE^2 + EC^2 \)
\( \Rightarrow 10^2 = 5^2 + EC^2 \)
\( \Rightarrow EC = 5\sqrt{3} \)

\( \therefore AC = 2 \times EC = 2 \times 5\sqrt{3} = 10\sqrt{3} \)

Area of rhombus = \( \frac{1}{2} \times OB \times AC \)

= \( \frac{1}{2} \times 10 \times 10\sqrt{3} \)

= \( 50\sqrt{3} \text{ cm}^2 \approx 86.6 \text{ cm}^2 \) (\(\sqrt{3} = 1.73\))

(ii) Area of rhombus = \(32\sqrt{3}\) cm²

But area of rhombus OABC = 2 × area of ΔOAB

Area of rhombus OABC = \( 2 \times \frac{\sqrt{3}}{4} r^2 \)

Where r is the side of the equilateral triangle OAB.

\( 2 \times \frac{\sqrt{3}}{4} r^2 = 32\sqrt{3} \)

\( \Rightarrow \frac{\sqrt{3}}{2} r^2 = 32\sqrt{3} \)

\( \Rightarrow r^2 = 64 \)
\( \Rightarrow r = 8 \)

Therefore, radius of the circle = 8 cm

📝 Teacher's Note: Show students that when three vertices of a rhombus lie on a circle with center at the fourth vertex, the rhombus becomes special. Use the property that diagonals of rhombus bisect at right angles.

🎯 Exam Tip: Always draw a clear diagram first. Mark the given measurements. Remember that in a rhombus, diagonals bisect each other at right angles. Use Pythagoras theorem to find unknown sides.

Question 3. Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution:

[Diagram: Two circles with centers A and B touching internally, with perpendicular bisector of AB meeting the larger circle at points P and Q]

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 - 3) cm = 2 cm.

Also, the common chord PQ is the perpendicular bisector of AB. Therefore, AC = CB = \( \frac{1}{2} \) AB = 1 cm

In right Δ ACP, we have \( AP^2 = AC^2 + CP^2 \)
\( \Rightarrow 5^2 = 1^2 + CP^2 \)
\( \Rightarrow CP^2 = 25 - 1 = 24 \)
\( \Rightarrow CP = \sqrt{24} = 2\sqrt{6} \) cm

Now, PQ = 2 CP
= \( 2 \times 2\sqrt{6} \) cm
= \( 4\sqrt{6} \) cm

📝 Teacher's Note: When two circles touch internally, the distance between centers equals the difference of radii, not the sum. Draw this clearly to avoid confusion.

🎯 Exam Tip: Write "internally touching circles: distance = |r₁ - r₂|" first. Then use the fact that perpendicular bisector creates right triangles. Show all steps clearly.

Question 4. Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of the angle BAC.
Solution:

[Diagram: Circle with center O, two equal chords AB and AC from point A, with angle bisector of angle BAC passing through center O]

Given: AB and AC are two equal chords of C (O, r).

To prove: Centre, O lies on the bisector of ∠ BAC.

Construction: Join BC. Let the bisector of ∠ BAC intersects BC in P.

Proof:

In Δ APB and Δ APC,

AB = AC (Given)

∠ BAP = ∠ CAP (Given)

AP = AP (Common)

∴ ΔAPB ≅ ΔAPC (SAS congruence criterion)

\( \Rightarrow \) BP = CP and ∠ APB = ∠ APC (CPCT)

∠ APB + ∠ APC = 180° (Linear pair)

=> 2 ∠ APB = 180° (∠ APB = ∠ APC)

\( \Rightarrow \) ∠ APB = 90°

Now, BP = CP and ∠ APB = 90°

∴ AP is the perpendicular bisector of chord BC.

\( \Rightarrow \) AP passes through the centre, O of the circle.

📝 Teacher's Note: This is a common theorem. Emphasize that equal chords from the same point create an isosceles triangle. The angle bisector becomes perpendicular bisector of the opposite side.

🎯 Exam Tip: Start with "Given" and "To Prove" clearly. Use SAS congruence. Write "perpendicular bisector of a chord passes through center" as the final step. This gets full marks.

Question 5. The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?
Solution:

[Diagram: Circle with diameter AB (20 cm), chord AC (12 cm), and perpendicular OL from center O to chord AC]

AB is the diameter and AC is the chord.

Draw OL ⊥ AC

Since OL ⊥ AC and hence it bisects AC, O is the centre of the circle.

Therefore, OA = 10 cm and AL = 6 cm

Now, in Rt.ΔOLA,

\( AO^2 = AL^2 + OL^2 \)
\( \Rightarrow 10^2 = 6^2 + OL^2 \)
\( \Rightarrow OL^2 = 100 - 36 = 64 \)
\( \Rightarrow OL = 8 \) cm

Therefore, chord is at a distance of 8 cm from the centre of the circle.

📝 Teacher's Note: Teach students that perpendicular from center to chord always bisects the chord. This creates a right triangle that we can solve using Pythagoras theorem.

🎯 Exam Tip: Always draw the perpendicular from center to chord. Write "perpendicular bisects chord" clearly. Use Pythagoras theorem: (radius)² = (half chord)² + (distance)².

Question 6. ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Solution:

[Diagram: Cyclic quadrilateral ABCD with BC parallel to AD, showing the given angles]

ABCD is a cyclic quadrilateral in which AD∥BC

∠ADC = 110°, ∠BAC = 50°
∠B + ∠D = 180°

(Sum of opposite angles of a quadrilateral)

\( \Rightarrow \) ∠B + 110° = 180°
\( \Rightarrow \) ∠B = 70°

Now in ΔABC,

∠BAC + ∠ABC + ∠ACB = 180°
\( \Rightarrow \) 50° + 70° + ∠ACB = 180°
\( \Rightarrow \) ∠ACB = 180° - 120° = 60°
∵ AD ∥ BC
∴ ∠DAC = ∠ACB = 60° (alternate angles)

Now in ΔADC,

∠DAC + ∠ADC + ∠DCA = 180°
\( \Rightarrow \) 60° + 110° + ∠DCA = 180°
\( \Rightarrow \) ∠DCA = 180° - 170° = 10°

📝 Teacher's Note: In cyclic quadrilaterals, opposite angles add up to 180°. When lines are parallel, use alternate angles property. Solve step by step using angle sum in triangles.

🎯 Exam Tip: Write "opposite angles of cyclic quadrilateral = 180°" first. Then use "alternate angles are equal" for parallel lines. Show each triangle angle calculation clearly.

Question 7. In the given figure, C and D are points on the semicircle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC
Solution:

[Diagram: Semicircle with diameter AB, points C and D on the semicircle, with given angles marked]

Since C and D are points on semicircle with AB as diameter:

∠ACB = 90° and ∠ADB = 90° (angles in semicircle)

Given: ∠BAD = 70° and ∠DBC = 30°

In ΔABD:
∠BAD + ∠ABD + ∠ADB = 180°
70° + ∠ABD + 90° = 180°
∠ABD = 20°

Now, ∠ABC = ∠ABD + ∠DBC = 20° + 30° = 50°

In ΔABC:
∠BAC + ∠ABC + ∠ACB = 180°
∠BAC + 50° + 90° = 180°
∠BAC = 40°

∠BDC = ∠BAC = 40° (angles in the same segment)

📝 Teacher's Note: Remember that angle in a semicircle is always 90°. Use the property that angles subtended by the same arc are equal (angles in same segment).

🎯 Exam Tip: Write "angle in semicircle = 90°" first. Then use "angles in same segment are equal". Mark all right angles clearly in your diagram. Show step-by-step angle calculations.

Question 8. In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.
Answer:
ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180°
\( \implies \) 3∠C + ∠C = 180°
\( \implies \) 4∠C = 180°
\( \implies \) ∠C = 45°
∵ ∠A = 3∠C
\( \implies \) ∠A = 3 × 45°
\( \implies \) ∠A = 135°

Similarly,
∴ ∠B + ∠D = 180°
\( \implies \) ∠B + 5∠B = 180°
\( \implies \) 6∠B = 180°
\( \implies \) ∠B = 30°
∵ ∠D = 5∠B
\( \implies \) ∠D = 5 × 30°
\( \implies \) ∠D = 150°

Hence, ∠A = 135°, ∠B = 30°, ∠C = 45°, ∠D = 150°
In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. We use this rule to find all angles by solving simple equations.

📝 Teacher's Note: Draw a quadrilateral inside a circle on the board. Show students that opposite angles (A and C, B and D) always add to 180°. This is the key rule for cyclic quadrilaterals.

🎯 Exam Tip: Always write "∠A + ∠C = 180°" and "∠B + ∠D = 180°" first. Then substitute the given relationships. Show all algebraic steps clearly.

[Diagram: This diagram shows a cyclic quadrilateral ABCD inscribed in a circle with all four vertices on the circle.]

Question 9. Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Answer:
Join AD.
AB is the diameter.
∴ ∠ADB = 90° (Angle in a semi-circle)
But, ∠ADB + ∠ADC = 180° (linear pair)
\( \implies \) ∠ADC = 90°

In △ABD and △ACD,
∠ADB = ∠ADC (each 90°)
AB = AC (Given)
AD = AD (Common)
∴ △ABD ≅ △ACD (RHS congruence criterion)
\( \implies \) BD = DC (C.P.C.T)

Hence, the circle bisects base BC at D.
In simple words: When we draw a circle using one equal side as diameter, it cuts the base exactly in half. This happens because of the 90° angle property in a semicircle.

📝 Teacher's Note: Use a cardboard isosceles triangle and draw a semicircle. Show students that the semicircle always cuts the base in the middle. This is a beautiful geometric property.

🎯 Exam Tip: Write "Angle in semicircle = 90°" as your first step. Then prove the two triangles are congruent using RHS. Finally write "BD = DC, so D bisects BC."

[Diagram: This diagram shows an isosceles triangle ABC with AB = AC, and a semicircle drawn with AB as diameter that intersects the base BC at point D.]

Question 10. Bisectors of vertex A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90° - \( \frac{1}{2} \)∠A
Answer:
Join ED, EF and DF. Also join BF, FA, AE and EC.
∠EBF = ∠ECF = ∠EDF...........(i) (angles in the same segment)

In cyclic quadrilateral AFBE,
∠EBF + ∠EAF = 180°...........(ii) (Sum of opposite angles)

Similarly in cyclic quadrilateral CEAF,
∠EAF + ∠ECF = 180°...........(iii)

Adding (ii) and (iii)
∠EBF + ∠ECF + 2∠EAF = 360°
\( \implies \) ∠EDF + ∠EDF + 2∠EAF = 360° (from (i))
\( \implies \) 2∠EDF + 2∠EAF = 360°
\( \implies \) ∠EDF + ∠EAF = 180°
\( \implies \) ∠EDF + ∠1 + ∠BAC + ∠2 = 180°
But ∠1 = ∠3 and ∠2 = ∠4
(angles in the same segment)
∴ ∠EDF + ∠3 + ∠BAC + ∠4 = 180°
But ∠4 = \( \frac{1}{2} \)∠C, ∠3 = \( \frac{1}{2} \)∠B
∴ ∠EDF + \( \frac{1}{2} \)∠B + ∠BAC + \( \frac{1}{2} \)∠C = 180°
\( \implies \) ∠EDF + \( \frac{1}{2} \)∠B + 2 × \( \frac{1}{2} \)∠A + \( \frac{1}{2} \)∠C = 180°
\( \implies \) ∠EDF + \( \frac{1}{2} \)(∠A + ∠B + ∠C) + \( \frac{1}{2} \)∠A = 180°
\( \implies \) ∠EDF + \( \frac{1}{2} \)(180°) + \( \frac{1}{2} \)∠A = 180°
\( \implies \) ∠EDF + 90° + \( \frac{1}{2} \)∠A = 180°
\( \implies \) ∠EDF = 180° - (90° + \( \frac{1}{2} \)∠A)
\( \implies \) ∠EDF = 180° - 90° - \( \frac{1}{2} \)∠A
\( \implies \) ∠EDF = 90° - \( \frac{1}{2} \)∠A
In simple words: When angle bisectors meet the circumcircle, they create special angle relationships. The angle at D depends on half of angle A.

📝 Teacher's Note: This is an advanced circle theorem. Focus on the concept that angle bisectors create equal angles in the same segment. Draw the diagram step by step with students.

🎯 Exam Tip: Start by writing "angles in same segment are equal." Use the fact that sum of angles in triangle = 180°. Show each step clearly with proper reasoning.

[Diagram: This diagram shows triangle ABC inscribed in a circle, with angle bisectors from each vertex meeting the circle at points D, E, and F, creating an inner triangle DEF.]

Question 11. In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.
Answer:
Join OB.
In △OBC,
BC = OD = OB (Radii of the same circle)
∴ ∠BOC = ∠BCO = 20°
and Ext.∠ABO = ∠BCO + ∠BOC
\( \implies \) Ext.∠ABO = 20° + 20° = 40°...........(i)

In △OAB,
OA = OB (Radii of the same circle)
∴ ∠OAB = ∠OBA = 40° (from (i))
∠AOB = 180° - ∠OAB - ∠OBA
\( \implies \) ∠AOB = 180° - 40° - 40° = 100°

Since DOC is a straight line
∴ ∠AOD + ∠AOB + ∠BOC = 180°
\( \implies \) ∠AOD + 100° + 20° = 180°
\( \implies \) ∠AOD = 180° - 120°
\( \implies \) ∠AOD = 60°
In simple words: We use the fact that radii are equal to find angles. Then we use the property that angles on a straight line add to 180°.

📝 Teacher's Note: Remind students that all radii of a circle are equal. This creates isosceles triangles which have equal base angles. Use this fact repeatedly in circle problems.

🎯 Exam Tip: Always mark equal radii first. Write "OA = OB = OC (radii)" at the start. Use isosceles triangle properties to find base angles. Show all angle calculations step by step.

[Diagram: This diagram shows a circle with center O, chord AB, and line segment DOC where point D is outside the circle and C is on the circle.]

Question 12. Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Answer:
Let ABC be a right triangle with ∠C = 90°.
Let r be the radius of incircle and R be the radius of circumcircle.
We need to prove: a + b + c = 2r + 4R

For a right triangle:
- Circumradius R = \( \frac{c}{2} \) where c is hypotenuse
- Inradius r = \( \frac{a + b - c}{2} \) where a, b are legs

From these formulas:
2r = a + b - c
4R = 2c

Adding: 2r + 4R = (a + b - c) + 2c = a + b + c

Therefore, perimeter = 2r + 4R
Or perimeter = diameter of incircle + twice diameter of circumcircle
In simple words: For a right triangle, there's a special relationship between its perimeter and the diameters of its two special circles (incircle and circumcircle).

📝 Teacher's Note: Draw a right triangle with both incircle and circumcircle. Show students that the circumcenter is at the midpoint of hypotenuse. The incenter is inside the triangle.

🎯 Exam Tip: Write the standard formulas for circumradius and inradius of right triangle first. Then substitute and simplify algebraically. Remember R = c/2 and r = (a+b-c)/2 for right triangles.

Question 13. P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Answer:

[Diagram: This diagram shows a circle with points A, P, and B on it. P is at the top of the arc AB. A tangent line ST is drawn at point P. There are also construction lines showing the center O and radii.]

Construction: Join AP and BP.

Given: P is the midpoint of arc APB. TPS is tangent at P.

To Prove: TPS || AB

Proof:
Since TPS is a tangent and PA is the chord of the circle,
\( \angle BPT = \angle PAB \) (angles in alternate segments)

But \( \angle PBA = \angle PAB \) (\( \because \) PA = PB)
\( \therefore \angle BPT = \angle PBA \)

But these are alternate angles
\( \therefore \) TPS || AB

In simple words: When P is in the middle of arc AB, the angles made by the tangent at P are equal to the angles made by the chord AB. This makes the tangent line parallel to the chord.

📝 Teacher's Note: Show students that when P is the midpoint of an arc, triangle PAB becomes isosceles. This makes the angle properties work perfectly. Use a string and pins to show this on a board.

🎯 Exam Tip: Always write "angles in alternate segments" when using the tangent-chord angle property. Also mention that PA = PB because P is the midpoint of the arc.

Question 14. In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent. Prove that the line NM produced bisects AB at P.
Answer:

[Diagram: This diagram shows two intersecting circles with MN as their common chord and AB as their common tangent line. The extended line MN meets AB at point P.]

Given: MN is the common chord of two circles. AB is their common tangent.

To Prove: Line NM produced bisects AB at P (i.e., AP = PB)

Proof:
From P, AP is the tangent and PMN is the secant for first circle.
\( \therefore AP^2 = PM \times PN \)......(i)

Again from P, PB is the tangent and PMN is the secant for second circle.
\( \therefore PB^2 = PM \times PN \)......(ii)

From (i) and (ii)
\( AP^2 = PB^2 \)
\( \Rightarrow AP = PB \)

Therefore, P is the midpoint of AB.

In simple words: When two circles share a chord and have a common tangent, the chord line always cuts the tangent exactly in half. This happens because of the equal power of point P with respect to both circles.

📝 Teacher's Note: Explain "power of a point" using simple language. From any outside point, if you draw a tangent and a secant to a circle, tangent squared equals the secant parts multiplied.

🎯 Exam Tip: Write the power of a point formula clearly: (tangent)² = (secant part 1) × (secant part 2). Use this for both circles to prove AP = PB.

Question 15. In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:
(i) ∠DBC
(ii) ∠BCP
(iii) ∠ADB
Answer:

[Diagram: This diagram shows a cyclic quadrilateral ABCD with BD as diameter. PQ is a tangent at point C. Various angles are marked.]

Given: ABCD is cyclic, PQ tangent at C, BD is diameter, ∠DCQ = 40°, ∠ABD = 60°

(i) Finding ∠DBC:
PQ is tangent and CD is a chord
\( \therefore \angle DCQ = \angle DBC \) (angles in the alternate segment)
\( \therefore \angle DBC = 40° \) (\( \because \angle DCQ = 40° \))

(ii) Finding ∠BCP:
\( \angle DCQ + \angle DCB + \angle BCP = 180° \)
\( \Rightarrow 40° + 90° + \angle BCP = 180° \) (\( \because \angle DCB = 90° \))
\( \Rightarrow \angle BCP = 180° - 130° = 50° \)

(iii) Finding ∠ADB:
In triangle ABD,
\( \angle BAD = 90° \), \( \angle ABD = 60° \)
\( \therefore \angle ADB = 180° - (90° + 60°) \)
\( \Rightarrow \angle ADB = 180° - 150° = 30° \)

In simple words: We use the rule that tangent-chord angles equal alternate segment angles. Also, angles in a semicircle are 90°. These two rules help us find all the required angles.

📝 Teacher's Note: Remind students that when BD is diameter, any angle subtended by BD on the circle is 90°. This is the angle in semicircle property. Very important for cyclic quadrilaterals.

🎯 Exam Tip: Always state which property you are using: "angle in alternate segment" or "angle in semicircle = 90°". Write the given angles clearly first, then solve step by step.

Question 16. The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90°
Answer:

[Diagram: This diagram shows a circle with center O. BCD is a tangent line at point C. Points A and C are on the circle, with various angles marked.]

Given: Circle with center O, BCD is tangent at C

To Prove: ∠ACD + ∠BAC = 90°

Construction: Join OC.

Proof:
BCD is the tangent and OC is the radius.
\( \therefore OC \perp BD \)
\( \Rightarrow \angle OCD = 90° \)
\( \Rightarrow \angle OCA + \angle ACD = 90° \)......(i)

But in triangle OCA,
OA = OC (radii of same circle)
\( \therefore \angle OCA = \angle OAC \)

Substituting in (i)
\( \angle OAC + \angle ACD = 90° \)
\( \Rightarrow \angle BAC + \angle ACD = 90° \)

In simple words: A tangent is always perpendicular to the radius at the point of contact. Using this fact and properties of isosceles triangles, we can prove the required angle sum is 90°.

📝 Teacher's Note: Draw the radius to the point of tangency first. Show students that this creates a 90° angle. Then use the isosceles triangle property (two radii are equal).

🎯 Exam Tip: Always join the center to the point of tangency when solving tangent problems. Write "tangent perpendicular to radius" as your first step. Then use "radii are equal" for isosceles triangles.

Question 17. ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that –
(i) AC × AD = AB²
(ii) BD² = AD × DC
Answer:

[Diagram: This diagram shows right triangle ABC with angle B = 90°. A circle is drawn with BC as diameter, meeting AC at point D. Point O is the center of the circle.]

Given: Triangle ABC with ∠B = 90°, circle with BC as diameter meets AC at D

(i) To Prove: AC × AD = AB²**

In triangle ABC,
∠B = 90° and BC is the diameter of the circle.
Therefore, AB is the tangent to the circle at B.
Now, AB is tangent and ADC is the secant
\( \therefore AB^2 = AD \times AC \)

(ii) To Prove: BD² = AD × DC**

In triangle ADB,
\( \angle ADB = 90° \) (angle in semicircle)
In triangle BDC,
\( \angle BDC = 90° \) (angle in semicircle)

From the intersecting chords theorem in the circle:
\( BD^2 = AD \times DC \)

In simple words: When we draw a circle on the hypotenuse of a right triangle, it creates special relationships. The original right angle helps us use tangent-secant properties and intersecting chord properties.

📝 Teacher's Note: Show that when BC is diameter, any angle subtended by BC on the circle is 90°. This makes AB tangent to the circle. Use the power of point A for the first part.

🎯 Exam Tip: For part (i), use tangent-secant formula: (tangent)² = (near secant part) × (whole secant). For part (ii), use the fact that BD is perpendicular to AC, making it a chord, then apply intersecting chords theorem.

Question 18. In the given figure AC = AE. Show that:
(i) CP = EP
(ii) BP = DP

[Diagram: A circle with external point A connected to the circle at points C and E. Points B, D, and P are also marked on the circle, with intersecting chords forming triangles.]

Answer:

Given: AC = AE

To Prove:
(i) CP = EP
(ii) BP = DP

Proof:

Part (i): To prove CP = EP

In triangles ΔADC and ΔABE,
∠ACD = ∠AEB (angles in the same segment)
AC = AE (Given)
∠A = ∠A (Common)

∴ ΔADC ≅ ΔABE (ASA Postulate)
⇒ AB = AD

but AC = AE
∴ AC - AB = AE - AD
⇒ BC = DE

In ΔBPC and ΔDPE,
∠C = ∠E (angles in the same segment)
BC = DE
∠CBP = ∠CDE (angles in the same segment)

∴ ΔBPC ≅ ΔDPE (ASA Postulate)
⇒ BP = DP and CP = PE (cpct)

Part (ii): BP = DP is already proved above.

In simple words: When two chords from outside point are equal, they create equal segments inside the circle. This happens because of congruent triangles.

📝 Teacher's Note: Draw the diagram clearly and mark all equal parts. Students often forget to use "angles in same segment are equal" property. Practice identifying congruent triangles step by step.

🎯 Exam Tip: Always write "Given", "To Prove", and "Proof" clearly. Use proper triangle congruence criteria names like ASA, SAS. Write "cpct" when using corresponding parts of congruent triangles.

Question 19. ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°. Calculate:
(i) ∠BEC
(ii) ∠BED

[Diagram: A cyclic pentagon ABCDE inscribed in a circle with center O. The pentagon shows equal sides AB = BC = CD and angle ABC = 120°.]

Answer:

Given:
AB = BC = CD and ∠ABC = 120°

Solution:

i) Join OC and OB.

AB = BC = CD and ∠ABC = 120°
∴ ∠BCD = ∠ABC = 120°

OB and OC are the bisectors of ∠ABC and ∠BCD respectively.
∴ ∠OBC = ∠BCO = 60°

In ΔBOC,
∠BOC = 180° - (∠OBC + ∠BCO)
⇒ ∠BOC = 180° - (60° + 60°)
⇒ ∠BOC = 180° - 120° = 60°

Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.
∴ ∠BEC = \( \frac{1}{2} \) ∠BOC = \( \frac{1}{2} \) × 60° = 30°

ii) In cyclic quadrilateral BCDE,
∠BED + ∠BCD = 180°
⇒ ∠BED + 120° = 180°
∴ ∠BED = 60°

In simple words: In a cyclic polygon, angles at center are double the angles at circumference. Also, opposite angles in a cyclic quadrilateral add up to 180°.

📝 Teacher's Note: Explain that equal chords subtend equal angles at center. Draw radii to show this clearly. Students often confuse inscribed angle theorem - angle at center is double the angle at circumference.

🎯 Exam Tip: Always join center to vertices when finding central angles. Write the inscribed angle theorem formula clearly. For cyclic quadrilaterals, opposite angles sum to 180°.

Question 20. In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30°, find:
(i) angle BCO
(ii) angle AOB
(iii) angle APB

[Diagram: A circle with center O, with tangents drawn from external point C touching the circle at points A and B. Point P is also marked on the circle.]

Answer:

Given: O is the centre of the circle and CA and CB are the tangents to the circle from C. Also, ∠ACO = 30°

P is any point on the circle. P and PB are joined.

To find:
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB

Proof:

(i) In ΔOAC and OBC,
OC = OC (common)
OA = OB (radius of the circle)
CA = CB (tangents to the circle)

∴ ΔOAC ≅ ΔOBC (SSS congruence criterion)
∴ ∠ACO = ∠BCO = 30°

(ii) ∴ ∠ACB = 30° + 30° = 60°
∴ ∠AOB + ∠ACB = 180°
⇒ ∠AOB + 60° = 180°
⇒ ∠AOB = 180° - 60°
⇒ ∠AOB = 120°

(iii) Arc AB subtends ∠AOB at the centre and ∠APB is in the remaining part of the circle
∴ ∠APB = \( \frac{1}{2} \) ∠AOB = \( \frac{1}{2} \) × 120° = 60°

In simple words: Tangents from outside point to a circle are equal. The angle between tangents and the central angle add up to 180°. Inscribed angle is half of central angle.

📝 Teacher's Note: Show students that tangents from external point are equal using compass. This creates congruent triangles. The quadrilateral CAOB has special properties - two right angles at A and B.

🎯 Exam Tip: Remember that tangent is perpendicular to radius at point of contact. Tangents from external point are equal. Use inscribed angle theorem: angle at circumference = half angle at center.

Question 21. ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.

[Diagram: Triangle ABC with three circles centered at vertices A, B, and C. The circles touch each other externally at points P, Q, and R.]

Answer:

Given: ABC is a triangle with AB = 10 cm, BC = 8 cm, AC = 6 cm. Three circles are drawn with centre A, B and C touch each other at P, Q and R respectively.

We need to find the radii of the three circles.

Let:
PA = AQ = x
QC = CR = y
RB = BP = z

∴ x + z = 10 .....(1)
z + y = 8 .....(2)
y + x = 6 .....(3)

Solving the equations:

From equation (1): z = 10 - x
From equation (3): y = 6 - x

Substituting in equation (2):
(10 - x) + (6 - x) = 8
16 - 2x = 8
2x = 8
x = 4

∴ y = 6 - 4 = 2
z = 10 - 4 = 6

Therefore:
Radius of circle with centre A = 4 cm
Radius of circle with centre B = 6 cm
Radius of circle with centre C = 2 cm

In simple words: When circles touch each other, the distance between their centers equals the sum of their radii. We get three equations from the three sides of the triangle and solve them.

📝 Teacher's Note: Explain that when two circles touch externally, distance between centers = sum of radii. Draw this clearly. Students should practice solving three simultaneous equations by substitution method.

🎯 Exam Tip: Always label the radii clearly as x, y, z. Write all three equations from the given side lengths. Solve step by step and check that x + y + z equals the semi-perimeter. Don't forget units in final answer.

Question 22. In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that –
i) \( \angle ONL + \angle OML = 180° \)
ii) \( \angle BAM = \angle BMA \)
iii) ALOB is a cyclic quadrilateral.

[Diagram: A square ABCD with diagonals AC and BD intersecting at O. Angle bisector from A meets BD at M, angle bisector from B meets AC at N and AM at L.]

Solution:

i) To prove: \( \angle ONL + \angle OML = 180° \)

\( \angle AOB = \angle AOD = 90° \)
In \( \triangle ANB \),
\( \angle ANB = 180° - (\angle NAB + \angle NBA) \)

\( \Rightarrow \angle ANB = 180° - \left(45° + \frac{45°}{2}\right) \) (NB is bisector of \( \angle ABD \))

\( \Rightarrow \angle ANB = 180° - 45° - \frac{45°}{2} = 135° - \frac{45°}{2} \)

But, \( \angle LNO = \angle ANB \) (vertically opposite angles)

\( \therefore \angle LNO = 135° - \frac{45°}{2} \) ......(i)

Now in \( \triangle AMO \),
\( \angle AMO = 180° - (\angle AOM + \angle OAM) \)

\( \Rightarrow \angle AMO = 180° - \left(90° + \frac{45°}{2}\right) \) (MA is bisector of \( \angle DAO \))

\( \Rightarrow \angle AMO = 180° - 90° - \frac{45°}{2} = 90° - \frac{45°}{2} \) ......(ii)

Adding (i) and (ii)

\( \angle LNO + \angle AMO = 135° - \frac{45°}{2} + 90° - \frac{45°}{2} \)

\( \Rightarrow \angle LNO + \angle AMO = 225° - 45° = 180° \)

\( \Rightarrow \angle ONL + \angle OML = 180° \)

ii) To prove: \( \angle BAM = \angle BMA \)

\( \angle BAM = \angle BAO + \angle OAM \)

\( \Rightarrow \angle BAM = 45° + \frac{45°}{2} = 67\frac{1°}{2} \)

and
\( \angle BMA = 180° - (\angle AOM + \angle OAM) \)

\( \Rightarrow \angle BMA = 180° - 90° - \frac{45°}{2} = 90° - \frac{45°}{2} = 67\frac{1°}{2} \)

\( \therefore \angle BAM = \angle BMA \)

iii) To prove: ALOB is a cyclic quadrilateral

In quadrilateral ALOB,
\( \angle ABO + \angle ALO = 45° + 90° + 45° = 180° \)
Therefore, ALOB is a cyclic quadrilateral.

In simple words: In a square, the diagonals cut each other at 90°. When we draw angle bisectors, they create special triangles. We use angle properties to prove these relationships step by step.

📝 Teacher's Note: Draw a square on the board and show how angle bisectors work. Students often forget that in a square, all angles are 90° and diagonals bisect at 90°. Show this clearly with colors.

🎯 Exam Tip: Always write "vertically opposite angles" and "angle bisector property" clearly. Mark all given angles on the diagram first. Write each step with proper reasons.

Question 23. The given figure shows a semicircle with centre O and diameter PQ. If PA = AB and \( \angle BOQ = 140° \); find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

[Diagram: A semicircle with center O and diameter PQ. Points A and B are on the semicircle, with PA = AB.]

Solution:

Join PB.

i) In cyclic quadrilateral PBCQ,

\( \angle BPQ + \angle BCQ = 180° \)
\( \Rightarrow \angle BPQ + 140° = 180° \)
\( \Rightarrow \angle BPQ = 40° \) ......(1)

Now in \( \triangle PBQ \),

\( \angle PBQ + \angle BPQ + \angle BQP = 180° \)
\( \Rightarrow 90° + 40° + \angle BQP = 180° \)
\( \Rightarrow \angle BQP = 50° \)

In cyclic quadrilateral PQBA,

\( \angle PQB + \angle PAB = 180° \)
\( \Rightarrow 50° + \angle PAB = 180° \)
\( \Rightarrow \angle PAB = 130° \)

ii) Now in \( \triangle PAB \),

\( \angle PAB + \angle APB + \angle ABP = 180° \)
\( \Rightarrow 130° + \angle APB + \angle ABP = 180° \)
\( \Rightarrow \angle APB + \angle ABP = 50° \)

But
\( \angle APB = \angle ABP \) (\( \because \) PA = PB
\( \therefore \angle APB = \angle ABP = 25° \)
\( \angle BAQ = \angle BPQ = 40° \)
\( \angle APB = 25° = \angle AQB \) (angles in the same segment)

\( \therefore \angle AQB = 25° \) ......(2)

iii) Arc AQ subtends \( \angle AOQ \) at the centre and \( \angle APQ \) at the remaining part of the circle.

We have,
\( \angle APQ = \angle APB + \angle BPQ \) ......(3)

From (1), (2) and (3), we have
\( \angle APQ = 25° + 40° = 65° \)

\( \therefore \angle AOQ = 2 \angle APQ = 2 \times 65° = 130° \)

Now in \( \triangle AOQ \),

\( \angle OAQ = \angle OQA \) (\( \because \) OA = OQ)
but
\( \angle OAQ + \angle OQA + \angle AOQ = 180° \)
\( \Rightarrow \angle OAQ + \angle OAQ + 130° = 180° \)
\( \Rightarrow 2 \angle OAQ = 50° \)
\( \Rightarrow \angle OAQ = 25° \)

\( \therefore \angle OAQ = \angle AQB \)

But these are alternate angles.
Hence, AO is parallel to BQ.

In simple words: In a semicircle, angles in the same segment are equal. When PA = AB, triangle PAB becomes isosceles. We use circle properties to find all angles step by step.

📝 Teacher's Note: Remind students that angles in a semicircle are 90°. Also show that when two sides are equal, the triangle is isosceles and base angles are equal.

🎯 Exam Tip: Always mark equal sides and equal angles on the diagram. Write "angles in same segment" and "alternate angles" clearly. These key phrases get you marks.

Question 24. The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate –
i) angle QTR
ii) angle QRP
iii) angle QRS
iv) angle STR

[Diagram: A circle with center O showing chords RS parallel to QT, with various points P, Q, R, S, T on the circle.]

Solution:

Join PQ, RQ and ST.

i)

\( \angle POQ + \angle QOR = 180° \)
\( \Rightarrow 100° + \angle QOR = 180° \)
\( \Rightarrow \angle QOR = 80° \)

Arc RQ subtends \( \angle QOR \) at the centre and \( \angle QTR \) at the remaining part of the circle.

\( \therefore \angle QTR = \frac{1}{2} \angle QOR \)

\( \Rightarrow \angle QTR = \frac{1}{2} \times 80° = 40° \)

ii) Arc QP subtends \( \angle QOP \) at the centre and \( \angle QRP \) at the remaining part of the circle.

\( \therefore \angle QRP = \frac{1}{2} \angle QOP \)

\( \Rightarrow \angle QRP = \frac{1}{2} \times 100° = 50° \)

iii) RS || QT

\( \therefore \angle SRT = \angle QTR \) (alternate angles)
but \( \angle QTR = 40° \)
\( \therefore \angle SRT = 40° \)
Now,
\( \angle QRS = \angle QRP + \angle PRT + \angle SRT \)
\( \Rightarrow \angle QRS = 50° + 20° + 40° = 110° \)

iv) Since RSTQ is a cyclic quadrilateral

\( \therefore \angle QRS + \angle QTS = 180° \) (sum of opposite angles)

\( \Rightarrow \angle QRS + \angle QTR + \angle STR = 180° \)
\( \Rightarrow 110° + 40° + \angle STR = 180° \)
\( \Rightarrow \angle STR = 30° \)

In simple words: When chords are parallel, they create alternate angles. We use the property that angle at center is double the angle at circumference. In cyclic quadrilaterals, opposite angles add up to 180°.

📝 Teacher's Note: Draw parallel lines and show alternate angles using different colors. Students often forget that opposite angles in a cyclic quadrilateral sum to 180°. Practice this with examples.

🎯 Exam Tip: Always write "alternate angles" when chords are parallel. Write "angle at center = 2 × angle at circumference" clearly. Mark the parallel symbol (||) in your diagram.

Question 25. In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:
i) ∠BAP = ∠ADQ
ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB.

[Diagram: A circle with center O. PAT is a tangent line touching the circle at point A. There is a chord BC parallel to the tangent line PAT. Points D and Q are on the circle, with CDQ forming a line segment through the circle.]

Solution:
i) Since PAT||BC
∴ ∠PAB = ∠ABC (alternate angles).......(i)

In cyclic quadrilateral ABCD,
Ext.∠ADQ = ∠ABC.......(ii)

from (i) and (ii)
∠PAB = ∠ADQ

ii) Arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.
∴ ∠AOB = 2∠ADB
⇒ ∠AOB = 2∠PAB (angles in alternate segments)
⇒ ∠AOB = 2∠ADQ (proved in (i) part)

iii)
∴ ∠BAP = ∠ADB (angles in alternate segments)
but
∠BAP = ∠ADQ (proved in (i) part)
∴ ∠ADQ = ∠ADB

In simple words: We use the fact that PAT is parallel to BC and properties of circles. When a tangent is parallel to a chord, it creates equal angles. We also use the rule that angles subtended by the same arc are equal.

📝 Teacher's Note: Draw the diagram step by step on the board. Show students how parallel lines create alternate angles. Then show how the same arc creates equal angles in the circle. This makes the proof easy to follow.

🎯 Exam Tip: Always write "alternate angles" and "angles in alternate segments" clearly. These are the key phrases examiners look for. Also mention "cyclic quadrilateral" when using that property.

Question 26. AB is a line segment and M is its midpoint. Three semicircles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semicircles. Show that: AB = 6 × r

[Diagram: A line segment AB with midpoint M. Three semicircles are drawn - one with diameter AM, one with diameter MB, and one large semicircle with diameter AB. All three semicircles are on the same side of line AB. A small circle with radius r touches all three semicircles from inside.]

Solution:
Let O, P and Q be the centers of the circle and semicircles.

Join OP and OQ.

OR = OS = r

and AP = PM = MQ = QB = \(\frac{AB}{4}\)

Now, OP = OR + RP = r + \(\frac{AB}{4}\) (since PM=RP=radii of same circle)

Similarly, OQ = OS + SQ = r + \(\frac{AB}{4}\)

OM = LM + OL = \(\frac{AB}{2}\) - r

Now in Rt. △OPM,
OP² = PM² + OM²
⇒ \(\left(r + \frac{AB}{4}\right)^2 = \left(\frac{AB}{4}\right)^2 + \left(\frac{AB}{2} - r\right)^2\)
⇒ \(r^2 + \frac{AB^2}{16} + \frac{rAB}{2} = \frac{AB^2}{16} + \frac{AB^2}{4} + r^2 - rAB\)
⇒ \(\frac{rAB}{2} = \frac{AB^2}{4} - rAB\)
⇒ \(\frac{AB^2}{4} = \frac{rAB}{2} + rAB\)
⇒ \(\frac{AB^2}{4} = \frac{3rAB}{2}\)
⇒ \(\frac{AB}{4} = \frac{3r}{2}\)
⇒ AB = \(\frac{3r}{2}\) × 4 = 6r

Hence AB = 6 × r

In simple words: We place the small circle so it touches all three semicircles. Using the Pythagorean theorem on the right triangle formed, we can find the relationship between AB and r. The answer comes out to be AB = 6r.

📝 Teacher's Note: This is a complex problem. Show students how to identify the right triangle OPM. Explain that when circles touch, the distance between centers equals sum of radii. Work through the algebra step by step.

🎯 Exam Tip: Set up coordinates carefully. Identify which triangles are right triangles. Use Pythagorean theorem correctly. Show all algebraic steps clearly - don't skip any steps in the calculation.

Question 27. TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.

[Diagram: A circle with center O. From external point T, two tangent lines TA and TB are drawn to the circle. The line OT passes through the center and intersects the circle at point P. Point A and B are the points where the tangents touch the circle.]

Solution:
Join PB.

In △ TAP and △ TBP,
TA = TB (tangent segments from an external points are equal in length)
Also, ∠ATP = ∠BTP (since OT is equally inclined with TA and TB) TP = TP (common)
⇒ △TAP ≅ △TBP (by SAS criterion of congruency)
⇒ ∠TAP = ∠TBP (corresponding parts of congruent triangles are equal)
But ∠TBP = ∠BAP (angles in alternate segments)
Therefore, ∠TAP = ∠BAP.
Hence, AP bisects ∠TAB.

In simple words: Since TA and TB are equal tangents, and OT passes through the center, everything is symmetric. This symmetry makes the triangles TAP and TBP equal. So the angles are also equal, which means AP bisects angle TAB.

📝 Teacher's Note: Emphasize the symmetry in this problem. When two tangents are drawn from an external point, everything is symmetric about the line joining the external point to the center. This is the key insight.

🎯 Exam Tip: Always mention "tangent segments from external point are equal" and "SAS congruency". Then use "corresponding parts of congruent triangles are equal". These are the key steps examiners look for.

Question 28. Two circles intersect in points P and Q. A secant passing through P intersects the circle in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.

[Diagram: Two circles intersecting at points P and Q. A line through P intersects one circle at A and the other circle at B. Tangent lines are drawn at points A and B, and these tangents meet at point T.]

Solution:
Join PQ.
AT is tangent and AP is a chord,
∴ ∠TAP = ∠AQP (angles in alternate segments).......(i)
Similarly, ∠TBP = ∠BQP.......(ii)
Adding (i) and (ii)
∠TAP + ∠TBP = ∠AQP + ∠BQP
⇒ ∠TAP + ∠TBP = ∠AQB...........(iii)
Now in △TAB,
∠ATB + ∠TAP + ∠TBP = 180°
⇒ ∠ATB + ∠AQB = 180°
Therefore, AQBT is a cyclic quadrilateral.
Hence, A, Q, B and T lie on a circle.

In simple words: We use the property that angles in alternate segments are equal. When we add up the angles, we find that opposite angles of quadrilateral AQBT add up to 180°. This means AQBT is a cyclic quadrilateral.

📝 Teacher's Note: The key insight is using alternate segment theorem twice, then adding the angles. Show students that when opposite angles of a quadrilateral add to 180°, the quadrilateral is cyclic.

🎯 Exam Tip: Write "angles in alternate segments" clearly. Then show that ∠ATB + ∠AQB = 180°. Always conclude with "Therefore, AQBT is a cyclic quadrilateral" to get full marks.

Question 29. Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle)

[Diagram: A regular pentagon ABCDE with vertices labeled A, B, C, D, E in order.]

Solution:
ABCDE is a regular pentagon.
∴ ∠BAE = ∠ABC = ∠BCD = ∠CDE = ∠DEA = \(\left(\frac{5-2}{5}\right) \times 180° = 108°\)

In △AED,
AE = ED (Sides of regular pentagon ABCDE)
∴ ∠EAD = ∠EDA

In △AED,
∠AED + ∠EAD + ∠EDA = 180°
⇒ 108° + ∠EAD + ∠EAD = 180°
⇒ 2∠EAD = 180° - 108° = 72°
⇒ ∠EAD = 36°

∴ ∠EDA = 36°

∠BAD = ∠BAE - ∠EAD = 108° - 36° = 72°

In quadrilateral ABCD,
∠BAD + ∠BCD = 108° + 72° = 180°

∴ ABCD is a cyclic quadrilateral

In simple words: In a regular pentagon, each interior angle is 108°. By calculating angles in triangles, we find that opposite angles in any set of four vertices add up to 180°. This means any four vertices lie on a circle.

📝 Teacher's Note: Start with the interior angle formula for regular polygons. Show how to break down the pentagon into triangles. The key is showing that opposite angles in any quadrilateral formed by four vertices sum to 180°.

🎯 Exam Tip: Always write the interior angle formula: \(\left(\frac{n-2}{n}\right) \times 180°\). Show that opposite angles sum to 180°. Conclude with "Therefore, ABCD is a cyclic quadrilateral" for any four vertices chosen.

Question 30. Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm. Calculate the length of CD.

Answer:
Given:
AB = 4 cm
BX = 6 cm
XD = 5 cm

To find: Length of CD

Solution:
When two chords of a circle intersect (or their extensions intersect), we use the theorem:
XA × XB = XC × XD

First, find XA:
XA = AB + BX = 4 + 6 = 10 cm

Now applying the theorem:
XA × XB = XC × XD
10 × 6 = XC × 5
60 = XC × 5
XC = 60 ÷ 5 = 12 cm

Therefore, CD = XC - XD = 12 - 5 = 7 cm

In simple words: When two chords meet outside a circle, there's a special rule. The products of the segments are equal. We use this rule to find the missing length.

📝 Teacher's Note: Draw a clear diagram showing X outside the circle with chords AB and CD extended to meet at X. Emphasize that XA × XB = XC × XD is the intersecting chords theorem (external case).

🎯 Exam Tip: Always write the theorem clearly: "XA × XB = XC × XD". Show all calculations step by step. Remember CD = XC - XD, not just XC. Units are important in the final answer.

Question 31. In the given figure, find TP if AT = 16 cm and AB = 12 cm.
Answer:
[Diagram: This diagram shows a circle with point T outside the circle. From T, there are two lines - one tangent line TP touching the circle at P, and one secant line TBA passing through points B and A on the circle. AT = 16 cm and AB = 12 cm are marked.]

PT is the tangent and TBA is the secant of the circle.

Therefore, \( TP^2 = TA \times TB \)

\( TP^2 = 16 \times (16-12) = 16 \times 4 = 64 = (8)^2 \)

Therefore, TP = 8 cm

In simple words: When a tangent and secant are drawn from the same outside point, there is a special rule. The tangent length squared equals the product of the whole secant and its outside part.

📝 Teacher's Note: Draw this on the board and show students the tangent-secant theorem formula. Make them remember: tangent squared = whole secant × outside part.

🎯 Exam Tip: Always write the theorem first: "PT is tangent, TBA is secant, so TP² = TA × TB". Then substitute values and solve step by step.

Question 32. In the following figure, A circle is inscribed in the quadrilateral ABCD. If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.
Answer:
[Diagram: This diagram shows a quadrilateral ABCD with a circle inscribed inside it (touching all four sides). The circle has center O with radius marked. BC = 38 cm, QB = 27 cm, DC = 25 cm, and AD is perpendicular to DC.]

From the figure we see that BQ = BR = 27 cm (since length of the tangent segments from an external point are equal)

As BC = 38 cm

\( \Rightarrow \) CR = CB - BR = 38 - 27

= 11 cm

Again,

CR = CS = 11 cm (length of tangent segments from an external point are equal)

Now, as DC = 25 cm

\( \therefore \) DS = DC - SC

= 25 - 11

= 14 cm

Now, in quadrilateral DSOP,

\( \angle \) PDS = 90° (given)

\( \angle \) OSD = 90° , \( \angle \) OPD = 90° (since tangent is perpendicular to the radius through the point of contact)

\( \Rightarrow \) DSOP is a parallelogram

\( \Rightarrow \) OP||SD and \( \Rightarrow \) PD||OS

Now, as OP = OS (radii of the same circle)

\( \Rightarrow \) OPDS is a square. \( \therefore \) DS = OP = 14 cm

\( \therefore \) radius of the circle = 14 cm

In simple words: When a circle is inside a quadrilateral touching all sides, the tangent lengths from each corner are equal. We use this rule to find missing lengths and then find the radius.

📝 Teacher's Note: Teach students that tangent segments from the same external point are always equal. This is the key property used in inscribed circle problems.

🎯 Exam Tip: Always mark equal tangent lengths with the same letter or symbol. Write "tangent segments from external point are equal" to show you know the property.

Question 33. In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that ∠AXB = 50° and ∠ABX = 70°. Calculate ∠BAY and ∠APY.
Answer:
[Diagram: This diagram shows a circle with diameter XY. Point A and B are on the circle. PQ is a tangent line at Y. Angles AXB = 50° and ABX = 70° are marked.]

In △ AXB,

\( \angle \) XAB + \( \angle \) AXB + \( \angle \) ABX = 180° [Triangle property]

\( \Rightarrow \angle \) XAB + 50° + 70° = 180°

\( \Rightarrow \angle \) XAB = 180° - 120° = 60°

\( \Rightarrow \angle \) XAY = 90° [Angle of semi-circle]

\( \therefore \angle \) BAY = \( \angle \) XAY - \( \angle \) XAB = 90° - 60° = 30°

and \( \angle \) BXY = \( \angle \) BAY = 30° [Angle of same segment]

\( \therefore \angle \) ACX = \( \angle \) BXY + \( \angle \) ABX [External angle = Sum of two interior angles]

= 30° + 70°

= 100°

also,

\( \angle \) XYP = 90° [Diameter ⊥ tangent]

\( \angle \) APY = \( \angle \) ACX - \( \angle \) CYP

\( \angle \) APY = 100° - 90°

\( \angle \) APY = 10°

In simple words: We use three main rules here: angles in a triangle add to 180°, angle in semicircle is 90°, and diameter is perpendicular to tangent.

📝 Teacher's Note: Remind students that angle in semicircle is always 90°. This is because the arc is exactly half the circle. Also, tangent is always perpendicular to radius.

🎯 Exam Tip: Write "angle in semicircle = 90°" and "diameter ⊥ tangent = 90°" clearly. These are the two key facts examiners look for in such problems.

Question 34. In the given figure, QAP is the tangent at point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°; find:
(i) ∠BAP
(ii) ∠ABD
(iii) ∠QAD
(iv) ∠BCD
Answer:
[Diagram: This diagram shows a circle with tangent QAP at point A. PBD is a straight line through the circle. Points C and D are also on the circle. Angles ACB = 36° and APB = 42° are marked.]

PAQ is a tangent and AB is a chord of the circle.

(i) \( \therefore \angle \) BAP = \( \angle \) ACB = 36° (angles in alternate segment)

(ii) In △APB,

Ext.\( \angle \) ABD = \( \angle \) APB + \( \angle \) BAP
\( \Rightarrow \) Ext.\( \angle \) ABD = 42° + 36° = 78°

(iii) \( \angle \) ADB = \( \angle \) ACB = 36° (angles in the same segment)

In △ABD,
\( \angle \) BAD + \( \angle \) ABD + \( \angle \) ADB = 180°
\( \angle \) BAD + 78° + 36° = 180°
\( \angle \) BAD = 180° - 114° = 66°

Now, \( \angle \) QAD = \( \angle \) QAB + \( \angle \) BAD
\( \angle \) QAD = 180° + 66° = 246°

But since we need the acute angle:
\( \angle \) QAD = 360° - 246° = 114°

Or more simply: \( \angle \) QAD = 180° - \( \angle \) BAD = 180° - 66° = 114°

(iv) \( \angle \) BCD = \( \angle \) BAD = 66° (angles subtended by same arc BD)

In simple words: We use the alternate segment theorem which says tangent-chord angle equals the angle in the alternate segment. Also, angles in the same segment are equal.

📝 Teacher's Note: The alternate segment theorem is very important. Draw the tangent and chord clearly, then show which angles are equal. Practice with different examples.

🎯 Exam Tip: Always write "alternate segment theorem" or "angles in same segment are equal" to show you know which property you're using. This gets you method marks.

Question 35. In the given figure, AB is the diameter. The tangent at C meets AB produced at Q. If ∠CAB = 34°, find i) ∠CBA ii) ∠CQB

[Diagram: A circle with diameter AB. Point C is on the circle. A tangent line at C extends to meet line AB (extended beyond B) at point Q.]

Answer:
i) Finding ∠CBA:
AB is diameter of circle.
\( \therefore \) ∠ACB = 90°
In triangle ABC,
∠A + ∠B + ∠C = 180°
\( \Rightarrow \) 34° + ∠CBA + 90° = 180°
\( \Rightarrow \) ∠CBA = 56°

ii) Finding ∠CQB:
QC is tangent to the circle
\( \therefore \) ∠CAB = ∠QCB
Angle between tangent and chord = angle in alternate segment
\( \therefore \) ∠QCB = 34°
ABQ is a straight line
\( \Rightarrow \) ∠ABC + ∠CBQ = 180°
\( \Rightarrow \) 56° + ∠CBQ = 180°
\( \Rightarrow \) ∠CBQ = 124°
Now,
∠CQB = 180° – ∠QCB – ∠CBQ
\( \Rightarrow \) ∠CQB = 180° – 34° – 124°
\( \Rightarrow \) ∠CQB = 22°

In simple words: When diameter makes a triangle with a point on circle, the angle at that point is always 90°. A tangent makes the same angle with a chord as the angle in the opposite segment.

📝 Teacher's Note: Draw a semicircle and show students that any angle from diameter is 90°. Then show how tangent angle equals the angle on opposite side of chord. Students remember this pattern easily.

🎯 Exam Tip: Always write "angle in semicircle = 90°" and "tangent-chord angle = alternate segment angle". These are key phrases examiners look for.

Question 36. In the given figure, O is the centre of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find: i) ∠BOD ii) ∠BPD

[Diagram: A circle with center O. Points A, B, C, D are on the circle. AB is parallel to CD. Tangents at B and D meet at external point P.]

Answer:
i) Finding ∠BOD:
∠BOD = 2∠BCD
\( \Rightarrow \) ∠BOD = 2 × 55° = 110°

ii) Finding ∠BPD:
Since BPDO is cyclic quadrilateral, opposite angles are supplementary.
\( \therefore \) ∠BOD + ∠BPD = 180°
\( \Rightarrow \) ∠BPD = 180° – 110° = 70°

In simple words: The angle at center is double the angle at circumference. When two tangents meet outside a circle, they form a quadrilateral with the center where opposite angles add up to 180°.

📝 Teacher's Note: Show students that central angle is always double the inscribed angle. Use a protractor to measure and verify. This helps them remember the relationship.

🎯 Exam Tip: Write "central angle = 2 × inscribed angle" clearly. Also write "opposite angles in cyclic quadrilateral are supplementary" for full marks.

Question 37. In the figure given below PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of: i) ∠QOP ii) ∠QCP

[Diagram: A circle with center O. Triangle PQR is shown with PQ = QR. Points P and R are on the circle. Point C is outside the circle with PC and CQ as tangents.]

Answer:
i) Finding ∠QOP:
PQ = RQ
\( \therefore \) ∠PRQ = ∠QPR (opposite angles of equal sides of a triangle)
\( \Rightarrow \) ∠PRQ + ∠QPR + 68° = 180°
\( \Rightarrow \) 2∠PRQ = 180° – 68°
\( \Rightarrow \) ∠PRQ = \( \frac{112°}{2} \) = 56°
Now, ∠QOP = 2∠PRQ (angle at the centre is double)
\( \Rightarrow \) ∠QOP = 2 × 56° = 112°

ii) Finding ∠QCP:
∠PQC = ∠PRQ (angles in alternate segments are equal)
∠QPC = ∠PRQ (angles in alternate segments)
\( \therefore \) ∠PQC = ∠QPC = 56° (\( \because \) ∠PRQ = 56° from i))
∠PQC + ∠QPC + ∠QCP = 180°
\( \Rightarrow \) 56° + 56° + ∠QCP = 180°
\( \Rightarrow \) ∠QCP = 68°

In simple words: In an isosceles triangle, the two base angles are equal. The center angle is double the angle made by the same arc at the circumference.

📝 Teacher's Note: Show students an isosceles triangle first. Mark the equal sides and equal angles. Then connect this to circle properties. This step-by-step approach helps weak students.

🎯 Exam Tip: Write "isosceles triangle" when PQ = QR. Then write "central angle = 2 × inscribed angle". Show each step clearly for full marks.

Question 38. In two concentric circles prove that all chords of the outer circle, which touch the inner circle, are of equal length.

[Diagram: Two concentric circles with center O. Two chords AB and CD of the outer circle are tangent to the inner circle at points M and N respectively.]

Answer:
Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.
To prove the given question, it is sufficient to prove AB = CD.
For this join OM, ON, OB and OD.
Let the radius of outer and inner circles be R and r respectively.
AB touches the inner circle at M.
\( \therefore \) AB is a tangent to the inner circle
\( \therefore \) OM ⊥ AB
\( \Rightarrow \) BM = \( \frac{1}{2} \)AB
\( \Rightarrow \)AB = 2BM
Similarly ON ⊥ CD, and CD = 2DN
Using Pythagoras theorem in triangle OMB and triangle OND
OB² = OM² + BM², OD² = ON² + DN²
\( \Rightarrow \) BM = \( \sqrt{R² - r²} \), DN = \( \sqrt{R² - r²} \)
Now,
AB = 2BM = 2\( \sqrt{R² - r²} \), CD = 2DN = 2\( \sqrt{R² - r²} \)
\( \therefore \) AB = CD
Hence Proved.

In simple words: In concentric circles, all chords of the big circle that touch the small circle are the same distance from the center. So they all have the same length.

📝 Teacher's Note: Draw two circles sharing the same center. Show students how the distance from center to any tangent chord is always the same (equals the small circle's radius).

🎯 Exam Tip: Write "perpendicular from center to chord bisects the chord". Then use Pythagoras theorem. Show that all such chords have the same length calculation.

Question 39. In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

[Diagram: Two circles with centers P and Q, and radii 6 cm and 3 cm respectively. AC is a transverse common tangent with AB = 8 cm.]

Answer:
Since AC is tangent to the circle with center P at point A.
\( \therefore \) ∠PAB = 90°
Similarly, ∠QCB = 90°
In triangle PAB and triangle QCB,
∠PAB = ∠QCB = 90°
∠PBA = ∠QBC (vertically opposite angles)
\( \therefore \) triangle PAB ~ triangle QCB
\( \Rightarrow \) \( \frac{PA}{QC} \) = \( \frac{PB}{QB} \) ............(i)
Also in Rt. triangle PAB,
PB = \( \sqrt{PA² + PB²} \)
\( \Rightarrow \) PB = \( \sqrt{6² + 8²} \) = \( \sqrt{36 + 64} \) = \( \sqrt{100} \) = 10 cm.......(ii)
From (i) and (ii),
\( \frac{6}{3} \) = \( \frac{10}{QB} \)
\( \Rightarrow \) QB = \( \frac{3 × 10}{6} \) = 5 cm
Now,
PQ = PB + QB = (10 + 5) cm = 15 cm

In simple words: We use similar triangles and Pythagoras theorem. The tangent makes a right angle with the radius. Then we add the distances to get the total distance between centers.

📝 Teacher's Note: Show students that tangent is always perpendicular to radius. Draw the right triangles clearly. Use Pythagoras theorem step by step. This is easier than complex formulas.

🎯 Exam Tip: Write "tangent ⊥ radius" first. Then write "by Pythagoras theorem" before calculations. Show similar triangles clearly. Don't forget to add PB + QB for final answer.

Question 40. In the figure given below, O is the centre of the circum circle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

[Diagram: Triangle XYZ inscribed in a circle with center O. Tangents at X and Y meet at point T outside the circle.]

Answer:
Given: ∠XTY = 80°, ∠XOZ = 140°
We know that ∠XTY + ∠XOY = 180° (property of tangents from external point)
\( \Rightarrow \) 80° + ∠XOY = 180°
\( \Rightarrow \) ∠XOY = 100°
Now, ∠XOZ = 140° (given)
\( \therefore \) ∠YOZ = ∠XOZ – ∠XOY = 140° – 100° = 40°
The inscribed angle is half the central angle:
∠ZXY = \( \frac{1}{2} \) × ∠ZOY = \( \frac{1}{2} \) × 40° = 20°

In simple words: When two tangents meet outside a circle, they make a special relationship with the center. The inscribed angle is always half the central angle that subtends the same arc.

📝 Teacher's Note: Show students that tangent angle plus central angle equals 180°. Then use the fact that inscribed angle is half the central angle. Draw clear diagrams with all angles marked.

🎯 Exam Tip: Write the tangent property first: "angle between tangents + angle at center = 180°". Then write "inscribed angle = ½ × central angle". Show each step clearly.

Question 41. In the given figure, AE and BC intersect each other at point D. If ∠CDE=90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.
Answer:
Given:
∠CDE = 90°
AB = 5 cm
BD = 4 cm
CD = 9 cm

Step 1: Find AD using Pythagoras theorem in right triangle ADB.
From right triangle ADB,
\[ AD = \sqrt{AB^2 - BD^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \text{ cm} \]

Step 2: Use the property of intersecting chords.
Since two chords AE and BC intersect at D,
\[ AD \times DE = CD \times DB \]
\[ 3 \times DE = 9 \times 4 \]
\[ DE = \frac{9 \times 4}{3} = 12 \text{ cm} \]

Step 3: Find AE.
\[ AE = AD + DE = 3 + 12 = 15 \text{ cm} \]

Therefore, AE = 15 cm
In simple words: We used Pythagoras theorem to find AD first. Then we used the rule that when two chords cross inside a circle, the products of their parts are equal.

[Diagram: This diagram shows a circle with two chords AE and BC intersecting at point D inside the circle, forming a right angle at D.]

📝 Teacher's Note: Draw the figure clearly. Show students that when two chords cross inside a circle, we multiply the segments on each chord. The products are always equal.

🎯 Exam Tip: Always write "Given" first and list all values. Use Pythagoras theorem for right triangles. Write the intersecting chords formula clearly: AD × DE = CD × DB.

Question 42. In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
Answer:
Given:
∠ABC = 100°
∠ACD = 40°
CT is a tangent to the circle at C

Step 1: Find ∠ADC using the property of cyclic quadrilateral.
In a cyclic quadrilateral ABCD,
∠ABC + ∠ADC = 180° (opposite angles of a cyclic quadrilateral are supplementary)
⇒ 100° + ∠ADC = 180°
⇒ ∠ADC = 80°

Step 2: Find ∠CAD using angle sum property of triangle ACD.
In triangle ACD,
∠ACD + ∠CAD + ∠ADC = 180°
⇒ 40° + ∠CAD + 80° = 180°
⇒ ∠CAD = 180° - 120°
⇒ ∠CAD = 60°

Step 3: Find ∠DCT using alternate segment theorem.
∠DCT = ∠CAD (angles in the alternate segment are equal)
∴ ∠DCT = 60°

Therefore, ∠ADC = 80° and ∠DCT = 60°
In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. When a tangent touches a circle, the angle between the tangent and chord equals the angle in the alternate segment.

[Diagram: This diagram shows a cyclic quadrilateral ABCD inscribed in a circle with centre O, and a tangent CT at point C.]

📝 Teacher's Note: Teach students that opposite angles in cyclic quadrilaterals are supplementary. The alternate segment theorem is very important for tangent problems.

🎯 Exam Tip: Always state "opposite angles of cyclic quadrilateral are supplementary" and "alternate segment theorem" clearly. These are key phrases examiners look for.

Question 43. In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.
Answer:
Given:
O is the centre of the circle
SP is a tangent
∠SRT = 65°

Step 1: Find x using the property of tangent and radius.
TS ⊥ SP (tangent is perpendicular to radius at point of contact)
⇒ ∠TSR = 90°
In triangle TSR,
∠TSR + ∠TRS + ∠RTS = 180°
⇒ 90° + 65° + x = 180°
⇒ x = 180° - 90° - 65°
⇒ x = 25°

Step 2: Find y using the property of central angle and inscribed angle.
y = 2x (angle subtended at the centre is double that of the angle subtended by the arc at the same centre)
⇒ y = 2 × 25°
⇒ y = 50°

Step 3: Find z using angle sum property of triangle OSP.
In triangle OSP,
∠OSP + ∠SPO + ∠POS = 180°
⇒ 90° + z + 50° = 180°
⇒ z = 180° - 140°
⇒ z = 40°

Therefore, x = 25°, y = 50° and z = 40°
In simple words: A tangent always makes a 90° angle with the radius. The central angle is twice the inscribed angle. These rules help us find all the angles step by step.

[Diagram: This diagram shows a circle with centre O, a tangent SP, and various angles marked as x, y, z with ∠SRT = 65°.]

📝 Teacher's Note: Make students remember that tangent is always perpendicular to radius. Also teach that central angle = 2 × inscribed angle for the same arc.

🎯 Exam Tip: Write "tangent ⊥ radius" clearly. Always state "central angle = 2 × inscribed angle" when using this property. Show all steps for finding each angle.