Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 17 Circles

Exercise 17.1

Answer 1.

(i)
Answer: AC = CB (Perpendicular from centre to a chord bisects the chord)

In right triangle ACO:
By Pythagoras theorem, \( OA^2 = OC^2 + AC^2 \)
\( 13^2 - 12^2 = AC^2 \)
\( AC^2 = 169 - 144 = 25 \)
AC = 5cm

Therefore, length of chord AB = 2AC = 2(5) = 10cm
In simple words: When a line from the centre of a circle meets a chord at right angles, it cuts the chord into two equal parts. We use Pythagoras theorem to find the unknown side.

[Diagram: Circle with center O, chord AB, perpendicular OC from center to chord, showing right triangle with measurements]

📝 Teacher's Note: Always draw the perpendicular from center to chord. This makes two right triangles. Students often forget that the perpendicular bisects the chord.

🎯 Exam Tip: Write "perpendicular from centre to chord bisects the chord" first. Then use Pythagoras theorem. Show all calculation steps clearly.

(ii)
Answer: AC = CB (Perpendicular from centre to a chord bisects the chord)

In right triangle ACO:
By Pythagoras theorem, \( OA^2 = OC^2 + AC^2 \)
\( AC^2 = (1.7)^2 - (1.5)^2 \)
\( AC^2 = 2.89 - 2.25 = 0.64 \)
AC = 0.8cm

Therefore, length of chord AB = 2AC = 2(0.8) = 1.6cm
In simple words: Same method as part (i). The perpendicular cuts the chord in half. We find half the chord length first, then double it.

📝 Teacher's Note: Make sure students are careful with decimal calculations. Show them how to subtract decimals step by step.

🎯 Exam Tip: Always write the final answer clearly. Double-check decimal calculations. Write units (cm) in your answer.

(iii)
Answer: BA = AC (Perpendicular from centre to a chord bisects the chord)

In right triangle OAB:
By Pythagoras theorem, \( OB^2 = OA^2 + AB^2 \)
\( AB^2 = 6.5^2 - 2.5^2 \)
\( AB^2 = 42.25 - 6.25 = 36 \)
AB = 6cm

Therefore, length of chord BC = 2AB = 2(6) = 12cm
In simple words: Here the perpendicular is BA instead of from center to middle of chord. We still use the same Pythagoras method to find the distance.

📝 Teacher's Note: Point out that the perpendicular can have different labels. The key is to identify the right triangle and use Pythagoras correctly.

🎯 Exam Tip: Identify which triangle is the right triangle first. Label the sides clearly before applying Pythagoras theorem.

Answer 2.
Answer: AD = DB = 1.6cm (Perpendicular from centre to a chord bisects the chord)

In right triangle ODA:
By Pythagoras theorem, \( OA^2 = OD^2 + AD^2 \)
\( OA^2 = 1.6^2 + 1.2^2 \)
\( OA^2 = 2.56 + 1.44 = 4 \)
OA = 2cm

Diameter(AP) = 2(OA) = 2(2) = 4cm
In simple words: We find the radius first using Pythagoras theorem. Then diameter is twice the radius. The radius goes from center to edge of circle.

[Diagram: Circle with center O, chord AB, perpendicular OD from center to chord, with diameter AP marked]

📝 Teacher's Note: Explain that diameter is always twice the radius. Show students how radius is the distance from center to any point on the circle.

🎯 Exam Tip: Write "Diameter = 2 × radius" clearly. Make sure to find radius first before calculating diameter.

Answer 3.
Answer: AF = FB = 8.4cm
And DE = EC (Perpendicular from centre to a chord bisects the chord)

In right triangle ODA:
By Pythagoras theorem, \( OA^2 = OF^2 + AF^2 \)
\( OA^2 = (11.2)^2 + (8.4)^2 \)
\( OA^2 = 125.44 + 70.56 = 196 \)
OA = 14cm

OA = OD = 14cm (radii of same circle)

Similarly, In triangle DEO
\( OD^2 = OE^2 + DE^2 \)
\( DE^2 = 14^2 - 8.4^2 \)
\( DE^2 = 196 - 70.56 = 125.44 \)
DE = 11.2cm

Therefore, length of chord DC = 2DE = 2(11.2) = 22.4cm
In simple words: Both chords belong to the same circle, so both use the same radius. We find the radius from one chord, then use it for the other chord.

[Diagram: Circle with center O, two chords AB and DC with their perpendiculars from center, showing the measurements]

📝 Teacher's Note: Emphasize that all radii of the same circle are equal. This is the key property students must remember for such problems.

🎯 Exam Tip: Always state "radii of same circle are equal" when using the radius from one calculation for another. This shows you understand the concept.

Answer 4.
Answer: AF = FB = 3cm
CE = ED = 7.2cm (Perpendicular from centre to a chord bisects the chord)

In right triangle AFO, By Pythagoras theorem:
\( OA^2 = OF^2 + AF^2 \)
\( OA^2 = (7.2)^2 + (3)^2 \)
\( OA^2 = 51.84 + 9 = 60.84 \)
OA = 7.8cm

OA = OC = 7.8cm (radii of same circle)

Similarly, In right triangle OFC:
\( OC^2 = OE^2 + EC^2 \)
\( OE^2 = (7.8)^2 - (7.2)^2 \)
\( OE^2 = 60.84 - 51.84 = 9 \)
OE = 3cm

Distance from centre of chord CD with length 14.4cm is 3cm.
In simple words: We find the radius using the first chord. Then we use that radius with the second chord to find how far the center is from that chord.

[Diagram: Circle with center O, two chords AB and CD with perpendiculars from center, showing all measurements]

📝 Teacher's Note: Make sure students understand what "distance from center to chord" means. It is always the perpendicular distance.

🎯 Exam Tip: The question asks for distance from center to chord, not the chord length. Read the question carefully and give exactly what is asked.

Answer 5.
Answer: AC = CB = 4cm (Perpendicular from centre to a chord bisects the chord)

In right △ ABO,
By Pythagoras theorem, \( OA^2 = OC^2 + AC^2 \)
\( OC^2 = 6^2 + 4^2 \)
OC = 36 - 16 = 20
\( OC^2 = 2\sqrt{5}cm \)

Perpendicular distance of chord from centre is \( 2\sqrt{5}cm \)

In simple words: We found how far the chord is from the center. We used Pythagoras theorem to find this distance.

[Diagram: Circle with center O, chord AB, and perpendicular OC from center to chord, showing right triangle OAC]

📝 Teacher's Note: Draw a circle and show how the perpendicular from center always cuts the chord into two equal parts. This is an important property students must remember.

🎯 Exam Tip: Always write "perpendicular from center bisects the chord" first. Then apply Pythagoras theorem. Show all steps clearly to get full marks.

Answer 6.
Answer: OA = OQ = 5cm (Radius of bigger circle)
PQ = 3cm (Radius of smaller circle)
OP = 2cm
Perpendicular bisector of OP, i.e. AB meets OP at M.

OM = MP = \( \frac{1}{2} \)OP = 1cm

In right △ OMA,
By Pythagoras theorem,
\( OA^2 = OM^2 + MA^2 \)
\( MA^2 = 5^2 - 1^2 \)
= 25 - 1
= 24

AM = \( 2\sqrt{6}cm \)

AM = MB = \( 2\sqrt{6} cm \)

AB = AM + MB = \( 2\sqrt{6} + 2\sqrt{6} = 4\sqrt{6} \)

In simple words: We found the length of chord AB. The chord cuts the line OP into two equal parts. We used Pythagoras theorem to find the answer.

[Diagram: Two concentric circles with centers O, showing chord AB that is perpendicular bisector of radius OP]

📝 Teacher's Note: Show students that a perpendicular bisector cuts a line into two equal parts. Use a ruler and compass to demonstrate this concept clearly.

🎯 Exam Tip: Write "perpendicular bisector cuts OP into two equal parts" first. Then find AM using Pythagoras theorem. Remember AB = 2 × AM.

Answer 7.
Answer: AP = PB = 3cm
CQ = QD = 6cm (Perpendicular from centre to a chord bisects the chord)
OA = OC = r (say)
Let OP = x, ∴ OQ = 3 - x

In right △ OQC,
By Pythagoras theorem,
\( OC^2 = OQ^2 + CQ^2 \)
\( r^2 = (3-x)^2 + 6^2 \) ----(1)

Similarly, In △ OPA,
\( OA^2 = AP^2 + PO^2 \)
\( r^2 = 3^2 + x^2 \) ----(2)

From (1) and (2)
\( (3-x)^2 + 6^2 = 3^2 + x^2 \)
-6x + 36 = 0
x = 6
from (2)
\( r^2 = 3^2 + 6^2 = 9 + 36 = 45 \)
r = \( 3\sqrt{5} \)

Thus, radius of the circle is \( 3\sqrt{5}cm \)

In simple words: Two chords cross each other. We used the fact that perpendicular from center cuts chords into equal parts. Then we solved equations to find the radius.

[Diagram: Circle showing two intersecting chords AB and CD with perpendiculars from center O to each chord]

📝 Teacher's Note: Draw two chords crossing inside a circle. Show how perpendiculars from center create right triangles. This helps students see the Pythagoras theorem application.

🎯 Exam Tip: Set up two equations using Pythagoras theorem. Make sure both equations have r² on one side. Solve by equating the expressions.

Answer 8.
Answer: CP = PO = 12cm
Let OA = OC = r (say)
Also, let OQ = x, ∴ OP = 17 - x

In right △ OPC,
By Pythagoras theorem,
\( OC^2 = OP^2 + PC^2 \)
\( r^2 = (17-x)^2 + 12^2 \) ----(1)

Similarly, In △ OQA,
\( OA^2 = AQ^2 + QO^2 \)
\( r^2 = 5^2 + x^2 \) ----(2)

From (1) and (2)
\( (17-x)^2 + 12^2 = 5^2 + x^2 \)
289 - 34x + 144 - 25 = 0
34x = 408
x = 12

From (2)
\( r^2 = 5^2 + 12^2 \)
25 + 144 = 169
r = 13

The radius of the circle is 13cm.

In simple words: We have two chords at right angles. We found the radius using Pythagoras theorem in the right triangles formed.

[Diagram: Circle with two perpendicular chords AB and CD intersecting, with perpendiculars drawn from center O]

📝 Teacher's Note: Emphasize that when two chords are perpendicular, they create right triangles. Students should identify these right triangles first before applying Pythagoras theorem.

🎯 Exam Tip: Identify the right triangles clearly. Set up two equations with r². Solve step by step and check your answer by substituting back.

Answer 9.
Answer: Given: AB = 18cm, MQ = 3cm
To find: PQ

OQ = OA = r cm(say)
∴ OM = OQ = MQ = (r - 3)cm
AM = MB = 9cm (PQ ⊥ AB)

In right △OMA,
\( OM^2 + MA^2 = OA^2 \)
⇒ \( (r-3)^2 + 9^2 = r^2 \)
⇒ \( r^2 - 6r + 9 + 81 = r^2 \)
⇒ 6r = 90
⇒ r = 15cm

PQ = 2r
(Perpendicular bisector of a chord passes through the centre of the circle)
PQ = 2(15)
PQ = 30cm

In simple words: PQ is a diameter because it passes through the center. We found the radius first, then doubled it to get the diameter.

[Diagram: Circle with diameter PQ passing through center O, intersecting chord AB at point M]

📝 Teacher's Note: Show that when a line from one side of circle to other passes through center, it is called diameter. Diameter = 2 × radius always.

🎯 Exam Tip: Write "PQ is diameter because it passes through center" clearly. Find radius first using Pythagoras theorem, then multiply by 2 for diameter.

Answer 10.
Answer:
Draw perpendiculars OR and OS to CD and AB respectively.

In triangle ORP and triangle OSP
OP = OP
OR = OS (Distance of equal chords from centre are equal)
\( \angle PRQ = \angle PSO \) (right angles)

Therefore, \( \triangle ORP \cong \triangle OSP \)
Hence, \( \angle RPO = \angle SPO \)
Thus OP bisects \( \angle CPB \).
In simple words: When two chords are equal, their distances from the center are also equal. This means the line from center to midpoint divides the angle equally.

[Diagram: This diagram shows a circle with center O, two chords AB and CD, and perpendiculars drawn from center O to both chords.]

📝 Teacher's Note: Draw equal chords on the board. Show students that distances from center are always equal for equal chords. This makes the proof easy to understand.

🎯 Exam Tip: Always write "equal chords are equidistant from centre" first. Then use congruent triangles (RHS). These are the key steps examiners want to see.

Answer 11.
Answer:
Given: PQ = RS
To Prove: TP = TR and TQ = TS
Construction: Draw ON ⊥ PQ and OM ⊥ RS

Proof: Since equal chords are equidistant from the circle therefore
PQ = RS \( \implies \) ON = OM ... (1)

Also perpendicular drawn from the centre bisects the chord.
So, PN = NQ = \( \frac{1}{2} \)PQ and RM = MS = \( \frac{1}{2} \)RS

But PQ = RS, we get
PN = RM ... (2)
And, NQ = MS ... (3)

Now in \( \triangle TMO \) and \( \triangle TNO \),
TO = TO (Common)
MO = NO (By (1))
\( \angle TMO = \angle TNO \) (Each 90 degrees)

Therefore, \( \triangle TMO \cong \triangle TNO \) (By RHS)
\( \implies \) TN = TM (By CPCT) ... (4)

Subtracting, (2) from (4), we get
TN - PN = TM - RM
\( \implies \)TP = TR

Adding (3) and (4), we get
TN + NQ = TM + MS
\( \implies \)TQ = TS

Hence Proved.
In simple words: When chords are equal, their distances from center are equal. This makes the triangles equal too. So the outside parts of chords from any point are also equal.

[Diagram: This diagram shows a circle with two equal chords PQ and RS, an external point T, and perpendiculars from center O to both chords.]

📝 Teacher's Note: Use colored chalk to show equal parts clearly. Students often get confused with all the equal segments. Make them practice drawing this step by step.

🎯 Exam Tip: Write all given equal parts clearly at the start. Use RHS congruence for triangles. Show each subtraction step clearly to get full marks.

Answer 12.
Answer:
Let QT be the diameter of \( \angle PQR \)

Since, PQ = QR
\( \therefore \) OM = ON

In \( \triangle OMQ \) and \( \triangle ONQ \)
OM = ON (equal chords are equidistant from the centre)
\( \angle OMQ = \angle ONQ \) (90° each)
OQ = OQ (common)

\( \triangle OMQ \cong \triangle ONQ \) (RHS)
\( \therefore \angle OQM = \angle OQN \) (CPCT)

Thus QT i.e. diameter of the circle bisects \( \angle PQR \)
In simple words: When two chords from one point are equal, the diameter through that point cuts the angle into two equal parts.

[Diagram: This diagram shows a circle with center O, two equal chords PQ and QR meeting at point Q, and diameter QT bisecting the angle PQR.]

📝 Teacher's Note: Draw this on board with compass. Show students that equal chords from same point always make equal angles with diameter. Use real compass to demonstrate.

🎯 Exam Tip: Write that equal chords are equidistant from center first. Then use RHS congruence. Finally conclude that angles are equal by CPCT.

Answer 13.
Answer:
AM = MB
CN = ND
\( \therefore \) OM ⊥ AB
and ON ⊥ CD (A line bisecting the chord and passing through the centre of the circle is perpendicular to the chord)
\( \therefore \angle OMA = \angle OND = 90° \) each

But these are alternate interior angles
\( \therefore \) AB || CD
In simple words: When two chords are cut into equal parts by lines from center, those chords are parallel to each other.

[Diagram: This diagram shows a circle with center O, two chords AB and CD, and perpendiculars from center O to both chords showing the chords are parallel.]

📝 Teacher's Note: Draw railway tracks to show parallel lines. Explain that when angles are equal like this, lines must be parallel. Students understand parallel lines from railway tracks.

🎯 Exam Tip: Write that perpendicular from center bisects chord. Then write about alternate interior angles being equal. This proves parallel lines.

Answer 14.
Answer:
Given: AB and CD are two chords of a circle with centre O.
AB || CD, M and N are midpoints of AB and CD respectively.
To prove: MN passes through centre O.
Construction: Join OM, ON, and through O, draw a straight line EF parallel to AB.

Proof: OM ⊥ AB (line joining the midpoint of a chord to the centre of a circle is perpendicular to it)
\( \angle AMO = 90° \)
\( \angle MOE = 90° \) [cointerior angle of \( \angle AMO \)]
\( \angle NOE = 90° \) [corresponding angle of \( \angle AMO \)]
\( \angle MOE + \angle NOE = 180° \)
MON is a straight line.
Hence, MN passes through centre O.
In simple words: When chords are parallel, the line joining their midpoints always goes through the center of circle.

[Diagram: This diagram shows a circle with parallel chords AB and CD, their midpoints M and N, and line MN passing through center O.]

📝 Teacher's Note: Use two parallel rulers to show parallel chords. Mark their midpoints and show the line goes through center. This visual helps students remember the concept.

🎯 Exam Tip: Write that perpendicular from center bisects chord. Use cointerior and corresponding angles with parallel lines. Show angles add to 180° for straight line.

Answer 15.
Answer:
Given: Two congruent circles with centre O and P. M is the mid-point of OP
To prove: Chord AB and CD are equal.
Construction: Draw OQ ⊥ AB and PR ⊥ CD.

Proof: In \( \triangle OQM \) and \( \triangle PRM \)
\( \angle OQM = \angle PRC \) (Each 90°)
OM = MP (As M is the mid-point)
\( \angle OMQ = \angle PMR \) (Vertically opposite angles)

Therefore, \( \triangle OQM \cong \triangle PRM \) (By AAS)
\( \implies \) OQ = PR (By CPCT)

Now the perpendicular distances of two chords in two congruent circles are equal, therefore chords are also equal.
\( \implies \) AB = CD.

Hence Proved.
In simple words: When circles are the same size and their centers are at specific distance, chords at those positions are always equal.

[Diagram: This diagram shows two congruent circles with centers O and P, point M as midpoint of OP, and equal chords AB and CD.]

📝 Teacher's Note: Draw two identical circles on board. Show how midpoint creates equal triangles. Students understand better when they see identical shapes.

🎯 Exam Tip: Write that circles are congruent first. Use AAS for triangle congruence. Then use the rule that equal distances from center mean equal chords.

Answer 16.
Answer:
Given: Two circles with centres O and P, and MN||OP||RQ
To prove: (i) MN = 2OP (ii) MN = RQ
Construction: OX ⊥ MN, PY ⊥ MN, OW ⊥ RZ, PZ ⊥ RQ

Proof: Since each angle of the quadrilateral XYZW is a right angle, XYZW is a rectangle.
Also, XYPO is a rectangle. ...(1)
Now, perpendicular drawn from the centre to the chord bisects the chord.
Therefore, MA = 2XA and AN = 2AY ...(2)
Now, MN = MA + AN = 2XA + 2AY [from (2)]
\( \implies \) MN = 2(XA + AY) = 2XY
\( \implies \) MN = 2OP [As XYPO is a rectangle, XY = OP] ...(3)
This proves part (i).
By similar arguments, we have RQ = 2OP ...(4)
Using (3) and (4), we get
MN = RQ.
This proves part (ii).

In simple words: When two parallel chords are drawn in two circles, they are equal in length. The distance between circle centres equals half the chord length.

[Diagram: This diagram shows two circles with centres O and P, with parallel chords MN and RQ, and perpendicular lines from centres to chords forming a rectangle.]

📝 Teacher's Note: Draw two circles on the board and show parallel chords. Explain that perpendiculars from centre always cut chords in half. This helps students see the rectangle clearly.

🎯 Exam Tip: Always write "perpendicular from centre bisects chord" as your first step. Show the rectangle formation clearly. These are key marks.

Answer 17.
Answer:
ABC is an equilateral triangle,
\( \therefore \) AB = AC
Also AN = MB (radii of same circle)
\( \implies \) NC = MB
In \( \triangle BNC \) and \( \triangle CMB \)
NC = MB (proved above)
\( \angle B = \angle C \) (60° each)
BC = BC (Common)
\( \therefore \triangle BNC \cong \triangle CMB \) (SAS)
\( \therefore \) BN = CM (CPCT)

In simple words: In an equilateral triangle, all sides are equal. When we draw circles from two corners, the triangles formed are equal by SAS rule.

📝 Teacher's Note: Show students that equilateral triangle has all angles 60°. Mark equal parts clearly on the diagram. Students often forget to mention "radii of same circle."

🎯 Exam Tip: Write "radii of same circle" clearly. State SAS congruence rule. Write CPCT for final step. These exact words get you marks.

Answer 18.
Answer:
In \( \triangle DAM \) and \( \triangle BAN \)
AN = AM (radii of same circle)
AD = AB (sides of square ABCD)
\( \angle DAM = \angle BAN \) (Common)
\( \therefore \triangle DAM \cong \triangle BAN \) (SAS)

In simple words: Both triangles have equal sides from the square and equal radii from the circle. The angle between them is same, so triangles are equal.

📝 Teacher's Note: Draw a square with a circle at corner A. Show that the two triangles have same parts. Students mix up which angles are equal - it's the angle at A that's common.

🎯 Exam Tip: Clearly mark "radii of same circle" and "sides of square." Write the common angle correctly. Use SAS rule and state it clearly.

Answer 19.
Answer:
M and N are mid points of equal chords AB and CD respectively.
ON ⊥ CD and OM ⊥ AB
\( \angle ONC = \angle OMA \) (90° each) ...(1) (A line bisecting the chord and passing through the centre of the circle is perpendicular to the chord)
AB = CD
ON = OM (equal chords are equidistant from the centre)
In \( \triangle MON \),
MO = NO
\( \therefore \angle ONM = \angle OMN \) ...(2)
Subtracting (2) from (1)
ONC - \( \angle ONM = \angle OMA - \angle OMN \)
\( \angle CNM = \angle AMN \)

In simple words: When two chords are equal, their distances from centre are equal. This makes the triangles equal and gives us equal angles.

[Diagram: This diagram shows a circle with centre O, two equal chords AB and CD, with M and N as midpoints, and perpendiculars from O to both chords.]

📝 Teacher's Note: Show students that equal chords are always same distance from centre. Draw clear perpendiculars from O to both chords. This is a key circle theorem.

🎯 Exam Tip: Write "equal chords are equidistant from centre" as the main theorem. Show that triangle MON is isosceles. This gets you full marks.

Answer 20.
Answer:
Join OX and OZ
In \( \triangle XOY \) and \( \triangle ZOY \)
OX = YZ (radii of same circle)
XY = YZ (given)
OY = OY (common)
\( \therefore \triangle XOY \cong \triangle ZOY \) (SSS)
\( \therefore \angle OYX = \angle OYZ \) (CPCT)
Hence, OY is the bisector of \( \angle XYZ \) passing through O

In simple words: When two chords from the same point are equal, the line from centre to that point cuts the angle in half.

[Diagram: This diagram shows a circle with centre O, with point Y on the circle and two equal chords XY and YZ, with lines drawn to centre O.]

📝 Teacher's Note: Draw two equal chords from the same point on circle. Show students that joining the point to centre makes an angle bisector. This is very useful in circle problems.

🎯 Exam Tip: State SSS congruence clearly. Write that OY bisects the angle. Use the words "angle bisector" - examiners look for this exact term.

Answer 21.
Answer:
Since ABC is an isosceles triangle, AOD is the perpendicular bisector of BC.
In triangle ADC, by Pythagoras theorem we have
AD² = AC² - DC² = 13² - 5² = 169 - 25 = 144
\( \therefore \) AD = 12 cm \( \implies \) AO + OD = 12 \( \implies \) AO = 12 - x (Assuming OD = x cm)
Again in triangle OBD,
O² = BD² + OD² = 25 + x² (As BD = 5 cm)
\( \implies \) (12 - x)² = 25 + x² (As AO = BO = radius)
\( \therefore \) 144 + x² - 24x = 25 + x²
\( \therefore \) - 24x = 25 - 144 = -119
\( \therefore \) x = 4.96 cm
\( \therefore \) AO = 12 - 4.96 = 7.04 cm

In simple words: We use Pythagoras theorem twice. First to find the height, then to find where the circle centre is on that height.

[Diagram: This diagram shows an isosceles triangle ABC inscribed in a circle with centre O, where AO is perpendicular to BC at point D, with measurements shown.]

📝 Teacher's Note: Show that in isosceles triangle, perpendicular from vertex bisects the base. Use Pythagoras step by step. Students often forget that AO = BO = radius.

🎯 Exam Tip: Always write "by Pythagoras theorem" before using it. Show all steps clearly. Remember that centre lies on the perpendicular bisector of any chord.

Exercise 17.2

Answer 1.
Answer: Since arc AB makes \( \angle AOB \) at the centre and \( \angle APB = 50° \) on the remaining part of the circle.

\( \angle AOB = 2\angle APB \)

\( \angle AOB = 2(50) \)

\( = 100° \)

AO = OB = x (radii of same circle)

In \( \triangle AOB \)

\( \angle AOB + \angle BAO + \angle ABO = 180 \)

180 + x + x = 180

2x = 80

x = 40

\( \therefore \angle OAB = 40° \)

In simple words: The angle at the centre is twice the angle at the edge. We used this rule and triangle properties to find the answer.

📝 Teacher's Note: Draw a circle on the board. Show how the centre angle is always double the edge angle. This is a very important circle theorem.

🎯 Exam Tip: Always write "angle at centre = 2 × angle at circumference" first. Then use triangle angle sum = 180°. Show all steps clearly.

Answer 2.
Answer: \( \angle AOC = 150° \)

Reflex \( \angle AOC = 360° - 150° = 210° \)

\( \angle ABC = \frac{1}{2} \) reflex \( \angle AOC = \frac{1}{2} (210°) \)

\( \angle ABC = 105° \)

In simple words: When the centre angle is bigger than 180°, we use the bigger angle (reflex angle). Then we take half of it to get the edge angle.

📝 Teacher's Note: Explain reflex angles by showing a clock. When hour hand goes from 12 to 5, the bigger angle is 210°. Students understand this easily.

🎯 Exam Tip: For angles bigger than 180°, always find the reflex angle first. Write "reflex angle = 360° - given angle". Then apply the theorem.

Answer 3.
Answer: BOC is the diameter of circle

\( \therefore \angle BOC = 180° \)

Since arc BC makes \( \angle BOA \) at the centre and \( \angle BAC \) on the remaining part of the circle

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

\( \therefore \angle BAC = \frac{1}{2}(180) \)

= 90°

In simple words: When one side of the triangle is the diameter, the angle opposite to it is always 90°. This is a special rule.

📝 Teacher's Note: Draw many triangles inside a circle with one side as diameter. All opposite angles are 90°. This amazes students and they remember it well.

🎯 Exam Tip: Write "angle in semicircle = 90°" when you see a diameter. This is worth easy marks in exams.

Answer 4.
Answer: Since arc BC makes \( \angle BOC \) at the centre and \( \angle BDC \) on the remaining part of the circle

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

\( \angle BDC = \angle BEC = \angle \frac{x}{2} \) (angles in the same segment)

\( \angle ADB = AEP = 180 - \angle \frac{x}{2} \)

Also, \( \angle BPC = \angle DPE = \angle y \) (vertically opposite)

In quadrilateral ADPE,

\( \angle ADP + \angle DEP + \angle PEA + \angle EAD = 360° \)

\( 180 - \angle \frac{x}{2} + \angle y + 180 - \angle \frac{x}{2} + z = 360° \)

-\( \angle x + \angle y + \angle z = 0 \)

\( \angle x = \angle y + \angle z \)

In simple words: We used circle theorems and properties of quadrilaterals. The angles are connected by this special relationship.

📝 Teacher's Note: This is a complex problem. Break it into small steps. Show each angle relationship separately before combining them.

🎯 Exam Tip: Label all angles clearly in the diagram first. Write "angles in same segment are equal" when you use it. Show every step.

Answer 5.
Answer: Let O be the centre of the circle on diameter AC of the circle

Since, EC make \( \angle EOC \) at the centre and \( \angle EBC \) on the remaining part of the circle

\( \therefore \angle EOC = 2\angle EBC \)

= 2(65)

= 130°

In \( \triangle EOC \),

\( \angle EOC + \angle OCE + \angle CEO = 180° \)

130 + x + x = 180° (OE = OC, \( \therefore \angle OEC = \angle OCE = x \))

2x = 50

x = 25

\( \angle OCE = \angle OEC = 25° \)

Also, \( \angle OCE = \angle CED = 25° \) (alternate interior angles)

In simple words: We used the centre angle theorem and isosceles triangle properties. OE and OC are equal radii, so base angles are equal.

📝 Teacher's Note: Remind students that radii are always equal. So triangles with two radii are always isosceles. Base angles are equal in isosceles triangles.

🎯 Exam Tip: When you see two radii in a triangle, immediately write "isosceles triangle" and "base angles are equal". This saves time.

Answer 6.
Answer: \( \angle AOB = q \)

Reflex \( \angle AOB = 360 - q \)

Since arc AB subtends reflex \( \angle AOB = (360 - q)° \) at the centre and \( \angle ACB \) on the remaining part of the circle.

\( \therefore \angle ACB = \frac{1}{2} \) (reflex \( \angle AOB \))

If OACB is a parallelogram

\( \angle AOB = \angle ACB \)

q = p

360 - 2p = p

3p = 360

P = 120°

In simple words: In a parallelogram, opposite angles are equal. We used this property along with circle theorems to find the angle.

📝 Teacher's Note: Draw a parallelogram and mark opposite angles with same color. Students see that opposite angles are always equal in parallelograms.

🎯 Exam Tip: Write "opposite angles of parallelogram are equal" when you use this property. Always check if you need reflex angle or normal angle.

Answer 7.
Answer: In \( \triangle PQR \),

PQ = PR

\( \therefore \angle PQR = \angle PRQ = 35° \)

Also, \( \angle PQR + \angle PRQ + \angle QPR = 180° \)

35 + 35 + \( \angle QPR = 180 \)

\( \angle QPR = 110° \)

In cyclic quadrilateral PQSR,

\( \angle QPR + \angle QSR = 180 \)

110 + \( \angle QSR = 180 \)

\( \angle QSR = 70 \)

Also, \( \angle QSR = \angle QTR = 70° \) (Angles in the same segment)

In simple words: In an isosceles triangle, base angles are equal. In a cyclic quadrilateral, opposite angles add up to 180°.

📝 Teacher's Note: Show that PQ = PR makes triangle isosceles. Then show opposite angles in cyclic quadrilateral always add to 180°. Use colored markers for clarity.

🎯 Exam Tip: Write "isosceles triangle" when two sides are equal. Write "opposite angles of cyclic quadrilateral = 180°" for the next step.

Answer 8.
Answer: In cyclic quadrilateral ABCD,

\( \angle BCD + \angle DAB = 180° \) (Opposite angles of a cyclic quadrilateral)

100 + \( \angle DAB = 180 \)

\( \angle DAB = 80° \)

In \( \triangle DAB \),

\( \angle DAB + \angle ABD + \angle BDA = 180° \)

80 + 70° + \( \angle BDA = 180° \)

\( \angle BDA = 30° \)

In simple words: In any four-sided shape inside a circle, opposite angles add up to 180°. Then we used triangle angle sum.

📝 Teacher's Note: Draw a cyclic quadrilateral and mark opposite angles. Show that they always add to 180°. This is different from normal quadrilaterals where they add to 360°.

🎯 Exam Tip: Always write "opposite angles of cyclic quadrilateral add to 180°" first. Then use triangle properties. Check your addition carefully.

Answer 22.
Answer:

[Diagram: Circle with triangle ABC inscribed, showing points B, N, M, O and various perpendicular lines and angle markings]

Given: AB = AC, \( \angle ABO = \angle CBO \)

To Prove: AB = BC

Construction: Draw ON \( \perp AB \) and OM \( \perp BC \)

Proof: In triangles BNO and BMO,

\( \angle NBO = \angle MBO \) (Given)

\( \angle BNO = \angle BMO \) (Each 90°)

BO = BO (Common)

Thus, \( \triangle BNO \cong \triangle BMO \) (By AAS)

\( \Rightarrow BN = BM \)

\( \Rightarrow 2BN = 2BM \) (Since perpendicular drawn from the centre bisects the chord)

\( \Rightarrow AB = BC \)

Hence Proved.

In simple words: We used the fact that perpendicular from center bisects the chord. When two chords have equal perpendicular distances from center, they are equal.

📝 Teacher's Note: Draw perpendiculars from center to chords on the board. Show that they always bisect the chords. Equal perpendicular distances mean equal chords.

🎯 Exam Tip: Always construct perpendiculars from center to chords when proving chord equality. Write "perpendicular from center bisects chord" clearly. Use AAS congruency.

Answer 9.
Answer: It is given that \( \angle AOC = 100° \)

Arc AC subtends \( \angle AOC \) at the centre of circle and \( \angle APC \) on the circumference of the circle.

\( \therefore \angle AOC = 2\angle APC \)

\( \Rightarrow \angle APC = \frac{100°}{2} = 50° \)

It can be seen that APCB is a cyclic quadrilateral.

\( \therefore \angle APC + \angle ABC = 180° \) (Sum of opposite angles of a cyclic quadrilateral)

\( \Rightarrow \angle ABC = 180° - 50° = 130° \)

Now, \( \angle ABC + \angle CBD = 180° \) (Linear pair angles)

\( \Rightarrow \angle CBD = 180° - 130° = 50° \)

In simple words: The angle at the centre is double the angle at the circumference for the same arc. Then we use the property that opposite angles of a cyclic quadrilateral add up to 180°.

📝 Teacher's Note: Draw a circle and mark the points clearly. Show students that the angle at centre is always double the angle at circumference. This is a key theorem to remember.

🎯 Exam Tip: Always state "angle at centre = 2 × angle at circumference" and "opposite angles of cyclic quadrilateral = 180°". These are the key formulas examiners look for.

Answer 10.
Answer: PQ is a diameter of the circle

\( \therefore \angle PRQ = 90° \) (angle in a semicircle is a right angle)

\( \angle RPQ = 40° \) (given)

In \( \triangle PQR \),

\( \angle PRQ + \angle RQP + \angle QPR = 180° \) (Angle sum property)

\( 90° + \angle RQP + 40° = 180° \)

\( \angle RQP = 50° \)

In simple words: When a line is drawn from any point on a circle to the ends of a diameter, it makes a 90° angle. Then we use the fact that all angles in a triangle add up to 180°.

📝 Teacher's Note: Show students a semicircle and draw lines from different points on the circle to the diameter ends. All will make 90° angles. This helps them remember the theorem.

🎯 Exam Tip: Always write "angle in a semicircle = 90°" clearly. This is the most important property. Then apply angle sum property of triangle.

Answer 11.
Answer: \( \angle B = 65° \) (given)

\( \angle B + \angle D = 180° \) (Opposite angles of a cyclic quadrilateral)

\( 65° + \angle D = 180° \)

\( \angle D = 115° \)

Also, AB || CD

\( \therefore \angle B + \angle C = 180° \) (Sum of angles on same side of transversal)

\( \angle C = 180° - 65° = 115° \)

Again, \( \angle A + \angle C = 180° \) (Opposite angles of a cyclic quadrilateral)

\( \angle A = 180° - 115° = 65° \)

In simple words: In a cyclic quadrilateral, opposite angles always add up to 180°. Also, when two parallel lines are cut by another line, angles on the same side add up to 180°.

📝 Teacher's Note: Use a simple example like a clock face to show how opposite points make 180° when you add the angles. This makes the concept clear for students.

🎯 Exam Tip: State both properties clearly: "opposite angles of cyclic quadrilateral = 180°" and "co-interior angles = 180°". Use both to find all angles.

Answer 12.
Answer:

[Diagram: A square ABCD is inscribed in a circle, with all four vertices touching the circle.]

m \( \angle A = 3 \) (m \( \angle C \))

\( \angle A + \angle C = 180° \) (Opposite angles of a cyclic quadrilateral)

\( 3\angle C + \angle C = 180° \)

\( 4\angle C = 180° \)

\( \angle C = 45° \)

m \( \angle A = 3 \)(m \( \angle C \))

= \( 3 × 45° \)

= \( 135° \)

m\( \angle A = 135° \)

In simple words: We know that opposite angles in a cyclic quadrilateral add up to 180°. We use this fact with the given condition that one angle is 3 times the other.

📝 Teacher's Note: Show students how to set up the equation when one quantity is a multiple of another. This type of problem appears often in geometry.

🎯 Exam Tip: Always write the equation clearly: if angle A = 3 × angle C, and A + C = 180°, then 3C + C = 180°. Show each step of solving.

Answer 13.
Answer:

[Diagram: A circle with points A, B, C, D, E marked on the circumference, with O at the center. Various chords and lines connect these points forming triangles.]

Arc AC subtends \( \angle AOC \) at the centre of circle and \( \angle ABC \) on the circumference of the circle.

\( \therefore \angle AOC = 2\angle ABC \) ... (1)

Similarly, \( \angle BOD \) and \( \angle DCB \) are the angles subtended by the arc DB at the centre and on the circumference of the circle respectively.

\( \therefore \angle BOD = 2\angle DCB \) ... (2)

Adding (1) and (2),

\( \angle AOC + \angle BOD = 2(\angle ABC + \angle DCB) \) ... (3)

In triangle ECB,

\( \angle AEC = \angle ECB + \angle EBC = \angle DCB + \angle ABC \)

From (3),

\( \angle AOC + \angle BOD = 2\angle AEC \)

Hence Proved.

In simple words: The angle at the centre is always double the angle at the circumference for the same arc. We use this property twice and then add the results to get the final proof.

📝 Teacher's Note: Draw the diagram step by step. First show one arc and its angles, then the second arc. Students understand better when they see the pattern twice.

🎯 Exam Tip: Always number your steps (1), (2), (3). Write "angle at centre = 2 × angle at circumference" for each arc. This gives you marks for method.

Answer 14.
Answer:

[Diagram: A circle with two intersecting chords. Points A, B, C, D are on the circle, with chords AB and CD intersecting at point P inside the circle.]

If two chords of a circle intersect internally then the products of the lengths of segments are equal, then

AP × BP = CP × DP ... (1)

But, AP = CP (Given) ... (2)

Then from (1) and (2), we have

BP = DP ... (3)

Adding (2) and (3),

AP + BP = CP + DP

\( \Rightarrow \) AB = CD

Hence Proved.

In simple words: When two chords cross inside a circle, we get equal products of their parts. If two parts are already equal, then the other two parts must also be equal, making the whole chords equal.

📝 Teacher's Note: Use two pencils crossing each other to show intersecting chords. Explain that the "crossing rule" gives us AP × BP = CP × DP. This visual helps students remember.

🎯 Exam Tip: Always state "intersecting chords theorem: AP × BP = CP × DP" first. Then use the given condition. Show each step clearly for full marks.

Answer 15.
Answer: \( \angle NYB = 50° \)

\( \angle YNB = 20° \)

In \( \triangle NYB \),

\( 20° + \angle NBY + 50° = 180° \)

\( \Rightarrow \angle NBY = 180° - 70° = 110° \)

Now, \( \angle MAN = \angle NBM = 110° \) (Angles in the same segment)

\( \angle MON = 2\angle MAN \) (Arc MN subtends \( \angle MON \) at centre and \( \angle MAN \) at remaining part of the circle)

\( \angle MON = 2(110°) = 220° \)

Reflex \( \angle MON = 360° - \angle MON \)

= \( 360° - 220° \)

Reflex \( \angle MON = 140° \)

In simple words: First we find all angles in the triangle using angle sum property. Then we use the fact that angles in the same segment are equal, and angle at centre is double the angle at circumference.

📝 Teacher's Note: Show students that "angles in same segment" means angles made by the same arc from different points on the circle. Draw this clearly on the board.

🎯 Exam Tip: Write "angles in same segment are equal" and "angle at centre = 2 × angle at circumference". Always mention "reflex angle" when the angle is greater than 180°.

Answer 16.
Answer: Given AP and AQ are diameters of circles with centre O and O¹ respectively.

\( \therefore \angle APB = 90° \) ---(1) (Angle in a semicircle is a right angle)

Similarly, \( \angle ABQ = 90° \) ---(2)

Adding (1) and (2)

\( \angle APB + \angle ABQ = 90° + 90° \)

\( \angle PBQ = 180° \)

Hence, PBQ is a straight line

\( \therefore \) P, B and Q are collinear.

In simple words: When you draw lines from any point on a circle to the ends of a diameter, you always get a 90° angle. Since we have two such angles that add up to 180°, the three points must be in a straight line.

📝 Teacher's Note: Draw two circles with diameters and show how the 90° angles add up. Use a ruler to show students that the three points really are in a straight line.

🎯 Exam Tip: Always state "angle in semicircle = 90°" for both circles. Then show that the angles add to 180°, which proves the points are collinear (in a straight line).

Answer 17.
Answer: AB and AC are diameters of circles with centre O and O¹ respectively.

\( \angle ADB = 90° \) ---(1) (Angle in a semi circle is a right angle)

Similarly, \( \angle ADC = 90° \) ---(2)

Adding (1) and (2)

\( \angle ADB + \angle ADC = 90 + 90 \)

\( \angle BDC = 180° \)

Hence, BDC is a straight line.
In simple words: When two circles share one point on their diameters, the angles made at any point on both circles add up to 180°. This makes all three points lie in a straight line.

📝 Teacher's Note: Draw two circles that touch each other. Show students that angles in semicircles are always 90°. When you add two 90° angles, you get 180° which means a straight line.

🎯 Exam Tip: Always write "angle in semicircle is 90°" as your reason. Then add the angles to show they equal 180°. This proves the points are collinear.

Answer 18.
Answer:
[Diagram: This diagram shows a circle with center O, triangle ABC inscribed in the circle, with point D inside forming various angles and chord relationships.]

BD = DC

AB be the diameter of the circle with centre O.

\( \angle ABD = 90° \) (Angles in a semicircle is a right triangle)

\( \angle ADC = 180° \) (linear pair)

\( \angle ADC = 180 - 90 = 90° \)

In \( \triangle ABD \) and \( \triangle ADC \)

\( \angle ABD = \angle ADC \) (90° each)

\( \angle BAD \) (Common)

\( \angle ABD = \angle ACD \) (RHS)

\( \therefore \triangle ABD \cong \triangle ADC \) (AA corollary)

\( \therefore BD = DC \) (CPCT)
In simple words: When AB is a diameter, angle ABD becomes 90°. Using triangle properties and the fact that angles in semicircle are 90°, we can prove that BD equals DC.

📝 Teacher's Note: Show students that diameter always creates 90° angles on the circle. Use this property to prove triangles are equal. Equal triangles have equal corresponding sides.

🎯 Exam Tip: Write "angle in semicircle = 90°" first. Then use triangle congruence rules like AA or RHS. Finally write "BD = DC by CPCT" for full marks.

Answer 19.
Answer:
[Diagram: This diagram shows a circle with center O and a rhombus ABCD where the diagonals of the rhombus intersect at point P, with the circle passing through certain vertices.]

We know that the diagonals of a rhombus bisect each other at right angles.

\( \angle APD = 90° \) - (1)

Also, AD is the diameter of the circle with centre O.

\( \angle APD = 90° \) - (2) (Angle in semi circle)

From (1) and (2), we get, The circle drawn with any side of a rhombus as a diameter, passes through point of intersection of its diagonals.
In simple words: In a rhombus, the diagonals always meet at 90°. Since angles in semicircle are also 90°, the meeting point always lies on the circle made with any side as diameter.

📝 Teacher's Note: Draw a rhombus and show that diagonals meet at 90°. Then draw a circle using any side as diameter. The intersection point will always be on this circle.

🎯 Exam Tip: State both properties: "diagonals of rhombus meet at 90°" and "angle in semicircle is 90°". Since both angles are 90°, the point lies on the circle.

Answer 20.
Answer: In cyclic quadrilateral ABCD,

\( \angle BAD + \angle BCD = 180° \) - (1)

Opposite angles of cyclic quadrilateral

Also, \( \angle BCD + \angle BCE = 180° \) - (2) (Linear pair)

From (1) and (2), we get

\( \angle BAD = \angle BCE \)

In \( \triangle EBC \) and \( \triangle EDA \)

\( \angle BAD = \angle BCE \) (proved above)

\( \angle BEC = \angle DEA \) (common)

\( \therefore \triangle EBC \sim \triangle EDA \) (AA corollary)
In simple words: In a cyclic quadrilateral, opposite angles add up to 180°. Using this property and linear pairs, we can prove that certain triangles formed are similar.

📝 Teacher's Note: Draw a cyclic quadrilateral and mark opposite angles. Show students that opposite angles always add to 180°. Use this to find equal angles in different triangles.

🎯 Exam Tip: Always start with "opposite angles of cyclic quadrilateral sum to 180°". Then use linear pair property. Finally prove triangle similarity using AA rule.

Answer 21.
Answer:
[Diagram: This diagram shows triangle ABC with a circle, where points D and E lie on the circle and various angle relationships are established to prove DE is parallel to BC.]

To prove DE || BC

In cyclic quadrilateral DECB

\( \angle DEC + \angle DBC = 180° \) - (1) (Opposite angles of cyclic quadrilateral)

Also, \( \angle AED + \angle DEC = 180° \) - (2) (Linear pair)

From (1) and (2), we get,

\( \angle DBC = \angle AED \) - (3)

But AC || (given)

\( \angle ABC = \angle ACB \) - (4) (angles opposite to equal sides of triangle)

From (3) and (4) \( \Rightarrow \angle AED = \angle ACB \)

Since these are corresponding angles,

DE || BC
In simple words: We use the property that opposite angles in a cyclic quadrilateral add to 180°. Combined with linear pair property, we can prove that DE and BC are parallel lines.

📝 Teacher's Note: Show students how to identify cyclic quadrilaterals in complex figures. Point out that when corresponding angles are equal, lines are parallel.

🎯 Exam Tip: Write "opposite angles of cyclic quadrilateral = 180°" first. Then use linear pair property. End with "corresponding angles are equal, so lines are parallel".

Answer 22.
Answer:
[Diagram: This diagram shows a cyclic quadrilateral ABCD inscribed in a circle, with various angle relationships and bisector properties illustrated.]

In cyclic quadrilateral ABCD

\( \angle A + \angle C = 180° \)

\( \frac{1}{2}\angle A + \frac{1}{2}\angle C = 90° \)

\( \angle EAB + \angle BCF = 90° \) - (1) (AE bisects \( \angle A \); CF bisects \( \angle C \))

Also,

\( \angle BCF = \angle BAF \) - (2) (Angles in the same segment)

Using (1) in (2) we get,

\( \angle EAB + \angle BAF = 90° \)

\( \angle FAE = 90° \)

EF is the diameter of the circle,

\( \therefore \) angle in a semi circle is a right angle
In simple words: When angle bisectors of opposite angles in a cyclic quadrilateral meet, they form a 90° angle. This means they lie on the diameter of the circle.

📝 Teacher's Note: Explain that angle bisectors divide angles into two equal parts. In cyclic quadrilaterals, this special property creates right angles at intersection points.

🎯 Exam Tip: Write "opposite angles of cyclic quadrilateral = 180°". Then show how half of each angle adds to 90°. Finally conclude with "angle in semicircle = 90°".

Answer 24.
Answer:
[Diagram: This diagram shows a circle with points A, B, C, D, E, F on the circumference, with AD, BE, and CF as angle bisectors of angles A, B, and C respectively.]

Since AD, BE and CF are bisectors of \( \angle A \), \( \angle B \) and \( \angle C \) respectively.

\( \angle 1 = \angle 2 = \frac{A}{2} \)

\( \angle 3 = \angle 4 = \frac{B}{2} \)

\( \angle 5 = \angle 6 = \frac{C}{2} \)

\( \angle ADE = \angle 3 \) ----(1)

Also \( \angle ADF = \angle 6 \) ----(2) (angles in the same segment)

Adding (1) and (2)

\( \angle ADE + \angle ADF = \angle 3 + \angle 6 \)

\( \angle D = \frac{1}{2}\angle B + \frac{1}{2}\angle C \)

\( \angle D = \frac{1}{2}(B + \angle C) = \frac{1}{2}(180 - \angle A) \) (\( \angle A + \angle B + \angle C = 180° \))

\( \angle D = 90 - \frac{1}{2}\angle A \)

Similarly,

\( \angle E = 90 - \frac{1}{2}\angle B \), \( \angle F = 90 - \frac{1}{2}\angle C \)
In simple words: When angle bisectors of a triangle meet the circumcircle, they create new angles. Each of these angles equals 90° minus half of the opposite angle from the original triangle.

📝 Teacher's Note: Draw a triangle with its circumcircle. Show how angle bisectors create equal angles in same segments. Use the property that angles in triangle add to 180°.

🎯 Exam Tip: Write "angles in same segment are equal" for each bisector. Then use "sum of angles in triangle = 180°" to derive the final formula. Show all algebraic steps clearly.

Exercise 17.3

Answer 1.
Answer:

[Diagram: Circle with center O, point P outside the circle, tangent line from P touching the circle at point A, and line OP drawn]

OA ⊥ AP (radius is perpendicular to tangent at the point of contact)
In right triangle OAP,
\( OP^2 = OA^2 + AP^2 \)
\( AP^2 = 5^2 - 3^2 \)
\( AP^2 = 25 - 9 \)
\( AP^2 = 16 \)
AP = 4cm
The length of the tangent is 4cm.
In simple words: We use the right triangle rule. The tangent makes a 90° angle with the radius. Then we use Pythagoras theorem to find the tangent length.

📝 Teacher's Note: Always draw the radius to the point where tangent touches. This makes a right angle. Students often forget to mention this property.

🎯 Exam Tip: Write "radius is perpendicular to tangent at point of contact" first. Then use Pythagoras theorem: \( OP^2 = OA^2 + AP^2 \).

Answer 2.
Answer:

[Diagram: Circle with center O, point P outside the circle, tangent line from P touching the circle at point A, and line OP drawn]

OA ⊥ AP (radius is perpendicular to tangent at the point of contact)
In right triangle OAP,
\( OP^2 = OA^2 + AP^2 \)
\( AP^2 = 17^2 + 15^2 \)
\( AP^2 = 289 - 225 \)
\( AP^2 = 64 \)
AP = 8
The radius of the circle is 8cm.
In simple words: We know the distance from center to outside point and the tangent length. Using the right triangle rule, we find the radius of the circle.

📝 Teacher's Note: Here we find radius when tangent length and distance from center are given. Make sure students understand which value they are finding.

🎯 Exam Tip: First identify what you need to find. Rearrange the formula: \( OA^2 = OP^2 - AP^2 \) when finding radius.

Answer 3.
Answer:
XP = XQ
AR = AP (Length of tangents drawn from an external point to a circle are equal)
BR = BQ equal

XP = XQ
XA + AP = XB + BR
XA + AR = XB + BR (Using (1))
In simple words: Tangents from the same outside point to a circle are always equal in length. This is a basic rule of circles.

📝 Teacher's Note: This is the tangent property. From any outside point, both tangent lines to the circle have the same length. Draw this clearly on the board.

🎯 Exam Tip: Always state: "Tangents from external point to a circle are equal" before using this property in proofs.

Answer 4.
Answer:

[Diagram: Circle with center O, external point P, tangent lines from P touching the circle at points A and B, with points K, M, N marked on the tangent lines]

KA = KM ---(1) (Length of tangents drawn from an external point to a circle are equal)
NM = NB circle are equal
KN = KM + MN
KN = KA + BM (Using (1))
In simple words: We use the tangent property twice. Each outside point has equal tangent lengths to the circle.

📝 Teacher's Note: Show students that K and N are both outside points. Each has its own equal tangents. This builds understanding step by step.

🎯 Exam Tip: Identify all external points first. Then apply the tangent property for each point separately. Number your steps clearly.

Answer 5.
Answer:

[Diagram: Circle with center O, external point P, tangent lines from P touching the circle at points A and B, with points C and D marked on the tangent lines]

PA = PB = 20 Units ---(1) (Length of tangents drawn from an external point to a circle are equal)
CQ = CA and DQ = DB
Perimeter of triangle PCD
= PC + CD + PD
= PC + CQ + QD + PD (Using (1))
= PA + PB
= 2PA
= 2(20)
= 40 Units
In simple words: The perimeter of the triangle equals twice the length of one tangent from the external point.

📝 Teacher's Note: This is a beautiful property. The perimeter always equals 2 times the tangent length. Students find this surprising and remember it well.

🎯 Exam Tip: Write the formula: Perimeter = 2 × tangent length from external point. This saves time in calculations.

Answer 6.
Answer:
To prove: AF + BD + CE = AE + BF + CD
Proof: AF = AE ---(1) (Length of tangents drawn from an external point to a circle are equal)
BD = BF ---(2)
CE = CD ---(3)
Adding (1), (2) and (3)
AF + BD + CE = AE + BF + CD
In simple words: Each vertex of the triangle is an outside point. The two tangents from each vertex are equal. Adding all proves the result.

📝 Teacher's Note: This applies to any triangle with an inscribed circle. Each vertex gives equal tangent segments. Very useful in geometry problems.

🎯 Exam Tip: Label the tangent points clearly. Apply the tangent property three times - once for each vertex of the triangle.

Answer 7.
Answer:
To prove: AQ = \( \frac{1}{2} \) (Perimeter of triangle ABC)

Proof: BQ = BR = 5 - r ---(1) (Lengths of tangents drawn from an external point to a circle are equal)
PC = CR = 12 - r ---(2)

Perimeter of triangle ABC = AB + BC + AC
= AB + BP + PC + AC
= AB + BQ + CR + AC Using (1)
= AQ + AR
= 2 AQ

2 AQ = Perimeter of triangle ABC

AQ = \( \frac{1}{2} \) (Perimeter of triangle ABC)
In simple words: One side of the triangle equals half the total perimeter. This happens when a circle touches all three sides of the triangle.

📝 Teacher's Note: This property is very useful. In any triangle with incircle, any tangent segment equals half the perimeter minus the opposite side length.

🎯 Exam Tip: Group the tangent segments properly. Show that the perimeter equals 2 times one tangent segment clearly.

Answer 8.
Answer:

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

Let the sides of parallelogram ABCD touch the circle at points P, Q, R and S.
AP = AS - (1)
PB = BQ - (2) (Length of tangents drawn from an external point to a circle are equal)
DR = DS - (3)
RC = CQ - (4)

Adding (1), (2), (3) and (4)
AP + PB + DR + RC = AS + BQ + DS + CQ
AB + CD = AD + BC
2 AB = 2 BC ⇒ AB = BC (Opposite sides of a parallelogram are equal)
∴ AB = BC = CD = DA,
Hence, ABCD is a rhombus.
In simple words: When a circle touches all sides of a parallelogram, opposite sides become equal. This makes it a rhombus (all sides equal).

📝 Teacher's Note: A parallelogram with an inscribed circle is always a rhombus. Students should remember this special case.

🎯 Exam Tip: Apply tangent property to all four vertices. Show that adjacent sides become equal, making it a rhombus.

Answer 9.
Answer:

[Diagram: Triangle PQR with an inscribed circle touching sides at points A, B, and T]

To prove: QT = TR
Proof: Let the circle touches sides PQ and PR at points A and B respectively.
PA = PB (Lengths of tangents drawn from an external point to a circle are equal)
AQ = QT
BR = TR

Given, PQ = PR
PA + AQ = PB + BR
AQ = BR (Using (1))
⇒ QT = TR
In simple words: In an equal-sided triangle, the circle touches the base at its middle point. So both parts of the base are equal.

📝 Teacher's Note: This shows the symmetry in isosceles triangles. The incircle divides the base into two equal parts.

🎯 Exam Tip: Use the given condition (PQ = PR) along with the tangent property. Show that the tangent segments on the base are equal.

Answer 10.
Answer:
In triangles AOP = triangle BOP
AP = PB (lengths of tangents drawn from and external point to a circle are equal)
OP = PO (common)
∠ PAO = ∠ PBO = 90° (radius is ⊥ to tangent at the point of contact)
∴ triangle AOP ≅ triangle BOP (By RHS)
triangle AOP = triangle BOP (By CPCT)

In triangles AMO and triangle BMO
AO = OB (radius of same circle)
∠ MOA = ∠ MOB (Proved above)
OM = MO (Common)
∴ triangle AMO ≅ triangle BMO (By CPCT)
∠ AMO = ∠ BMO
∠ AMO + ∠ BMO = 180°
∴ 2 ∠ AMO = 180°
∠ BMO = ∠ AMO = 90°

Hence, OP is the perpendicular bisector of AB.
In simple words: The line from the center through the external point cuts the line joining the tangent points at 90°. It also cuts it into two equal parts.

📝 Teacher's Note: This is an important theorem. The line from center to external point is the perpendicular bisector of the chord joining tangent points.

🎯 Exam Tip: Use congruent triangles to prove this. First prove triangles AOP and BOP are congruent, then use CPCT to get equal angles.

Answer 11.
Answer:
[Diagram: Circle with center point P and four points A, B, C, D on the circumference forming intersecting chords]

Let DP = x cm

In \( \triangle APC \) and \( \triangle DPB \)

\( \angle PAC = \angle PDB \) (angles in the same segment)
\( \angle APC = \angle DPB \) (vertically opposite angle)

\( \implies \triangle APC \sim \triangle DPB \) (AA corollary)

\( \frac{AP}{DP} = \frac{PC}{PB} \) (similar sides of similar triangles)

\( \frac{5}{x} = \frac{2.5}{3} \)


\( \implies x = \frac{15}{2.5} = \frac{150}{25} = 6 \text{ cm} \)

📝 Teacher's Note: When two chords cross inside a circle, they make similar triangles. This is because angles in the same part of the circle are equal. Students should mark the angles first.

🎯 Exam Tip: Write "AA corollary" to show triangles are similar. Then write the ratio of sides clearly. Cross multiply to find the answer. Show all steps.

Answer 12.
Answer:
Let TQ = x cm

In \( \triangle PTR \) and \( \triangle STQ \)

\( \angle TPR = \angle TSQ \) (angles in the same segment)
\( \angle PTR = \angle STQ \) (vertically opposite angles)

\( \implies \angle PTR = \angle STQ \) (AA corollary)

\( \frac{PT}{ST} = \frac{TR}{TQ} \) (similar sides of similar triangles)

\( \frac{18}{6} = \frac{12}{x} \)

= x = 4

= TQ = 4 cm

📝 Teacher's Note: This is the same rule as Answer 11. When chords cross, they make similar triangles. Teach students to identify which sides match which sides in the ratio.

🎯 Exam Tip: Always write the ratio with matching sides. PT matches ST, and TR matches TQ. Don't mix them up or you will get wrong answer.

Answer 13.
Answer:
[Diagram: Circle with center O, points A, B, C, D on circumference, and external point P with lines drawn to the circle]

Let OD = OC = r (say)

PO = 14.5, CP = r + 14.5

PD = 14.5 - r

In \( \triangle BPD \) and \( \triangle APC \)

\( \angle BPD = \angle APC \) (Common)
\( \angle ABD + \angle DBP = 180° \) ---(1) (Linear pair)

Also, \( \angle ABD + \angle ACD = 180° \) ---(2) (Opposite angles of a cyclic quadrilateral)

From (1) and (2)
\( \angle DBP = \angle ACD \)

\( \implies \triangle BPD \sim \triangle CPA \) (AA corollary)

\( \frac{8}{r + 14.5} = \frac{4.5 - r}{15} \)

120° = \( 14.5^2 - r^2 \)

r² = 210.25 - 120
r² = 90.25
r = 9.50

Radius of the circle is 9.5cm.

📝 Teacher's Note: This problem uses the property that a point outside a circle makes similar triangles with the secants. Students must be careful with the distances - some are added, some subtracted.

🎯 Exam Tip: Draw the diagram clearly. Mark all the given lengths. Write "cyclic quadrilateral" when you use the 180° property. Show each step of calculation.

Answer 14.
Answer:
Let PT = x cm

Since, PAB is a secant and PT is a tangent to the given circle, we have,

PA · PB = \( PT^2 \)

\( \implies 4 \cdot 9 = PT^2 \)

\( \implies PT^2 = 36 \)

\( \implies PT = 6 \text{ cm} \)

📝 Teacher's Note: When a tangent and secant are drawn from the same outside point, the tangent squared equals the secant parts multiplied. This is a standard circle theorem.

🎯 Exam Tip: Write the formula first: "tangent² = secant part 1 × secant part 2". Then substitute values. Always take positive square root for length.

Answer 15.
Answer:
Let PT = x cm

Since, PAB is a secant and PT is a tangent to the given circle, we have,

PA · PB = \( PT^2 \)

\( \implies 4 \cdot 9 = PT^2 \)

\( \implies PT^2 = 36 \)

\( \implies PT = 6 \text{ cm} \)

📝 Teacher's Note: This is exactly the same type as Answer 14. The tangent-secant theorem is very important. Practice more problems like this.

🎯 Exam Tip: The formula is always the same: tangent² = product of secant segments. Learn this formula by heart.

Answer 16.
Answer:
Let PT = x cm

Since, PAB is a secant and PT is a tangent to the given circle, we have,

PA · PB = \( PT^2 \)

\( \implies 4 \cdot 9 = PT^2 \)

\( \implies PT^2 = 36 \)
\( \implies PT = 6 \text{ cm} \)

📝 Teacher's Note: Again the same theorem. Students should recognize this pattern quickly. The numbers might change but the method stays the same.

🎯 Exam Tip: Look for keywords "tangent" and "secant from external point". This tells you which theorem to use. Write the theorem name for extra marks.

Answer 17.
Answer:
Let OD = OC = x cm (radius of same circle)

Since, PCD is a secant and PT is a tangent to the given circle, we have,

PC · PD = \( PT^2 \)

3 · (3 + 2x) = \( 6^2 \)

\( \implies 9 + 6x = 36 \)

\( \implies 6x = 27 \)

\( \implies x = \frac{27}{6} = \frac{9}{2} \)

Radius of the circle is \( \frac{9}{2} \) cm, diameter is 9cm

📝 Teacher's Note: Here we had to find the radius first. The secant length is PC + CD, where CD is the diameter (2 times radius). Be careful with the algebra.

🎯 Exam Tip: When the problem asks for radius, remember that diameter = 2 × radius. Set up your equation carefully and solve step by step.

Answer 18.
Answer:
\( R_1 = 4 \text{ cm}, R_2 = 12 \text{ cm} \)

PQ = 15cm

\( AB^2 = PQ^2 + (R_2 - R_1)^2 \)

\( \implies AB^2 = 15^2 + (12 - 4)^2 \)
\( \implies AB^2 = 225 + 64 \)
\( \implies AB^2 = 289 \)
\( \implies AB = 17 \text{ cm} \)

The diameter between the centre is 17cm

📝 Teacher's Note: This uses Pythagoras theorem for two circles. The distance between centers, the difference of radii, and the common tangent form a right triangle.

🎯 Exam Tip: Draw the diagram with both circles and the tangent line. Mark all the given measurements. Use Pythagoras: (distance between centers)² = (common tangent)² + (difference of radii)²

Answer 19.
Answer:
To find: PQ
\( R_1 = 3 \text{ cm}, R_2 = 8 \text{ cm} \)
AB = 13cm
\( PQ^2 = AB^2 - (R_2 - R_1)^2 \)
\( \implies PQ^2 = 13^2 - (8 - 3)^2 \)
\( \implies PQ^2 = 169 - 25 \)
\( \implies PQ^2 = 144 \)
\( \implies PQ = 12 \text{ cm} \)

Length of direct common tangent is 12cm

📝 Teacher's Note: This is the reverse of Answer 18. Now we know the distance between centers and need to find the tangent length. Same formula, different unknown.

🎯 Exam Tip: Rearrange the Pythagoras formula to find what you need. Here: (tangent length)² = (distance between centers)² - (difference of radii)²

Answer 21.
Answer:
In right \( \triangle BAC \),

\( BC^2 = AC^2 + AB^2 \)
\( AC^2 = 13^2 - 5^2 \)
\( AC^2 = 169 - 25 \)
\( AC^2 = 144 \)
AC = 12

Let OP = OQ = r (say) (radius of same circle)

\( \angle OQP = \angle OPQ = 90° \) (radius is ⊥ to tangent at the point of contact)

\( \implies \) OPAQ is a square.

AQ = AP = OP = OQ = r

BQ = BR = 5 - r ---(1) {length of tangents drawn from an external point

PC = CR = 12 - r -(2) to a circle are equal}

BC = CR + BR

13 = 12 - r + 5 - r [ from (1) and (2)]

2r = 4

r = 2

Thus, radius of the circle is 2cm.

📝 Teacher's Note: This problem combines Pythagoras theorem with the inscribed circle properties. The tangent lengths from external points are always equal.

🎯 Exam Tip: First find the third side using Pythagoras. Then use the property that tangent segments from the same external point are equal. Set up equations carefully.

Answer 22.
Answer:
\( \angle OAP = \angle OBP = 90° \) (radius is ⊥ to tangent at the point of contact)

In right \( \triangle OAP \),

\( OP^2 = OA^2 + AP^2 \)
\( OP^2 = 5^2 + 12^2 = 25 + 144 = 169 \)
OP = 13cm



In right \( \triangle OBP \),

\( OP^2 = OB^2 + BP^2 \)
\( BP^2 = 13^2 - 3^2 \)
\( BP^2 = 169 - 9 = 160 \)
BP = \( 4\sqrt{10} \) cm

📝 Teacher's Note: When tangents are drawn from an external point, the radius to each tangent point is perpendicular to that tangent. This creates right triangles.

🎯 Exam Tip: Mark the right angles at the points where radius meets tangent. Use Pythagoras theorem for each right triangle. Leave square roots in simplest form unless asked to find decimal.