## Chapter 10 Circles Ex 10.2

**Question 1.**

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

(a) 7 cm Sol.

(b) 12 cm

(c) 15 cm

(d) 24.5 cm **Solution:**

**Question 2.**

In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

(a) 60°

(b) 70°

(c) 80°

(d) 90° **Solution:**

∠OPT = 90°

∠OQT = 90°

∠POQ = 110°

TPOQ is a quadrilateral,

∴ ∠PTQ + ∠POQ = 180° ⇒ ∠PTQ + 110° = 180°

⇒∠PTQ = 180°- 110° = 70°

Hence, correct option is (b).

**Question 3.**

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to

(a) 50°

(b) 60°

(c) 70°

(d) 80° **Solution:**

In AOAP and AOBP

OA = OB [Radii]

PA = PB

[Lengths of tangents from an external point are equal]

OP = OP [Common]

∴ ∆OAP ≅ ∆OBP [SSS congruence rule]

∠AOB + ∠APB = 180° ⇒ ∠AOB + 80° = 180°

⇒∠AOB = 180° – 80° = 100°

From eqn. (i), we get

⇒∠POA = 12 x 100° = 50°

Hence, correct option is (a)

**Question 4.**

Prove that the tangents drawn at the ends of a diameter of a circle are parallel. **Solution:**

AB is a diameter of the circle, p and q are two tangents.

OA ⊥ p and OB ⊥ q

∠1 = ∠2 = 90°

⇒ p || q ∠1 and ∠2 are alternate angles]

**Question 5.**

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

XY tangent to the circle C(0, r) at B and AB ⊥ XY. Join OB.

∠ABY = 90° [Given]

∠OBY = 90°

[Radius through point of contact is perpendicular to the tangent]

∴ ∠ABY + ∠OBY = 180° ⇒ ABOiscollinear

∴ AB passes through centre of the circle.

**Question 6.**

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. **Solution:**

OA = 5 cm, AP = 4 cm OP = Radius of the circle

∠OPA = 90° [Radius and tangent are perpendicular]

**Question 7.**

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. **Solution:**

Radius of larger circle = 5 cm Radius of smaller circle = 3 cm

OP ⊥ AB

[Radius of circle is perpendicular to the tangent]

AB is a chord of the larger circle

**Question 8.** A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

**Solution:**

AP = AS … (i)

[Lengths of tangents from an external point are equal]

BP = BQ … (ii)

CR = CQ … (iii)

DR = DS … (iv)

Adding equations (i), (ii), (iii) and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS

⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Hence proved.

**Question 9.**

In figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

**Solution:**

Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X’Y’ at B To Prove: ∠AOB = 90° with point of contact C intersects XY at A and X’Y’ at B

To Prove: ∠AOB = 90°

Construction: Join OA, OB and OC

Proof: In ∆AOP and ∆AOC

**Question 10.**

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. **Solution:**

PA and PB are two tangents, A and B are the points of contact of the tangents.

**Question 11.**

Prove that the parallelogram circumscribing a circle is a rhombus. **Solution:**

Parallelogram ABCD circumscribing a circle with centre O.

OP ⊥ AB and OS ⊥ AD

**Question 12.**

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 (see figure). Find the sides AB and AC.

**Solution:**

BD = 8 cm and DC = 6 cm

BE = BD = 8 cm

CD = CF = 6 cm

Let AE = AF = x cm

In ∆ABC, a = 6 + 8 = 14 cm

b = (x + 6) cm

c = (x + 8) cm

**Question 13.**

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

AB touches at P.

BC, CD and DA touch the circle at Q, R and S.

Construction: Join OA, OB, OC, OD and OP, OQ, OR, OS.

Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.