Class 10 Mathematics · Chapter 10

10. Circles

Step-by-step solutions to Exercise 10.1 & 10.2. NCERT Mathematics Textbook for Class X (CBSE).

Exercise 10.1

1How many tangents can a circle have?

A tangent can be drawn at every point of a circle. A circle has infinitely many points.

A circle can have infinitely many tangents.

2Fill in the blanks:

(i) A tangent to a circle intersects it in _{_}_{_}_{_} point (s).

one

(ii) A line intersecting a circle in two points is called a _{_}_{_}_{_} .

secant

(iii) A circle can have _{_}_{_}_{_} parallel tangents at the most.

two

(iv) The common point of a tangent to a circle and the circle is called _{_}_{_}_{_} .

point of contact

3A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is : (A) 12 cm (B) 13 cm (C) 8.5 cm (D) √119 cm.

The tangent at P is perpendicular to the radius OP. So, ∠OPQ = 90°.

PQ² = OQ² – OP² = 144 – 25 = 119 ⇒ PQ = √119 cm

(D) √119 cm

4Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Steps of construction:

Draw a circle with centre O and a line l outside the circle.
Draw OP perpendicular to l, meeting the circle at P.
At P, draw a line m perpendicular to OP. Line m touches the circle only at P. So, m is a tangent and m || l.
Draw another line n parallel to l, passing between O and P. It cuts the circle at two points A and B. So, n is a secant.

l is the given line. m is the tangent at P and n is the secant cutting the circle at A and B. Both are parallel to l.

Exercise 10.2

1From 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 (B) 12 cm (C) 15 cm (D) 24.5 cm

Let P be the point of contact. The radius OP is perpendicular to the tangent PQ.

OP² = OQ² – PQ² = 625 – 576 = 49 ⇒ OP = 7 cm

(A) 7 cm

2In Fig. 10.11, 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°

∠OPT = ∠OQT = 90° (radius ⊥ tangent)

In quadrilateral OPTQ:

∠PTQ = 360° – 90° – 90° – 110° = 70°

(B) 70°

3If 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°

∠OAP = ∠OBP = 90° ; ∠AOB = 360° – 90° – 90° – 80° = 100°

OP bisects ∠AOB (ΔOAP ≅ ΔOBP by RHS).

∠POA = 12 × 100° = 50°

(A) 50°

4Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Given: AB is a diameter of a circle with centre O. PQ and RS are tangents at A and B respectively.

To prove: PQ || RS.

Proof: The radius is perpendicular to the tangent at the point of contact.

OA ⊥ PQ ⇒ ∠OAP = 90° ; OB ⊥ RS ⇒ ∠OBS = 90°

So, ∠BAP = ∠ABS = 90°

These are alternate angles made by the transversal AB with PQ and RS.

Hence, PQ || RS

5Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Given: AB is a tangent to a circle with centre O at the point P. PX is perpendicular to AB at P.

To prove: PX passes through O.

Proof: By Theorem 10.1, the radius OP is perpendicular to the tangent AB at P. So, OP ⊥ AB.

Also, PX ⊥ AB at P (given).

At a point on a line, only one perpendicular can be drawn to it. So, PX and PO are the same line.

Hence, the perpendicular PX passes through the centre O.

6The 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.

Let P be the point of contact. Then OP ⊥ AP.

OP² = OA² – AP² = 25 – 16 = 9 ⇒ OP = 3 cm

Radius of the circle 3 cm

7Two 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.

AB is a chord of the larger circle touching the smaller circle at P.

AB touches the smaller circle at P. So, OP ⊥ AB.

The perpendicular from the centre to a chord bisects the chord. So, AP = PB.

AP² = OA² – OP² = 25 – 9 = 16 ⇒ AP = 4 cm

AB = 2 × AP = 8 cm

Length of the chord 8 cm

8A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

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

The lengths of tangents drawn from an external point to a circle are equal.

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

Adding these:

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

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

Hence, AB + CD = AD + BC

9In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle 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°.

Let XY touch the circle at P and X′Y′ touch it at Q. Join OC.

In ΔOPA and ΔOCA:

OP = OC (radii) ; AP = AC (tangents from A) ; OA = OA (common)

So, ΔOPA ≅ ΔOCA (SSS). Hence, ∠POA = ∠COA ... (1)

Similarly, ΔOQB ≅ ΔOCB. Hence, ∠QOB = ∠COB ... (2)

XY || X′Y′, so POQ is a diameter (a straight line).

∠POA + ∠COA + ∠COB + ∠QOB = 180°

Using (1) and (2): 2∠COA + 2∠COB = 180° ⇒ ∠COA + ∠COB = 90°

So, ∠AOB = 90°

10Prove 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.

Given: PA and PB are tangents from an external point P to a circle with centre O.

To prove: ∠APB + ∠AOB = 180°.

Proof: The radius is perpendicular to the tangent at the point of contact.

∠OAP = 90° and ∠OBP = 90°

In quadrilateral OAPB, the sum of all angles is 360°.

∠APB + ∠AOB + 90° + 90° = 360° ⇒ ∠APB + ∠AOB = 180°

Hence, ∠APB and ∠AOB are supplementary.

11Prove that the parallelogram circumscribing a circle is a rhombus.

Let ABCD be a parallelogram circumscribing a circle.

As proved in Q.8, for a quadrilateral circumscribing a circle:

AB + CD = AD + BC

In a parallelogram, opposite sides are equal: AB = CD and AD = BC.

2AB = 2AD ⇒ AB = AD

So, AB = BC = CD = DA.

Hence, ABCD is a rhombus.

12A 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 Fig. 10.14). Find the sides AB and AC.

The incircle touches BC at D, CA at E and AB at F. BD = 8 cm, DC = 6 cm, radius = 4 cm.

Tangents from an external point are equal.

BF = BD = 8 cm ; CE = CD = 6 cm ; let AF = AE = x cm

AB = x + 8 ; AC = x + 6 ; BC = 14

Semi-perimeter s = (x + 8) + (x + 6) + 142 = x + 14

By Heron’s formula:

Area of ΔABC = √s(s – a)(s – b)(s – c) = √(x + 14)(x)(6)(8) = √48x(x + 14)

Also, area of ΔABC = area(OBC) + area(OCA) + area(OAB) = 12 × 4 × (perimeter) = 4(x + 14)

√48x(x + 14) = 4(x + 14) ⇒ 48x(x + 14) = 16(x + 14)² ⇒ 3x = x + 14 ⇒ x = 7

AB = 15 cm and AC = 13 cm

13Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Let ABCD circumscribe a circle with centre O, touching AB, BC, CD and DA at P, Q, R and S.

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

In ΔOAP and ΔOAS: OP = OS, AP = AS, OA = OA. So, ΔOAP ≅ ΔOAS (SSS) and ∠AOP = ∠AOS.

Similarly, ∠BOP = ∠BOQ, ∠COQ = ∠COR and ∠DOR = ∠DOS.

The angles round the point O add up to 360°.

2(∠AOP + ∠BOP + ∠COR + ∠DOR) = 360°

(∠AOP + ∠BOP) + (∠COR + ∠DOR) = 180°

∠AOB + ∠COD = 180°

Similarly, ∠BOC + ∠AOD = 180°.

Hence, opposite sides subtend supplementary angles at the centre.

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