Class 10 - Chapter 10: Circles

Exercise 10.2

Q1. 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. Find the radius of the circle.

Given:

OQ = 25 cm

QP = 24 cm (tangent)

OP = radius of the circle

Since radius drawn to the point of contact is perpendicular to the tangent,

∠OPQ = 90°

Applying Pythagoras Theorem in right △OPQ:

OQ² = OP² + PQ²

25² = OP² + 24²

625 = OP² + 576

OP² = 49

OP = 7 cm

Therefore, the radius of the circle is 7 cm.

Correct Option: (A) 7 cm

Q2. In 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:

Given:

∠POQ = 110°

TP and TQ are tangents.

Radius is perpendicular to the tangent at the point of contact.

Therefore,

∠OPT = 90°

∠OQT = 90°

In quadrilateral OPTQ:

∠OPT + ∠OQT + ∠POQ + ∠PTQ = 360°

90° + 90° + 110° + ∠PTQ = 360°

290° + ∠PTQ = 360°

∠PTQ = 70°

Therefore, ∠PTQ = 70°.

Correct Option: (B) 70°

Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to:

Given:

∠APB = 80°

OA ⊥ PA

OB ⊥ PB

In quadrilateral OAPB:

∠A + ∠B + ∠APB + ∠AOB = 360°

90° + 90° + 80° + ∠AOB = 360°

260° + ∠AOB = 360°

∠AOB = 100°

Now, OA = OB (radii)

PA = PB (tangents from the same external point)

Therefore, triangles OAP and OBP are congruent.

Hence,

OP bisects ∠AOB.

∠POA = 100°/2

= 50°

Therefore, ∠POA = 50°.

Correct Option: (A) 50°

Q4. Prove 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.

Let the tangents at A and B be lines l and m respectively.

We know that:

A tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore,

OA ⊥ l

OB ⊥ m

Since OA and OB lie on the same straight line AB,

both tangents l and m are perpendicular to the same line AB.

Lines perpendicular to the same line are parallel.

Therefore,

l ∥ m

Hence, the tangents drawn at the ends of a diameter are parallel.

Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Given:

A tangent touches the circle at point P.

Let a line be drawn through P perpendicular to the tangent.

We know that:

The radius drawn to the point of contact of a tangent is perpendicular to the tangent.

Therefore, OP is perpendicular to the tangent at P.

Since only one perpendicular can be drawn through a point to a given line,

the perpendicular drawn at P must coincide with OP.

As OP passes through the centre O,

the perpendicular at the point of contact passes through the centre.

Hence proved.

Q6. 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.

Given:

OA = 5 cm

AP = 4 cm (tangent)

OP = radius

Since radius is perpendicular to the tangent,

∠OPA = 90°

In right △OPA:

OA² = OP² + AP²

5² = r² + 4²

25 = r² + 16

r² = 9

r = 3 cm

Radius of the circle = 3 cm

Q7. 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.

Radius of larger circle = 5 cm

Radius of smaller circle = 3 cm

Since the chord of the larger circle touches the smaller circle,

the distance of the chord from the common centre O is 3 cm.

Let AB be the chord and M be its midpoint.

OM = 3 cm

OA = 5 cm

OM ⊥ AB

In right △OMA:

OA² = OM² + AM²

5² = 3² + AM²

25 = 9 + AM²

AM² = 16

AM = 4 cm

Therefore,

AB = 2 × AM

AB = 8 cm

Length of the chord = 8 cm

Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Let the circle touch:

• AB at P
• BC at Q
• CD at R
• DA at S

Tangents drawn from the same external point are equal.

Therefore,

AP = AS ..........(1)

BP = BQ ..........(2)

CQ = CR ..........(3)

DR = DS ..........(4)

Now,

AB = AP + PB

BC = BQ + QC

CD = CR + RD

AD = AS + SD

Adding AB and CD:

AB + CD

= (AP + PB) + (CR + RD)

= (AS + BQ) + (CQ + DS)

= (AS + SD) + (BQ + QC)

= AD + BC

Therefore,

AB + CD = AD + BC

Hence proved.

Q9. In 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 P and Q be the points of contact of tangents XY and X'Y' respectively.

Since tangents from an external point are equal,

AP = AC

BQ = BC

Also,

OP ⊥ AP

OC ⊥ AC

Therefore, triangles OPA and OCA are congruent.

Hence,

OA bisects ∠PAC.

Similarly,

OB bisects ∠QBC.

Since XY ∥ X'Y',

∠PAC + ∠QBC = 180°

Taking half of both sides:

∠OAB + ∠ABO = 90°

In △AOB,

∠AOB = 180° − 90°

∠AOB = 90°

Hence proved.

Q10. 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.

Given:

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

Join OA and OB.

We know:

OA ⊥ PA

OB ⊥ PB

Therefore,

∠OAP = 90°

∠OBP = 90°

In quadrilateral OAPB:

∠OAP + ∠OBP + ∠AOB + ∠APB = 360°

90° + 90° + ∠AOB + ∠APB = 360°

180° + ∠AOB + ∠APB = 360°

∠AOB + ∠APB = 180°

Therefore,

The angle between the two tangents is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Hence proved.

Q11. Prove that the parallelogram circumscribing a circle is a rhombus.

Given:

ABCD is a parallelogram which circumscribes a circle.

We know that a quadrilateral circumscribing a circle satisfies:

AB + CD = BC + AD

Since ABCD is a parallelogram,

AB = CD

BC = AD

Substituting these values:

AB + AB = BC + BC

2AB = 2BC

AB = BC

But in a parallelogram:

AB = CD and BC = AD

Therefore,

AB = BC = CD = AD

Hence all four sides are equal.

A parallelogram having all sides equal is a rhombus.

Therefore, the parallelogram circumscribing a circle is a rhombus.

Hence proved.

Q12. 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.

Given:

Radius of the circle = 4 cm

BD = 8 cm

DC = 6 cm

Therefore,

BC = BD + DC

BC = 8 + 6 = 14 cm

Let:

AF = AE = x cm

BF = BD = 8 cm

CE = CD = 6 cm

Then,

AB = AF + BF = x + 8

AC = AE + CE = x + 6

Semi-perimeter:

s = (AB + BC + AC)/2

= [(x + 8) + 14 + (x + 6)]/2

= (2x + 28)/2

= x + 14

Area of triangle:

= r × s

= 4(x + 14)

Using Heron's Formula:

Area = √[s(s-a)(s-b)(s-c)]

= √[(x+14)(x)(8)(6)]

= √[48x(x+14)]

Equating the two expressions:

4(x + 14) = √[48x(x + 14)]

Squaring both sides:

16(x + 14)² = 48x(x + 14)

(x + 14) = 3x

14 = 2x

x = 7

Therefore,

AB = x + 8 = 15 cm

AC = x + 6 = 13 cm

AB = 15 cm and AC = 13 cm

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Given:

ABCD is a quadrilateral circumscribing a circle with centre O.

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

Join OP, OQ, OR and OS.

Since radii are perpendicular to tangents at the points of contact,

∠OPA = ∠OPB = 90°

∠OQB = ∠OQC = 90°

∠ORC = ∠ORD = 90°

∠OSD = ∠OSA = 90°

In quadrilateral OPBQ:

∠POQ + ∠B = 180°

∠POQ = 180° − ∠B

Similarly,

∠QOR = 180° − ∠C

∠ROS = 180° − ∠D

∠SOP = 180° − ∠A

Now,

∠POR = ∠POQ + ∠QOR

= (180° − B) + (180° − C)

= 360° − (B + C)

Since in a quadrilateral,

A + B + C + D = 360°

Therefore,

B + C = 360° − (A + D)

Hence,

∠POR = A + D

Similarly,

∠QOS = B + C

Therefore,

∠POR + ∠QOS

= (A + D) + (B + C)

= 360°

Thus the angles subtended at the centre by opposite sides are supplementary.

Hence proved.