CBSE Class 10 Circles Sure Shot Questions Set 01

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PRACTICE QUESTIONS
CLASS X: CHAPTER - 10
CIRCLES

1. Prove that “The tangent at any point of a circle is perpendicular to the radius through the point of contact”.

Answer:
1. Given: a circle with centre O and a tangent XY touching the circle at the point P.
2. To prove: OP ⟂ XY.
3. Take any point Q on XY other than P and join OQ.
4. The tangent touches the circle only at P, so Q lies outside the circle.
5. So OQ is greater than the radius OP, that is, OQ > OP.
6. This is true for every point Q on XY other than P, so OP is the shortest distance from O to the line XY.
7. The shortest distance from a point to a line is the perpendicular, so OP ⟂ XY.
8. Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Teacher's Note:
a) The proof rests on the fact that the tangent meets the circle at exactly one point.
b) Do not forget the construction: take another point Q on the tangent.
c) The shortest distance from a point to a line is the perpendicular.

2. Prove that “The lengths of tangents drawn from an external point to a circle are equal.”

Answer:
1. Given: a circle with centre O. PA and PB are two tangents drawn from the external point P, touching the circle at A and B.
2. To prove: PA = PB.
3. Construction: join OA, OB and OP.
4. Since a tangent is perpendicular to the radius through the point of contact, ∠OAP = ∠OBP = 90°.
5. In right triangles OAP and OBP: OA = OB (radii of the same circle), OP = OP (common) and ∠OAP = ∠OBP = 90°.
6. So △OAP ≅ △OBP by the RHS congruence rule.
7. Hence, PA = PB (by CPCT).

Teacher's Note:
a) Use RHS congruence for right triangles: hypotenuse and one side.
b) Mention that the tangent is perpendicular to the radius, as it gives the right angles.

3. Prove that “The centre lies on the bisector of the angle between the two tangents drawn from an external point to a circle.”

Answer:
1. Given: PA and PB are tangents from the external point P to a circle with centre O, touching it at A and B.
2. To prove: O lies on the bisector of ∠APB.
3. Join OA, OB and OP. Then ∠OAP = ∠OBP = 90°, as a tangent is perpendicular to the radius at the point of contact.
4. In △OAP and △OBP: OA = OB (radii), OP = OP (common) and PA = PB (tangents from an external point).
5. So △OAP ≅ △OBP by the SSS (or RHS) congruence rule.
6. Hence, ∠APO = ∠BPO (by CPCT), so OP bisects ∠APB.
7. Therefore, the centre O lies on the bisector of the angle between the two tangents.

Teacher's Note:
a) Join the centre to the external point and to both points of contact.
b) The key facts are equal radii, equal tangent lengths and the common side OP.

4. Find the length of the tangent drawn to a circle of radius 3 cm, from a point distant 5 cm from the centre.

Answer:
1. Let O be the centre, r = OT = 3 cm, and P the external point with OP = 5 cm. The tangent PT is perpendicular to the radius OT.
2. In right triangle OTP, PT² = OP² - OT² = 5² - 3² = 25 - 9 = 16.
3. PT = 4 cm.
4. Hence, the length of the tangent is 4 cm.

Teacher's Note:
a) Tangent length = square root of (distance² - radius²).
b) This is a 3-4-5 right triangle.

5. A point P is at a distance 13 cm from the centre C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.

Answer:
1. The tangent PT is perpendicular to the radius CT, so △CTP is right-angled at T.
2. CP² = CT² + PT², so CT² = 13² - 12² = 169 - 144 = 25.
3. CT = 5 cm.
4. Hence, the radius of the circle is 5 cm.

Teacher's Note:
a) The tangent is perpendicular to the radius at the point of contact.
b) Here CP is the hypotenuse, so it is the largest side.

6. From appoint Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre of the circle is 25 cm. Find the radius of the circle.

Answer:
1. Let O be the centre and QT the tangent, so OT ⟂ QT and QT = 24 cm, OQ = 25 cm.
2. OT² = OQ² - QT² = 25² - 24² = 625 - 576 = 49.
3. OT = 7 cm.
4. Hence, the radius of the circle is 7 cm.

Teacher's Note:
a) The hypotenuse is the distance from the centre to the point.
b) This is a 7-24-25 triple.

7. The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.

Answer:
1. Let O be the centre, OT = 6 cm and PT = 8 cm with OT ⟂ PT.
2. OP² = OT² + PT² = 36 + 64 = 100, so OP = 10 cm.
3. OP meets the circle at a nearest point N, with ON = 6 cm.
4. PN = OP - ON = 10 - 6 = 4 cm.
5. Hence, the distance of P from the nearest point of the circle is 4 cm.

Teacher's Note:
a) The nearest point lies on the line OP.
b) Remember to subtract the radius from OP.
c) The distance from the centre (10 cm) is not the answer.

8. Prove that in two concentric circles, the chord of the bigger circle, which touches the smaller circle is bisected at the point of contact.

Answer:
1. Given: two concentric circles with centre O. AB is a chord of the bigger circle which touches the smaller circle at P.
2. To prove: AP = PB.
3. Join OP. Since AB is a tangent to the smaller circle at P, OP ⟂ AB.
4. For the bigger circle, AB is a chord and OP is the perpendicular from the centre to the chord.
5. The perpendicular from the centre of a circle to a chord bisects the chord.
6. Hence, AP = PB, and the chord is bisected at the point of contact.

Teacher's Note:
a) Two facts are needed: tangent ⟂ radius, and perpendicular from the centre bisects the chord.
b) Draw the radius OP to the point of contact as the first step.

9. △PQR circumscribes a circle of radius r such that angle Q = 90°, PQ = 3 cm and QR = 4 cm. Find r.

Answer:
1. In right triangle PQR, PR² = PQ² + QR² = 3² + 4² = 25, so PR = 5 cm.
2. For a right-angled triangle with the right angle at Q, the incircle radius is r = \( \frac{PQ + QR - PR}{2} \) = \( \frac{3 + 4 - 5}{2} \) = 1 cm.
3. Check by area: area = \( \frac{1}{2} \) × 3 × 4 = 6 cm² and semi-perimeter s = \( \frac{3+4+5}{2} \) = 6 cm, so r = area ÷ s = 1 cm.
4. Hence, r = 1 cm.

Teacher's Note:
a) For a right triangle, the incircle radius is (sum of the legs - hypotenuse) ÷ 2.
b) Verify with area = r × semi-perimeter.

10. Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:
1. Given: ABCD is a parallelogram circumscribing a circle, that is, its sides touch the circle at P, Q, R and S.
2. To prove: ABCD is a rhombus, that is, AB = BC = CD = DA.
3. The tangents from an external point are equal, so AP = AS, BP = BQ, CR = CQ and DR = DS.
4. Adding, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ), that is, AB + CD = AD + BC.
5. In a parallelogram, AB = CD and AD = BC. So 2AB = 2AD, which gives AB = AD.
6. Hence, AB = BC = CD = DA, so ABCD is a rhombus.

Teacher's Note:
a) The key result is AB + CD = AD + BC for a quadrilateral circumscribing a circle.
b) Opposite sides of a parallelogram are equal.
c) Equal adjacent sides in a parallelogram make it a rhombus.

OR

If all the sides of a parallelogram touch the circle, show that the parallelogram is a rhombus.

Answer:
1. Let ABCD be a parallelogram whose sides AB, BC, CD and DA touch the circle at P, Q, R and S.
2. The tangents from an external point are equal: AP = AS, BP = BQ, CR = CQ and DR = DS.
3. Adding, AB + CD = AD + BC.
4. Since AB = CD and AD = BC in a parallelogram, 2AB = 2BC, so AB = BC.
5. Hence, all four sides are equal and ABCD is a rhombus.

Teacher's Note:
a) A parallelogram with equal adjacent sides is a rhombus.
b) Use the equal tangent lengths from each vertex.

11. ABC is an isosceles triangle in which AB = AC, circumscribed about a circle. Show that BC is bisected at the point of contact.

Answer:
1. Let the circle touch BC, CA and AB at D, E and F respectively.
2. The tangents drawn from an external point to a circle are equal: AE = AF, BF = BD and CE = CD.
3. Since AB = AC, subtract the equal parts: AB - AF = AC - AE, so BF = CE.
4. Hence, BD = BF = CE = CD.
5. Therefore, BC is bisected at the point of contact D.

Teacher's Note:
a) Write the three pairs of equal tangents first.
b) Subtracting AF = AE from AB = AC gives BF = CE.

12. In Fig., a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle.

Answer:
1. The tangents from an external point are equal, so DR = DS = 5 cm.
2. AR = AD - DR = 23 - 5 = 18 cm, so AQ = AR = 18 cm.
3. BQ = AB - AQ = 29 - 18 = 11 cm.
4. OQ ⟂ AB, OP ⟂ BC and ∠B = 90°, so OPBQ is a square. Hence, r = OQ = BQ = 11 cm.
5. Hence, the radius of the circle is 11 cm.

Teacher's Note:
a) Use equal tangents from D, then from A, to reach BQ.
b) A quadrilateral with three right angles and equal adjacent sides is a square.

13. ABCD is a quadrilateral such that ∠D = 90°. A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, find r.

Answer:
1. The tangents from B are equal: BQ = BP = 27 cm.
2. CQ = BC - BQ = 38 - 27 = 11 cm, so CR = CQ = 11 cm.
3. DR = CD - CR = 25 - 11 = 14 cm.
4. OR ⟂ CD, OS ⟂ DA and ∠D = 90°, so ORDS is a square. Hence, r = DR = 14 cm.
5. Hence, the radius of the circle is 14 cm.

Teacher's Note:
a) Move around the quadrilateral from B to C to D using equal tangents.
b) The radius equals DR because ORDS is a square.

14. An isosceles triangle ABC is inscribed in a circle. If AB = AC = 13 cm and BC = 10 cm, find the radius of the circle.

Answer:
1. Draw AD ⟂ BC. In an isosceles triangle, D is the midpoint of BC, so BD = 5 cm.
2. AD² = AB² - BD² = 169 - 25 = 144, so AD = 12 cm.
3. The centre O of the circle lies on AD. Let the radius be R, so OA = OB = R and OD = 12 - R.
4. In right triangle OBD: OB² = OD² + BD², so R² = (12 - R)² + 25.
5. R² = 144 - 24R + R² + 25, so 24R = 169 and R = \( \frac{169}{24} \) cm ≈ 7.04 cm.
6. Hence, the radius of the circle is \( \frac{169}{24} \) cm (about 7.04 cm).

Teacher's Note:
a) The centre lies on the altitude from A, which is the perpendicular bisector of BC.
b) Check: R = \( \frac{abc}{4 \times \text{area}} = \frac{13 \times 13 \times 10}{4 \times 60} \) gives the same value.

15. Two tangents TP and TQ are drawn from a external point T to a circle with centre O, as shown in fig. If they are inclined to each other at an angle of 100° then what is the value of ∠POQ ?

Answer:
1. The radii OP and OQ are perpendicular to the tangents, so ∠OPT = ∠OQT = 90°.
2. In quadrilateral OPTQ, the sum of the angles is 360°.
3. ∠POQ = 360° - 90° - 90° - 100° = 80°.
4. Hence, ∠POQ = 80°.

Teacher's Note:
a) ∠POQ and the angle between the tangents are supplementary: 80° + 100° = 180°.
b) The quadrilateral OPTQ has two right angles.

16. The incircle of △ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = CD.

Answer:
1. The tangents from an external point are equal: AE = AF, BF = BD and CE = CD.
2. Given AB = AC. Subtracting AF = AE: AB - AF = AC - AE, so BF = CE.
3. So BD = BF = CE = CD.
4. Hence, BD = CD.

Teacher's Note:
a) The same result as Q11: the point of contact bisects the base.
b) Do not use BD = CD at the start; it is what you must prove.

17. XP and XQ are tangents from X to the circle with O, R is a point on the circle and a tangent through R intersect XP and XQ at A and B respectively. Prove that XA + AR = XB + BR.

Answer:
1. XP and XQ are tangents from the external point X, so XP = XQ.
2. AP and AR are tangents from A, so AP = AR. BQ and BR are tangents from B, so BQ = BR.
3. XP = XA + AP = XA + AR (since AP = AR).
4. XQ = XB + BQ = XB + BR (since BQ = BR).
5. Since XP = XQ, we get XA + AR = XB + BR.
6. Hence, proved.

Teacher's Note:
a) Use equal tangents from three points: X, A and B.
b) Replace AP with AR and BQ with BR in XP and XQ.

18. A circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7cm and CD = 4 cm. Find AD.

Answer:
1. A circle touching all four sides is inscribed in the quadrilateral, so AB + CD = AD + BC.
2. 6 + 4 = AD + 7, so AD = 10 - 7 = 3 cm.
3. Hence, AD = 3 cm.

Teacher's Note:
a) For a quadrilateral circumscribing a circle, the sums of opposite sides are equal.
b) Prove the property from the equal tangents if asked.

19. TP and TQ are tangents to a circle with centre O at P and Q respectively. PQ = 8cm and radius of circle is 5 cm. Find TP and TQ.

Answer:
1. OT is the perpendicular bisector of PQ. Let it meet PQ at M, so PM = 4 cm.
2. In right triangle OMP: OM = \( \sqrt{25 - 16} \) = 3 cm.
3. In right triangle OPT, PM ⟂ OT, so OP² = OM × OT, that is, 25 = 3 × OT, so OT = \( \frac{25}{3} \) cm.
4. TP² = OT² - OP² = \( \frac{625}{9} \) - 25 = \( \frac{400}{9} \), so TP = \( \frac{20}{3} \) cm.
5. TQ = TP = \( \frac{20}{3} \) cm.
6. Hence, TP = TQ = \( \frac{20}{3} \) cm (about 6.67 cm).

Teacher's Note:
a) OT bisects PQ at right angles; this gives OM.
b) Use OP² = OM × OT in the right triangle OPT, with PM as the altitude.
c) Tangents from T are equal, so TQ = TP.

20. In the below figure PT is tangent to a circle with centre O, PT = 36 cm, AP = 24 cm. Find the radius of the circle.

Answer:
1. Let the radius be r. Then OT = OA = r and OP = OA + AP = r + 24.
2. The tangent PT is perpendicular to OT, so △OTP is right-angled at T.
3. OP² = OT² + PT², so (r + 24)² = r² + 36².
4. r² + 48r + 576 = r² + 1296, so 48r = 720.
5. r = 15 cm.
6. Hence, the radius of the circle is 15 cm.

Teacher's Note:
a) OP is the hypotenuse because the right angle is at T.
b) The r² terms cancel; this leaves a linear equation.

21. In the below figure, find the actual length of sides of △OTP.

Answer:
1. The tangent TP is perpendicular to the radius OT, so △OTP is right-angled at T.
2. OP² = OT² + TP², so (x + 2)² = (x - 6)² + (x + 1)².
3. x² + 4x + 4 = x² - 12x + 36 + x² + 2x + 1, so x² - 14x + 33 = 0.
4. x² - 11x - 3x + 33 = 0, so (x - 11)(x - 3) = 0, giving x = 11 or x = 3.
5. For x = 3, OT = x - 6 = -3 would be negative, which is not possible. So x = 11.
6. OT = 11 - 6 = 5 cm, TP = 11 + 1 = 12 cm and OP = 11 + 2 = 13 cm.
7. Hence, the sides of △OTP are OT = 5 cm, TP = 12 cm and OP = 13 cm.

Teacher's Note:
a) Reject the root that makes a side negative.
b) Check: 5² + 12² = 13².

22. In the above sided figure, find the value of x.

Answer:
1. The figure shows a quadrilateral GDEF circumscribed about a circle, with GF = 4 cm, GD = 6 cm and FE = 7 cm.
2. For a quadrilateral circumscribing a circle, the sums of opposite sides are equal: GD + FE = GF + DE.
3. 6 + 7 = 4 + x, so x = 13 - 4 = 9.
4. Hence, x = 9 cm.

Teacher's Note:
a) The figure gives three sides, and the fourth side x follows from the property of opposite sides.
b) The tangent-length labels are read as side lengths, as in the similar Q18.

23. Find the perimeter of DEFG.

Answer:
1. The tangents from an external point are equal: GP = GQ = 5 cm, FP = FS = 7 cm, ES = ER = 5 cm and DR = DQ = 4 cm.
2. The perimeter is the sum of the sides: DE + EF + FG + GD.
3. Each side is the sum of two tangent segments, so the perimeter = 2(GQ + FP + ES + DR).
4. = 2(5 + 7 + 5 + 4) = 2 × 21 = 42 cm.
5. Hence, the perimeter of DEFG is 42 cm.

Teacher's Note:
a) Every tangent segment appears twice in the perimeter.
b) Check by adding the four sides: GD = 9, DE = 9, EF = 12 and FG = 12, which total 42 cm.

24. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

Answer:
1. Let ∠PTQ = θ. Tangents from an external point are equal, so TP = TQ.
2. In △TPQ, ∠TPQ = ∠TQP = \( \frac{180^{\circ} - \theta}{2} \) = 90° - \( \frac{\theta}{2} \).
3. The radius OP is perpendicular to the tangent TP, so ∠OPT = 90°.
4. ∠OPQ = ∠OPT - ∠TPQ = 90° - (90° - \( \frac{\theta}{2} \)) = \( \frac{\theta}{2} \).
5. So θ = 2∠OPQ, that is, ∠PTQ = 2∠OPQ.
6. Hence, proved.

Teacher's Note:
a) Use equal tangents to find the base angles of the isosceles triangle TPQ.
b) OP ⟂ TP is the link between ∠OPQ and ∠TPQ.
c) The result also says ∠PTQ + ∠POQ = 180°.

25. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

Answer:
1. Join OT; it is the perpendicular bisector of PQ. Let it meet PQ at M, so PM = 4 cm.
2. OM = \( \sqrt{OP^{2} - PM^{2}} = \sqrt{25 - 16} \) = 3 cm.
3. In right triangle OPT, PM is the altitude to the hypotenuse OT, so OP² = OM × OT: 25 = 3 × OT, so OT = \( \frac{25}{3} \) cm.
4. TP² = OT² - OP² = \( \frac{625}{9} \) - 25 = \( \frac{400}{9} \).
5. TP = \( \frac{20}{3} \) cm.
6. Hence, TP = \( \frac{20}{3} \) cm (about 6.67 cm).

Teacher's Note:
a) Same method as Q19: the perpendicular bisector OT gives OM.
b) The altitude property OP² = OM × OT saves a long calculation.

26. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Answer:
1. Given: a circle with centre O and a tangent XY touching it at P. PQ is the line perpendicular to XY at P.
2. To prove: PQ passes through the centre O.
3. Suppose PQ does not pass through O, and join OP.
4. The tangent is perpendicular to the radius through the point of contact, so OP ⟂ XY.
5. Then both PQ and OP would be perpendicular to XY at the same point P. This is impossible, because only one perpendicular can be drawn to a line at a given point on it.
6. So OP and PQ must be the same line.
7. Hence, the perpendicular to the tangent at the point of contact passes through the centre.

Teacher's Note:
a) Use proof by contradiction.
b) The result says that the normal to a tangent at the point of contact is a radius line.

27. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer:
1. Let O be the centre and AT the tangent, with OA = 5 cm and AT = 4 cm. OT ⟂ AT.
2. OT² = OA² - AT² = 25 - 16 = 9.
3. OT = 3 cm.
4. Hence, the radius of the circle is 3 cm.

Teacher's Note:
a) The tangent is perpendicular to the radius at the point of contact.
b) The distance from the centre is the hypotenuse.

28. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:
1. Let O be the centre and AB the chord of the larger circle touching the smaller circle at P. Then OP = 3 cm and OP ⟂ AB.
2. In right triangle OPA, OA = 5 cm (radius of the larger circle).
3. AP² = OA² - OP² = 25 - 9 = 16, so AP = 4 cm.
4. OP bisects the chord, so AB = 2 × AP = 8 cm.
5. Hence, the length of the chord is 8 cm.

Teacher's Note:
a) The perpendicular from the centre bisects the chord.
b) Do not stop at AP = 4 cm; the chord is twice that.

29. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC

Answer:
1. Let the sides AB, BC, CD and DA of the quadrilateral touch the circle at P, Q, R and S respectively.
2. The tangents drawn from an external point to a circle are equal: AP = AS, BP = BQ, CR = CQ and DR = DS.
3. Adding the four equations: (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ).
4. AB + CD = AD + BC.
5. Hence, proved.

Teacher's Note:
a) Write the four pairs of equal tangents first.
b) Add them so that the pairs form whole sides.

30. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer:
1. Let PA and PB be the tangents from the external point P to a circle with centre O, touching it at A and B.
2. To prove: ∠APB + ∠AOB = 180°.
3. The radii OA and OB are perpendicular to the tangents, so ∠OAP = ∠OBP = 90°.
4. In quadrilateral OAPB, the sum of the angles is 360°: ∠APB + ∠AOB + 90° + 90° = 360°.
5. ∠APB + ∠AOB = 180°.
6. Hence, the angle between the tangents and the angle subtended at the centre are supplementary.

Teacher's Note:
a) This uses the angle sum of a quadrilateral and tangent ⟂ radius.
b) Compare Q15, where ∠POQ = 80° for an angle of 100°.

31. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer:
1. Let the circle with centre O touch the sides AB, BC, CD and DA of the quadrilateral ABCD at P, Q, R and S. Join OA, OB, OC, OD, OP, OQ, OR and OS.
2. The tangents from an external point are equal. In △OAP and △OAS: OP = OS (radii), AP = AS and OA = OA. So △OAP ≅ △OAS and ∠AOP = ∠AOS. Call this angle 1.
3. In the same way, ∠BOP = ∠BOQ (angle 2), ∠COQ = ∠COR (angle 3) and ∠DOR = ∠DOS (angle 4).
4. The angles around the point O add up to 360°: 2(∠1 + ∠2 + ∠3 + ∠4) = 360°, so ∠1 + ∠2 + ∠3 + ∠4 = 180°.
5. (∠1 + ∠2) + (∠3 + ∠4) = 180°, that is, ∠AOB + ∠COD = 180°.
6. Similarly, ∠BOC + ∠DOA = 180°.
7. Hence, opposite sides subtend supplementary angles at the centre.

Teacher's Note:
a) OA, OB, OC and OD bisect the angles formed at the centre by the radii to the points of contact.
b) The total angle at O is 360°, so half of it is 180°.
c) Prove congruence for each pair of right triangles with equal tangents.

OR

A circle touches all the four sides a quadrilateral ABCD. Prove that the angles subtended at the centre of the circle by the opposite sides are supplementary.

Answer:
1. Let the circle with centre O touch AB, BC, CD and DA at P, Q, R and S. Join O to A, B, C, D and to the points of contact.
2. The triangles OAP and OAS are congruent (OP = OS, AP = AS, OA common), so ∠AOP = ∠AOS. In the same way ∠BOP = ∠BOQ, ∠COQ = ∠COR and ∠DOR = ∠DOS.
3. The sum of the angles at O is 360°, so 2(∠AOP + ∠BOP + ∠COQ + ∠DOR) = 360°, which gives ∠AOP + ∠BOP + ∠COQ + ∠DOR = 180°.
4. So ∠AOB + ∠COD = 180°. In the same way, ∠BOC + ∠DOA = 180°.
5. Hence, the angles subtended at the centre by opposite sides are supplementary.

Teacher's Note:
a) Each pair of equal tangents gives a pair of equal angles at the centre.
b) Half of 360° is 180°.

32. PA and PB are the two tangents to a circle with centre O in which OP is equal to the diameter of the circle. Prove that △APB is an equilateral triangle.

Answer:
1. Let the radius be r, so OP = 2r and OA = r.
2. OA ⟂ PA, so in right triangle OAP, sin∠OPA = \( \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2} \).
3. So ∠OPA = 30°. Similarly, ∠OPB = 30°.
4. ∠APB = ∠OPA + ∠OPB = 60°.
5. PA = PB (tangents from an external point), so ∠PAB = ∠PBA = \( \frac{180^{\circ} - 60^{\circ}}{2} \) = 60°.
6. All three angles of △APB are 60°.
7. Hence, △APB is an equilateral triangle.

Teacher's Note:
a) The sine of 30° is one half; this is the key step.
b) OP bisects ∠APB, as shown in Q3.

33. Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the center of the circle.

Answer:
1. Let AB and CD be two parallel tangents to a circle with centre O, touching it at P and Q. Let a third tangent XY touch the circle at R and meet AB at X and CD at Y. Join OX, OY and OR.
2. The tangents from X are equal and OP = OR, so △OPX ≅ △ORX. Hence, ∠POX = ∠ROX. Let ∠PXO = ∠RXO = a.
3. In the same way, ∠RYO = ∠QYO = b.
4. AB ∥ CD, so the interior angles on the same side add up to 180°: ∠PXY + ∠QYX = 180°, that is, 2a + 2b = 180°, so a + b = 90°.
5. In △XOY, ∠XOY = 180° - (a + b) = 180° - 90° = 90°.
6. Hence, the intercept XY subtends a right angle at the centre.

Teacher's Note:
a) Use the fact that OX and OY bisect the angles at X and Y.
b) The co-interior angles between parallel lines add to 180°.

34. If PQ and RS are two parallel tangents to a circle with centre O and another tangent X, with point of contact C intersects PQ at A and RS at B. Prove that ∠AOB = 90°.

Answer:
1. Join OA, OB and OC. Let the tangents PQ and RS touch the circle at M and N.
2. In △OMA and △OCA: OM = OC (radii), AM = AC (tangents from A) and OA is common, so they are congruent. Hence, ∠OAM = ∠OAC = a.
3. In the same way, ∠OBN = ∠OBC = b.
4. PQ ∥ RS, so ∠MAB + ∠NBA = 180°, that is, 2a + 2b = 180°, so a + b = 90°.
5. In △AOB, ∠AOB = 180° - (a + b) = 90°.
6. Hence, ∠AOB = 90°.

Teacher's Note:
a) This is the same result as Q33 with different labels.
b) Name the points of contact on PQ and RS before using the equal tangents.

35. The incircle of △ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = DC.

Answer:
1. The tangents from an external point are equal: AF = AE, BF = BD and CE = CD.
2. Given AB = AC, subtract AF = AE: AB - AF = AC - AE, so BF = CE.
3. So BD = BF = CE = DC.
4. Hence, BD = DC.

Teacher's Note:
a) Same as Q11 and Q16.
b) Start from the equal tangents.

36. Two tangents PA and PB are drawn to the circle with center O, such that ∠APB = 120°. Prove that OP = 2AP.

Answer:
1. OP bisects ∠APB, so ∠OPA = \( \frac{120^{\circ}}{2} \) = 60°.
2. OA ⟂ PA, so △OAP is right-angled at A.
3. In △OAP, cos 60° = \( \frac{AP}{OP} \), and cos 60° = \( \frac{1}{2} \).
4. So \( \frac{AP}{OP} = \frac{1}{2} \), which gives OP = 2AP.
5. Hence, proved.

Teacher's Note:
a) The line from the centre to the external point bisects the angle between the tangents.
b) Use the cosine of the angle between the tangent and OP.

37. A circle is touching the side BC of △ABC at P and is touching AB and AC when produced at Q and R respectively. Prove that AQ = ½ (Perimeter of △ABC).

Answer:
1. The tangents from A are equal: AQ = AR.
2. The tangents from B are equal: BQ = BP. The tangents from C are equal: CR = CP.
3. Perimeter of △ABC = AB + BC + CA = AB + BP + PC + CA.
4. Replace BP with BQ and PC with CR: perimeter = (AB + BQ) + (CR + CA) = AQ + AR.
5. Since AQ = AR, perimeter = 2AQ.
6. Hence, AQ = ½ (perimeter of △ABC).

Teacher's Note:
a) This is the excircle result: the tangent length from A is the semi-perimeter.
b) Break BC into BP + PC and use equal tangents.

38. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Answer:
1. Let the circle touch AB at F and AC at E, and let AE = AF = x. Then BF = BD = 8 cm and CE = CD = 6 cm.
2. AB = x + 8, AC = x + 6 and BC = 14 cm. The semi-perimeter s = \( \frac{(x+8) + 14 + (x+6)}{2} \) = x + 14.
3. By Heron's formula, area = \( \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(x)(8)(6)} \), since s - 14 = x, s - (x + 6) = 8 and s - (x + 8) = 6.
4. Also area = r × s = 4(x + 14).
5. So 16(x + 14)² = 48x(x + 14), which gives x + 14 = 3x and x = 7.
6. AB = 7 + 8 = 15 cm and AC = 7 + 6 = 13 cm.
7. Hence, AB = 15 cm and AC = 13 cm.

Teacher's Note:
a) Use area = r × s together with Heron's formula.
b) Check: the sides 13, 14 and 15 have area 84 = 4 × 21.
c) Equal tangents from each vertex are the starting point.

39. In figure, chords AB and CD of the circle intersect at O. OA = 5cm, OB = 3cm and OC = 2.5cm. Find OD.

Answer:
1. When two chords of a circle intersect, the products of the segments are equal: OA × OB = OC × OD.
2. 5 × 3 = 2.5 × OD, so OD = \( \frac{15}{2.5} \) = 6 cm.
3. Hence, OD = 6 cm.

Teacher's Note:
a) This is the intersecting chords theorem.
b) Write the product of the two parts of each chord.

40. In figure. Chords AB and CD intersect at P. If AB = 5cm, PB = 3cm and PD = 4cm. Find the length of CD.

Answer:
1. Here P lies outside the circle on the extensions of the chords, so PA × PB = PC × PD.
2. PA = PB + BA = 3 + 5 = 8 cm.
3. 8 × 3 = PC × 4, so PC = 6 cm.
4. CD = PC - PD = 6 - 4 = 2 cm.
5. Hence, CD = 2 cm.

Teacher's Note:
a) For external intersection, use the product of the whole secant and its outer part.
b) PC is the whole secant; subtract PD to get the chord CD.

41. In the figure, ABC is an isosceles triangle in which AB = AC. A circle through B touches the side AC at D and intersect the side AB at P. If D is the midpoint of side AC, Then AB = 4AP.

Answer:
1. AC is a tangent to the circle at D and ABP is a secant, so AD² = AP × AB.
2. D is the midpoint of AC, so AD = \( \frac{AC}{2} \) = \( \frac{AB}{2} \) (as AB = AC).
3. So \( \frac{AB^{2}}{4} \) = AP × AB, which gives AP = \( \frac{AB}{4} \).
4. Hence, AB = 4AP. Proved.

Teacher's Note:
a) The tangent-secant relation is AD² = AP × AB.
b) Replace AC with AB, as AB = AC.

42. In the figure. Find the value of AB Where PT = 5cm and PA = 4cm.

Answer:
1. PT is a tangent and PAB is a secant, so PT² = PA × PB.
2. 25 = 4 × PB, so PB = \( \frac{25}{4} \) = 6.25 cm.
3. AB = PB - PA = 6.25 - 4 = 2.25 cm.
4. Hence, AB = 2.25 cm.

Teacher's Note:
a) PB is the whole secant.
b) Subtract PA from PB to get the chord AB.

43. In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6cm, BC = 7cm and CD = 4cm. Find AD.

Answer:
1. The quadrilateral circumscribes a circle, so AB + CD = AD + BC.
2. 6 + 4 = AD + 7, so AD = 3 cm.
3. Hence, AD = 3 cm.

Teacher's Note:
a) Sums of opposite sides are equal for a tangential quadrilateral.
b) This is the same as Q18.

44. Prove that “If a line touches a circle and from the point of contact a chord is drawn, the angle which this chord makes with the given line are equal respectively to the angles formed in the corresponding alternate segments.”

Answer:
1. Given: a circle with centre O, a tangent PQ touching it at A, and a chord AB. C is a point in the alternate segment (on the other side of AB from the angle BAQ).
2. To prove: ∠BAQ = ∠ACB and ∠BAP = ∠AEB (E in the other segment).
3. Construction: draw the diameter AD and join BD.
4. AD ⟂ PQ (tangent ⟂ radius), so ∠DAQ = 90°. Hence, ∠BAQ = 90° - ∠BAD.
5. ∠ABD = 90° (angle in a semicircle), so in △ABD, ∠ADB = 90° - ∠BAD.
6. So ∠BAQ = ∠ADB.
7. ∠ADB = ∠ACB (angles in the same segment on chord AB), so ∠BAQ = ∠ACB.
8. ∠BAP = 180° - ∠BAQ = 180° - ∠ACB = ∠AEB, as AEBC is a cyclic quadrilateral (E is a point on the other arc).
9. Hence, the angles between the tangent and the chord are equal to the angles in the corresponding alternate segments.

Teacher's Note:
a) Draw the diameter through the point of contact; this is the key construction.
b) Use the angle in a semicircle and the angles in the same segment.
c) The second equality follows from the linear pair and the cyclic quadrilateral.

45. Prove that “If a line is drawn through an end point of a chord of a circle so that the angle formed by it with the chord is equal to the angle subtend by chord in the alternate segment, then the line is a tangent to the circle.”

Answer:
1. Given: a chord AB of a circle with centre O, and a line PAQ such that ∠BAQ = ∠ACB, where C is a point in the alternate segment.
2. To prove: PQ is a tangent at A.
3. Draw the diameter AD and join BD.
4. ∠ADB = ∠ACB (angles in the same segment), so ∠ADB = ∠BAQ.
5. ∠ABD = 90° (angle in a semicircle), so in △ABD, ∠BAD + ∠ADB = 90°.
6. So ∠BAD + ∠BAQ = 90°, that is, ∠DAQ = 90°.
7. So PQ ⟂ AD at A, where AD is a diameter, which means PQ is perpendicular to the radius at A.
8. Hence, PQ is a tangent to the circle at A.

Teacher's Note:
a) This is the converse of the alternate segment theorem.
b) A line perpendicular to a radius at its end on the circle is a tangent.

46. In figure. l and m are two parallel tangents at A and B. The tangent at C makes an intercept DE between the tangent l and m. Prove that ∠DFE = 90°

Answer:
1. Here F is the centre of the circle. Join FA, FB and FC.
2. The tangents from D are equal and FA = FC, so △FAD ≅ △FCD. Hence, ∠AFD = ∠CFD (call it a) and ∠ADF = ∠CDF.
3. Similarly, △FCE ≅ △FBE, so ∠CFE = ∠BFE (call it b).
4. A, F and B lie on a straight line (the diameter), since l ∥ m and the radii FA and FB are perpendicular to them.
5. ∠AFD + ∠CFD + ∠CFE + ∠BFE = 180°, so 2a + 2b = 180° and a + b = 90°.
6. ∠DFE = ∠CFD + ∠CFE = a + b = 90°.
7. Hence, ∠DFE = 90°.

Teacher's Note:
a) Tangents at the ends of a diameter are parallel.
b) The angles on a straight line AFB add up to 180°.

47. In figure, a circle is inscribed in a △ABC having sides AB = 12 cm, BC = 8cm and AC = 10cm. Find AD, BE and CF.

Answer:
1. Let AD = AF = x, BD = BE = y and CE = CF = z (tangents from the same point are equal).
2. AB = x + y = 12, BC = y + z = 8 and AC = x + z = 10.
3. Adding, 2(x + y + z) = 30, so x + y + z = 15.
4. x = 15 - 8 = 7, y = 15 - 10 = 5 and z = 15 - 12 = 3.
5. Hence, AD = 7 cm, BE = 5 cm and CF = 3 cm.

Teacher's Note:
a) Equal tangents from each vertex give three simple equations.
b) Check: AD + BD = 7 + 5 = 12 and BE + CE = 5 + 3 = 8.

OR

A circle is inscribed in a △ABC having sides 8 cm, 10 cm and 12 cm as shown in fig. Find AD, BE and CF.

Answer:
1. Taking the sides as AB = 12 cm, BC = 8 cm and AC = 10 cm (the sides are not labelled in the figure).
2. Let AD = AF = x, BD = BE = y and CE = CF = z. Then x + y = 12, y + z = 8 and x + z = 10.
3. Adding, x + y + z = 15, so x = 7, y = 5 and z = 3.
4. Hence, AD = 7 cm, BE = 5 cm and CF = 3 cm.

Teacher's Note:
a) The figure does not label the sides, so the same assignment as the first version is used.
b) Different assignments of the sides give different tangent lengths, but the method is the same.

48. If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B, prove that OP is the perpendicular bisector of AB.

Answer:
1. Let OP meet AB at C. Join OA and OB.
2. PA = PB (tangents from an external point) and OA = OB (radii).
3. So P and O are both equidistant from A and B, so both lie on the perpendicular bisector of AB.
4. Alternatively: in △APC and △BPC, PA = PB, ∠APC = ∠BPC (OP bisects ∠APB) and PC is common, so △APC ≅ △BPC. Hence, AC = BC and ∠ACP = ∠BCP = 90° (they form a linear pair).
5. Hence, OP is the perpendicular bisector of AB.

Teacher's Note:
a) Points equidistant from two points lie on the perpendicular bisector.
b) The line through O and P is the axis of symmetry of the figure.

49. If △ABC is isosceles with AB = AC, Prove that the tangent at A to the circumcircle of △ABC is parallel to BC.

Answer:
1. Let DAE be the tangent at A to the circumcircle, with D on the side of B.
2. By the alternate segment theorem, ∠DAB = ∠ACB (the angle in the alternate segment).
3. AB = AC, so ∠ABC = ∠ACB.
4. So ∠DAB = ∠ABC. These are alternate interior angles for the lines DE and BC with transversal AB.
5. Hence, DE ∥ BC, that is, the tangent at A is parallel to BC.

Teacher's Note:
a) Use the alternate segment theorem to link the tangent angle with the angle in the triangle.
b) Equal alternate angles give parallel lines.

50. Two circles intersect at A and B. From a point P on one of these circles, two lines segments PAC and PBD are drawn intersecting the other circles at C and D respectively. Prove that CD is parallel to the tangent at P.

Answer:
1. Let TPT' be the tangent at P to the first circle (through P, A, B).
2. By the alternate segment theorem in the first circle, ∠TPA = ∠PBA.
3. A, C, D and B lie on the second circle, so ACDB is a cyclic quadrilateral. Its exterior angle at B is equal to the interior opposite angle: ∠PBA = ∠ACD.
4. So ∠TPA = ∠ACD, that is, ∠TPC = ∠PCD.
5. These are alternate angles for the lines TP and CD with transversal PC.
6. Hence, CD ∥ TP, that is, CD is parallel to the tangent at P.

Teacher's Note:
a) Use the alternate segment theorem in the first circle and the cyclic quadrilateral in the second.
b) The exterior angle of a cyclic quadrilateral equals the interior opposite angle.

51. Two circles intersect in points P and Q. A secant passing through P intersects the circles at A an B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, T and B are concyclic.

Answer:
1. Join AQ and BQ. In the circle through A, P, Q, the tangent at A and the chord AP make the angle ∠TAP, and by the alternate segment theorem ∠TAB = ∠AQP.
2. In the circle through B, P, Q, the tangent at B gives ∠TBA = ∠BQP.
3. Add: ∠AQB = ∠AQP + ∠PQB = ∠TAB + ∠TBA.
4. In △ATB, ∠TAB + ∠TBA = 180° - ∠ATB.
5. So ∠AQB = 180° - ∠ATB, that is, ∠AQB + ∠ATB = 180°.
6. The opposite angles Q and T of the quadrilateral AQBT are supplementary.
7. Hence, A, Q, T and B are concyclic.

Teacher's Note:
a) If a pair of opposite angles of a quadrilateral is supplementary, it is cyclic.
b) The alternate segment theorem is used twice, once for each circle.

52. In the given figure TAS is a tangent to the circle, with centre O, at the point A. If ∠OBA = 32°, find the value of x and y.

Answer:
1. OA = OB (radii), so ∠OAB = ∠OBA = 32°.
2. OA ⟂ TS (tangent ⟂ radius), so ∠OAS = 90°.
3. x = ∠BAS = ∠OAS - ∠OAB = 90° - 32° = 58°.
4. y = ∠ACB is the angle in the alternate segment, so y = ∠BAS = 58°.
5. Check: ∠AOB = 180° - 32° - 32° = 116°, and the angle at the circumference is half of it: y = 58°.
6. Hence, x = 58° and y = 58°.

Teacher's Note:
a) The angle between the tangent and the chord equals the angle in the alternate segment.
b) Check the answer with the central angle theorem.

53. In the given figure. PT is a tangent and PAB is a secant to a circle. If the bisector of ∠ATB intersect AB in M, Prove that: (i) ∠PMT = ∠PTM (ii) PT = PM

Answer:
1. Let ∠ATM = ∠BTM = θ (TM bisects ∠ATB) and ∠TBA = β.
2. By the alternate segment theorem, ∠PTA = ∠TBA = β.
3. ∠PTM = ∠PTA + ∠ATM = β + θ.
4. ∠PMT is the exterior angle of △BMT, so ∠PMT = ∠MBT + ∠BTM = β + θ.
5. So ∠PMT = ∠PTM. This proves (i).
6. In △PTM, the sides opposite equal angles are equal, so PT = PM. This proves (ii).
7. Hence, proved.

Teacher's Note:
a) Use the alternate segment theorem to get ∠PTA = ∠TBA.
b) An exterior angle equals the sum of the two interior opposite angles.

54. In the adjoining figure, ABCD is a cyclic quadrilateral. AC is a diameter of the circle. MN is tangent to the circle at D, ∠CAD = 40°, ∠ACB = 55° Determine ∠ADM and ∠BAD

Answer:
1. AC is a diameter, so ∠ADC = 90° and ∠ABC = 90° (angles in a semicircle).
2. In △ADC, ∠ACD = 180° - 90° - 40° = 50°.
3. By the alternate segment theorem, ∠ADM = ∠ACD = 50°.
4. In △ABC, ∠BAC = 180° - 90° - 55° = 35°.
5. ∠BAD = ∠BAC + ∠CAD = 35° + 40° = 75°.
6. Hence, ∠ADM = 50° and ∠BAD = 75°.

Teacher's Note:
a) The angle in a semicircle is 90°.
b) Check: ∠BAD + ∠BCD = 75° + 105° = 180°, as for a cyclic quadrilateral.

55. The diagonals of a parallelogram ABCD intersect at E. Show that the circumcircle of △ADE and △BCE touch each other at E.

Answer:
1. Let PEQ be the tangent at E to the circumcircle of △ADE, with P on the side of A and Q on the side of C. A, E and C are collinear (E is on the diagonal AC).
2. By the alternate segment theorem, ∠PEA = ∠EDA.
3. AD ∥ BC, so ∠EDA = ∠EBC (alternate angles, with transversal BD).
4. ∠PEA = ∠CEQ (vertically opposite angles).
5. So ∠CEQ = ∠EBC. This says the angle between the line EQ and the chord EC of the circle BCE is equal to the angle ∠EBC in the alternate segment.
6. By the converse of the alternate segment theorem, EQ touches the circumcircle of △BCE at E.
7. So the line PQ is a common tangent at E to both circles.
8. Hence, the two circles touch each other at E.

Teacher's Note:
a) Prove that the same line is tangent to both circles at E.
b) The converse of the alternate segment theorem is used in the last step.

56. A circle is drawn with diameter AB interacting the hypotenuse AC of right triangle ABC at the point P. Show that the tangent to the circle at P bisects the side BC.

Answer:
1. ∠APB = 90° (angle in a semicircle), so BP ⟂ AC, and ∠BPC = 90°.
2. BC ⟂ AB at B, and AB is a diameter, so BC is a tangent to the circle at B.
3. Let the tangent at P meet BC at Q. The tangents from Q are equal: QP = QB, so ∠QPB = ∠QBP.
4. ∠QPC = 90° - ∠QPB and ∠QCP = 90° - ∠PBC (in the right triangle BPC). Also ∠QBP = ∠PBC (Q lies on BC).
5. So ∠QPC = ∠QCP, which gives QC = QP.
6. QB = QP = QC, so Q is the midpoint of BC.
7. Hence, the tangent at P bisects BC.

Teacher's Note:
a) BC is a tangent at B because it is perpendicular to the diameter AB.
b) Equal tangents and equal angles give QB = QC.

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

Answer:
1. Let the radii of the outer and inner circles be R and r, with centre O.
2. Let AB be any chord of the outer circle touching the inner circle at P. Then OP = r and OP ⟂ AB.
3. In right triangle OPA, AP² = OA² - OP² = R² - r².
4. OP bisects the chord, so AB = 2AP = \( 2\sqrt{R^{2} - r^{2}} \).
5. This value depends only on R and r, which are fixed.
6. Hence, all such chords are of equal length.

Teacher's Note:
a) The answer does not depend on the position of the chord.
b) Same method as Q8 and Q28.

58. If AB, AC, PQ are tangents in below figure and AB = 5 cm, find the perimeter of △APQ.

Answer:
1. The tangents from A are equal: AB = AC = 5 cm.
2. The tangents from P are equal: PX = PB, and the tangents from Q are equal: QX = QC.
3. Perimeter of △APQ = AP + PQ + AQ = AP + (PX + XQ) + AQ = AP + PB + QC + AQ.
4. = (AP + PB) + (AQ + QC) = AB + AC = 5 + 5 = 10 cm.
5. Hence, the perimeter of △APQ is 10 cm.

Teacher's Note:
a) Break PQ into PX + XQ at the point of contact.
b) Perimeter = AB + AC = twice the tangent length from A.

59. If PA and PB are tangents from an outside point P, such that PA = 10 cm and ∠APB = 60°. Find the length of chord AB.

Answer:
1. PA = PB = 10 cm (tangents from an external point).
2. In △PAB, ∠PAB = ∠PBA = \( \frac{180^{\circ} - 60^{\circ}}{2} \) = 60°.
3. All angles are 60°, so △PAB is equilateral.
4. So AB = PA = 10 cm.
5. Hence, the length of the chord AB is 10 cm.

Teacher's Note:
a) An isosceles triangle with a 60° angle is equilateral.
b) Equal tangents make △PAB isosceles.

60. From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of △PCD.

Answer:
1. Take C on PA and D on PB. The tangents from P are equal: PA = PB = 14 cm.
2. The tangents from C are equal: CE = CA. The tangents from D are equal: DE = DB.
3. Perimeter of △PCD = PC + CD + PD = PC + (CE + ED) + PD = PC + CA + DB + PD.
4. = (PC + CA) + (PD + DB) = PA + PB = 14 + 14 = 28 cm.
5. Hence, the perimeter of △PCD is 28 cm.

Teacher's Note:
a) Split CD at the point of contact E.
b) The perimeter equals PA + PB.

61. Prove that the tangents at the extremities of any chord make equal angles with the chord.

Answer:
1. Let AB be a chord of a circle with centre O, and let the tangents at A and B meet at T.
2. TA = TB (tangents from an external point).
3. In △TAB, the angles opposite equal sides are equal: ∠TAB = ∠TBA.
4. So the tangents at the extremities of the chord make equal angles with the chord.
5. Hence, proved.

Teacher's Note:
a) The tangents from T are equal, so △TAB is isosceles.
b) This also follows from the alternate segment theorem.

62. From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

Answer:
1. Let OP meet AB at C. Join OA and OB.
2. PA = PB (tangents from an external point) and OA = OB (radii). So both O and P are equidistant from A and B.
3. So both O and P lie on the perpendicular bisector of AB, which means the line OP is the perpendicular bisector of AB.
4. Also, in △APC and △BPC: PA = PB, ∠APC = ∠BPC and PC is common, so they are congruent, giving AC = BC and ∠ACP = ∠BCP = 90°.
5. Hence, OP is the perpendicular bisector of AB.

Teacher's Note:
a) Same as Q48.
b) Either of the two methods is accepted.

63. The radius of the incircle of a triangle is 4 cm and the segments into which one side divided by the point of contact are 6 cm and 8 cm. Find the other two sides of the triangle.

Answer:
1. Let the side BC be divided at D into BD = 8 cm and DC = 6 cm. Let AE = AF = x be the tangents from A.
2. AB = x + 8, AC = x + 6 and BC = 14 cm, so s = x + 14.
3. Area = \( \sqrt{(x+14)(x)(8)(6)} \) (Heron) and area = r × s = 4(x + 14).
4. 16(x + 14)² = 48x(x + 14), so x + 14 = 3x, giving x = 7.
5. AB = 15 cm and AC = 13 cm.
6. Hence, the other two sides are 15 cm and 13 cm.

Teacher's Note:
a) Same method as Q38.
b) Verify: the triangle 13, 14, 15 has area 84 = 4 × 21.

64. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that △APB is an equilateral triangle.

Answer:
1. Let the radius be r, so OP = 2r and OA = r.
2. OA ⟂ PA, so in right triangle OAP, sin∠OPA = \( \frac{r}{2r} = \frac{1}{2} \), so ∠OPA = 30°.
3. OP bisects ∠APB, so ∠APB = 60°.
4. PA = PB, so ∠PAB = ∠PBA = 60°.
5. All three angles are 60°.
6. Hence, △APB is an equilateral triangle.

Teacher's Note:
a) Same as Q32.
b) The sine of 30° is one half.

65. In fig. ABC is a right triangle right angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.

Answer:
1. In right triangle ABC, AC² = AB² + BC² = 64 + 36 = 100, so AC = 10 cm.
2. Let the incircle touch AB, BC and CA at P, Q and R. OPBQ is a square, since ∠B = 90° and OP ⟂ AB, OQ ⟂ BC. So BP = BQ = r.
3. AP = AR = 8 - r and CQ = CR = 6 - r.
4. AC = AR + RC = (8 - r) + (6 - r) = 14 - 2r = 10, so r = 2 cm.
5. Check by area: area = \( \frac{1}{2} \) × 6 × 8 = 24 and s = \( \frac{6+8+10}{2} \) = 12, so r = \( \frac{24}{12} \) = 2 cm.
6. Hence, the radius of the incircle is 2 cm.

Teacher's Note:
a) For a right triangle, r = (AB + BC - AC) ÷ 2.
b) Check using area = r × s.

66. In the below figure, △ABC is circumscribed, find the value of x.

Answer:
1. The tangents from an external point are equal: AP = AQ = 4 cm and BP = BR = 6 cm.
2. AC = 12 cm, so QC = AC - AQ = 12 - 4 = 8 cm and CR = CQ = 8 cm.
3. x = BC = BR + RC = 6 + 8 = 14 cm.
4. Hence, x = 14 cm.

Teacher's Note:
a) Use equal tangents from A, then from C, and add at B.
b) Check: s = 18, so the sides 10, 12 and 14 satisfy the triangle inequality.

67. In the above right-sided figure, quadrilateral ABCD is circumscribed, find the value of x.

Answer:
1. The tangents from D are equal: DS = DR = 3 cm. DC = 7 cm, so CR = DC - DR = 7 - 3 = 4 cm.
2. The tangents from C are equal: CQ = CR = 4 cm.
3. The tangents from B are equal: BQ = BP = 5 cm.
4. x = BC = BQ + QC = 5 + 4 = 9 cm.
5. Hence, x = 9 cm.

Teacher's Note:
a) Move from D to C to B using equal tangents.
b) The labels 3 cm, 7 cm and 5 cm are read as DS, DC and BP.

68. In the below figure, quadrilateral ABCD is circumscribed, find the perimeter of quadrilateral ABCD.

Answer:
1. The tangents from each vertex are equal: AP = AS = 4, DS = DR = 3, CR = CQ = 5 and BP = BQ = 6 (cm).
2. AB = AP + PB = 4 + 6 = 10 cm, BC = BQ + QC = 6 + 5 = 11 cm, CD = CR + RD = 5 + 3 = 8 cm and DA = DS + SA = 3 + 4 = 7 cm.
3. Perimeter = 10 + 11 + 8 + 7 = 36 cm.
4. Hence, the perimeter of ABCD is 36 cm.

Teacher's Note:
a) Perimeter = 2 × (sum of the four tangent lengths) = 2 × 18.
b) Check by adding all four sides.

69. In the above right sided figure, quadrilateral ABCD is circumscribed and AD ⟂ DC, find the value of x if the radius of incircle is 10 cm.

Answer:
1. The figure shows the right angle at B, so AB ⟂ BC. (The printed condition "AD ⟂ DC" does not agree with the figure, so the right angle at B is used.) Since OP ⟂ AB and OQ ⟂ BC, OPBQ is a square, so BP = BQ = 10 cm.
2. Let AP = AS = a, DS = DR = d and CQ = CR = c. From AD = a + d = 27 and DC = c + d = 38: c = a + 11 and d = 27 - a.
3. For a quadrilateral with an incircle of radius r, the half-angle at a vertex satisfies tan(half-angle) = r ÷ (tangent length from that vertex), and the four half-angles add up to 180°. With B/2 = 45°: A/2 + C/2 + D/2 = 135°.
4. Applying tan to this relation with r = 10 gives the equation a³ - 6a² - 737a + 1830 = 0. The only root with 0 < a < 27 is a ≈ 2.45 cm.
5. So c = a + 11 ≈ 13.45 cm.
6. x = BC = BQ + QC = 10 + c ≈ 23.45 cm.
7. Hence, x ≈ 23.45 cm.

Teacher's Note:
a) The figure and the printed data are not neat; this value uses the right angle shown at B.
b) If the right angle were at D, then DS = DR = 10 cm, but the remaining data would not fix x.
c) Tangent lengths and the incircle radius together fix the angles of the quadrilateral.

70. If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Answer:
1. Draw AD ⟂ BC. The perpendicular bisector of BC passes through the centre O, so O lies on the line AD. Let AD = h and BD = y.
2. In right triangle ABD: y² + h² = 36.
3. The centre O lies on AD and OB = OA = 9 cm, so OD = |9 - h|. In right triangle OBD: y² + (9 - h)² = 81.
4. Subtracting: y² + h² - 18h + 81 = 81, so 36 - 18h = 0 and h = 2 cm.
5. y² = 36 - 4 = 32, so y = 4√2 cm and BC = 8√2 cm.
6. Area = \( \frac{1}{2} \) × BC × AD = \( \frac{1}{2} \) × 8√2 × 2 = 8√2 cm².
7. Hence, the area of the triangle is 8√2 cm² (about 11.31 cm²).

Teacher's Note:
a) The triangle is obtuse at A, since the circumradius (9 cm) is large compared with AB (6 cm).
b) Check: R = \( \frac{abc}{4 \times \text{area}} = \frac{6 \times 6 \times 8\sqrt{2}}{4 \times 8\sqrt{2}} \) = 9 cm.

71. A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the △ABC.

Answer:
1. OP ⟂ AP, so in right triangle OPA: AP² = OA² - OP² = 169 - 25 = 144, so AP = 12 cm and AQ = 12 cm.
2. The tangents from B are equal: BP = BR. The tangents from C are equal: CQ = CR.
3. Perimeter of △ABC = AB + BC + CA = AB + (BR + RC) + CA = AB + BP + CQ + CA.
4. = (AB + BP) + (AC + CQ) = AP + AQ = 12 + 12 = 24 cm.
5. Hence, the perimeter of △ABC is 24 cm.

Teacher's Note:
a) Find the tangent length AP first using the Pythagoras theorem.
b) The perimeter of the cut-off triangle is twice the tangent length from A.

72. The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA

Answer:
1. AB is a diameter, so ∠ACB = 90° (angle in a semicircle).
2. ∠PCB = ∠PCA - ∠ACB = 110° - 90° = 20°.
3. By the alternate segment theorem, ∠CAB = ∠PCB = 20°.
4. In △ABC, ∠CBA = 180° - ∠ACB - ∠CAB = 180° - 90° - 20° = 70°.
5. Hence, ∠CBA = 70°.

Teacher's Note:
a) The angle in a semicircle is 90°.
b) The alternate segment theorem gives ∠CAB directly.

73. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.

Answer:
1. ∠APB = 90° (angle in a semicircle), so BP ⟂ AC, and ∠BPC = 90°.
2. BC ⟂ AB at B, and AB is a diameter, so BC is a tangent to the circle at B.
3. Let the tangent at P meet BC at Q. The tangents from Q are equal: QP = QB, so ∠QPB = ∠QBP.
4. In the right triangle BPC, ∠QPC = 90° - ∠QPB and ∠QCP = 90° - ∠PBC. Also, ∠QBP = ∠PBC.
5. So ∠QPC = ∠QCP, which gives QC = QP = QB.
6. Hence, Q is the midpoint of BC, that is, the tangent at P bisects BC.

Teacher's Note:
a) Same as Q56.
b) BC touches the circle at B because it is perpendicular to the diameter AB.

74. AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects extended AB at a point D. Prove that BC = BD.

Answer:
1. AB is a diameter, so ∠ACB = 90°. In △ABC, ∠ABC = 180° - 90° - 30° = 60°.
2. By the alternate segment theorem, ∠BCD = ∠BAC = 30°.
3. ∠CBD = 180° - ∠ABC = 120° (linear pair).
4. In △BCD, ∠BDC = 180° - 120° - 30° = 30°.
5. So ∠BCD = ∠BDC = 30°, and the sides opposite equal angles are equal: BC = BD.
6. Hence, BC = BD.

Teacher's Note:
a) The sides opposite equal angles are equal.
b) Use the angle in a semicircle and the alternate segment theorem.

75. In the below figure from an external point A, tangents AB and AC are drawn to a circle. PQ is a tangent to the circle at X. If AC = 15 cm, find the perimeter of the triangle APQ.

Answer:
1. The tangents from A are equal: AB = AC = 15 cm.
2. The tangents from P are equal: PX = PB. The tangents from Q are equal: QX = QC.
3. Perimeter of △APQ = AP + PQ + AQ = AP + (PX + XQ) + AQ = AP + PB + QC + AQ.
4. = (AP + PB) + (AQ + QC) = AB + AC = 15 + 15 = 30 cm.
5. Hence, the perimeter of △APQ is 30 cm.

Teacher's Note:
a) Same method as Q58.
b) The perimeter is twice the tangent length from A.

Question. In the above question (no. 3) what is the ratio of AM : BM?
(a) 5 : 6
(b) 3 : 2
(c) 1 : 1
(d) can't be determined
Answer: (c) 1 : 1

Question. If the diagonals of a cyclic quadrilateral are equal, then the quadrilateral is
(a) rhombus
(b) square
(c) rectangle
(d) none of the options
Answer: (c) rectangle

Question. The quadrilateral formed by angle bisectors of a cyclic quadrilateral is a:
(a) rectangle
(b) square
(c) parallelogram
(d) cyclic quadrilateral
Answer: (d) cyclic quadrilateral

Question. Three circles touch each other externally. The distance between their centre is 5 cm, 6 cm and 7 cm. Find the radii of the circles :
(a) 2 cm, 3 cm, 4 cm
(b) 3 cm, 4 cm, 1 cm
(c) 1 cm, 2.5 cm, 3.5 cm
(d) 1 cm, 2 cm, 4 cm
Answer: (a) 2 cm, 3 cm, 4 cm

Question. O and O' are the centres of two circles which touch each other externally at P. AB is a common tangent. Find \( \angle APO \):
(a) 90°
(b) 120°
(c) 60°
(d) data insufficient
Answer: (a) 90°

Question. In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is :
(a) 23 cm
(b) 30 cm
(c) 15 cm
(d) none of the options
Answer: (b) 30 cm

Question. A circle has two parallel chords of lengths 6 cm and 8 cm. If the chords are 1 cm apart and the centre is on the same side of the chords, then a diameter of the circle is of length:
(a) 5 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
Answer: (d) 10 cm

Question. Three equal circles of unit radius touch each other. Then, the area of the circle circumscribing the three circles is :
(a) \( 6\pi (2 + \sqrt{3})^2 \)
(b) \( \frac{\pi}{6}(2 + \sqrt{3})^2 \)
(c) \( \frac{\pi}{3}(2 + \sqrt{3})^2 \)
(d) \( 3\pi (2 + \sqrt{3})^2 \)
Answer: (c) \( \frac{\pi}{3}(2 + \sqrt{3})^2 \)

Question. Through any given set of four points P,Q, R, S it is possible to draw :
(a) atmost one circle
(b) exactly one circle
(c) exactly two circles
(d) exactly three circles
Answer: (a) atmost one circle

Question. The number of common tangents that can be drawn to two given circles is at the most :
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4

Question. AB and CD are two parallel of a circle such that AB=10 cm and CD = 24 cm. The chords are on opposite sides of the centre and the distance between them is 17 cm. Find the radius of the circle
(a) 11 cm
(b) 12 cm
(c) 13 cm
(d) 14 cm
Answer: (c) 13 cm

Question. An equilateral triangle has side \( 2\sqrt{3} \) cm. The radius of its circumcircle will be
(a) 2 cm
(b) \( \sqrt{3} \) cm
(c) 3 cm
(d) 4 cm
Answer: (a) 2 cm