Chapter 10: Circles - Solutions

Exercise 10.1

Exercise 10.2

Key Theorems

Exercise 10.1

1. How many tangents can a circle have?

A circle can have infinitely many tangents.

2. Fill in the blanks:

(i) A tangent to a circle intersects it in one point(s).

(ii) A line intersecting a circle in two points is called a secant.

(iii) A circle can have two parallel tangents at the most.

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

3. A 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:

Using the Pythagorean theorem:

OP = 5 cm (radius)

OQ = 12 cm

PQ² = OQ² - OP² = 12² - 5² = 144 - 25 = 119

PQ = √119 cm

Therefore, the correct option is (D) √119 cm.

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

To solve this:

Draw a circle with center O.
Draw a line AB outside the circle.
Draw a line CD parallel to AB that touches the circle at exactly one point (tangent).
Draw another line EF parallel to AB that cuts the circle at two points (secant).

Fig: A circle with a tangent and a secant parallel to a given line

Exercise 10.2

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:

Let O be the center of the circle, and let the tangent touch the circle at point P.

Given: QP = 24 cm (length of tangent), OQ = 25 cm

Since the tangent is perpendicular to the radius at the point of contact, ∠OPQ = 90°.

Using Pythagoras theorem in right triangle OPQ:

OP² + QP² = OQ²

OP² + 24² = 25²

OP² + 576 = 625

OP² = 625 - 576 = 49

OP = √49 = 7 cm

Therefore, the radius of the circle is 7 cm.

The correct option is (A) 7 cm.

2. 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:

In quadrilateral OPTQ:

∠OPT = 90° (tangent perpendicular to radius)

∠OQT = 90° (tangent perpendicular to radius)

∠POQ = 110° (given)

Sum of angles in a quadrilateral = 360°

So, ∠PTQ = 360° - (90° + 90° + 110°) = 360° - 290° = 70°

Therefore, ∠PTQ = 70°

The correct option is (B) 70°.

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:

Given: ∠APB = 80°

Since PA and PB are tangents from an external point P, OP bisects ∠APB.

So, ∠APO = ∠BPO = 80°/2 = 40°

In right triangle OAP (since tangent is perpendicular to radius):

∠OAP = 90°

So, ∠POA = 180° - (90° + 40°) = 180° - 130° = 50°

Therefore, ∠POA = 50°

The correct option is (A) 50°.

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

Let AB be a diameter of a circle with center O.

Let PA and QB be tangents at points A and B respectively.

Since tangent is perpendicular to radius at point of contact:

PA ⊥ OA and QB ⊥ OB

But OA and OB are parts of the same line AB (diameter).

So, PA ⊥ AB and QB ⊥ AB

If two lines are perpendicular to the same line, they are parallel to each other.

Therefore, PA ∥ QB

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

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

Let a tangent PQ touch the circle at point P.

We know that the radius through the point of contact is perpendicular to the tangent.

So, OP ⊥ PQ, where O is the center of the circle.

But there can be only one perpendicular to a line at a given point on it.

Therefore, the perpendicular to the tangent at the point of contact must pass through the center of the circle.

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.

Let O be the center of the circle, and let the tangent from A touch the circle at P.

Given: OA = 5 cm, AP = 4 cm

Since tangent is perpendicular to radius, ∠OPA = 90°

Using Pythagoras theorem in right triangle OPA:

OP² + AP² = OA²

OP² + 4² = 5²

OP² + 16 = 25

OP² = 25 - 16 = 9

OP = √9 = 3 cm

Therefore, the radius of the circle is 3 cm.

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.

Let O be the common center of both circles.

Let AB be the chord of the larger circle that touches the smaller circle at P.

Then OP is the radius of the smaller circle, so OP = 3 cm.

OA is the radius of the larger circle, so OA = 5 cm.

Since AB is tangent to the smaller circle at P, OP ⊥ AB.

In right triangle OPA:

AP² = OA² - OP² = 5² - 3² = 25 - 9 = 16

AP = √16 = 4 cm

Since OP is perpendicular to AB from the center, it bisects AB.

So, AB = 2 × AP = 2 × 4 = 8 cm

Therefore, the length of the chord is 8 cm.

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

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

From a point outside a circle, the lengths of tangents to the circle are equal.

So:

AP = AS ...(1) (tangents from A)

BP = BQ ...(2) (tangents from B)

CR = CQ ...(3) (tangents from C)

DR = DS ...(4) (tangents from D)

Adding equations (1), (2), (3), and (4):

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

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

AB + CD = AD + BC

Hence proved.

9. 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°.

Join OC.

In triangles OPA and OCA:

OP = OC (radii)

OA = OA (common)

∠OPA = ∠OCA = 90° (tangent perpendicular to radius)

So, ΔOPA ≅ ΔOCA (RHS congruence)

Therefore, ∠POA = ∠COA ...(1)

Similarly, in triangles OQB and OCB:

OQ = OC (radii)

OB = OB (common)

∠OQB = ∠OCB = 90° (tangent perpendicular to radius)

So, ΔOQB ≅ ΔOCB (RHS congruence)

Therefore, ∠QOB = ∠COB ...(2)

Now, POQ is a straight line (since XY and X'Y' are parallel tangents).

So, ∠POA + ∠AOC + ∠COB + ∠BOQ = 180°

From (1) and (2):

∠AOC + ∠AOC + ∠BOC + ∠BOC = 180°

2(∠AOC + ∠BOC) = 180°

∠AOC + ∠BOC = 90°

But ∠AOC + ∠BOC = ∠AOB

Therefore, ∠AOB = 90°

Hence proved.

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.

Let PA and PB be two tangents from an external point P to a circle with center O.

Let A and B be the points of contact.

We need to prove that ∠APB + ∠AOB = 180°

In quadrilateral OAPB:

∠OAP = 90° (tangent perpendicular to radius)

∠OBP = 90° (tangent perpendicular to radius)

So, ∠OAP + ∠OBP = 180°

Sum of angles in a quadrilateral = 360°

So, ∠AOB + ∠APB + 180° = 360°

Therefore, ∠AOB + ∠APB = 180°

Hence proved.

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

Let ABCD be a parallelogram circumscribing a circle.

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

Since lengths of tangents from an external point to a circle are equal:

AP = AS ...(1)

BP = BQ ...(2)

CR = CQ ...(3)

DR = DS ...(4)

Adding equations (1), (2), (3), and (4):

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

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

AB + CD = AD + BC

But in a parallelogram, AB = CD and AD = BC

So, 2AB = 2BC ⇒ AB = BC

Therefore, all sides of the parallelogram are equal.

Hence, ABCD is a rhombus.

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

Let the circle touch sides AB, BC, and CA at points E, D, and F respectively.

Given: BD = 8 cm, DC = 6 cm, radius = 4 cm

Since lengths of tangents from an external point to a circle are equal:

AE = AF = x (say)

BE = BD = 8 cm

CF = CD = 6 cm

Now, sides of triangle ABC:

AB = AE + BE = x + 8

BC = BD + DC = 8 + 6 = 14

AC = AF + CF = x + 6

Semi-perimeter, s = (AB + BC + AC)/2 = (x + 8 + 14 + x + 6)/2 = (2x + 28)/2 = x + 14

Area of triangle ABC = r × s = 4 × (x + 14)

Also, by Heron's formula:

Area = √[s(s - AB)(s - BC)(s - AC)]

= √[(x + 14)(x + 14 - x - 8)(x + 14 - 14)(x + 14 - x - 6)]

= √[(x + 14)(6)(x)(8)]

= √[48x(x + 14)]

Equating both expressions for area:

4(x + 14) = √[48x(x + 14)]

Squaring both sides:

16(x + 14)² = 48x(x + 14)

Dividing both sides by 16(x + 14) (since x + 14 ≠ 0):

x + 14 = 3x

14 = 2x

x = 7

Therefore, AB = x + 8 = 7 + 8 = 15 cm

AC = x + 6 = 7 + 6 = 13 cm

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

Let ABCD be a quadrilateral circumscribing a circle with center O.

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

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

Since tangent is perpendicular to radius at point of contact:

∠OPA = ∠OSA = 90°

∠OQB = ∠OPB = 90°

∠ORC = ∠OQC = 90°

∠OSD = ∠ORD = 90°

In quadrilateral OPAQ:

∠OPA + ∠PAQ + ∠AQO + ∠QOP = 360°

90° + ∠PAQ + 90° + ∠QOP = 360°

∠PAQ + ∠QOP = 180° ...(1)

Similarly, in quadrilateral OQCR:

∠QCR + ∠ROQ = 180° ...(2)

In quadrilateral ORDS:

∠RDS + ∠SOR = 180° ...(3)

In quadrilateral OSPA:

∠SPA + ∠AOS = 180° ...(4)

Adding equations (1), (2), (3), and (4):

(∠PAQ + ∠QCR + ∠RDS + ∠SPA) + (∠QOP + ∠ROQ + ∠SOR + ∠AOS) = 720°

But ∠PAQ + ∠QCR + ∠RDS + ∠SPA = 360° (sum of angles of quadrilateral ABCD)

And ∠QOP + ∠ROQ + ∠SOR + ∠AOS = 360° (angles around point O)

So, 360° + 360° = 720°, which is true.

Now, ∠AOB = ∠QOP, ∠BOC = ∠ROQ, ∠COD = ∠SOR, ∠DOA = ∠AOS

From (1): ∠PAQ + ∠AOB = 180°

But ∠PAQ is the angle between sides AB and AD, which is ∠BAD

So, ∠BAD + ∠AOB = 180°

Similarly, from (2): ∠QCR + ∠BOC = 180°

But ∠QCR is the angle between sides BC and CD, which is ∠BCD

So, ∠BCD + ∠BOC = 180°

Hence, opposite sides of the quadrilateral subtend supplementary angles at the center.

Key Theorems

Theorem 10.1

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

Proof of Theorem 10.1

Given: A circle with center O and a tangent XY to the circle at point P.

To prove: OP is perpendicular to XY.

Take a point Q on XY other than P and join OQ.

The point Q must lie outside the circle (otherwise XY would be a secant).

Therefore, OQ > OP (radius).

Since this happens for every point on XY except P, OP is the shortest distance from O to XY.

Therefore, OP is perpendicular to XY.

Theorem 10.2

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

Proof of Theorem 10.2

Given: A circle with center O, a point P outside the circle, and two tangents PQ and PR from P to the circle.

To prove: PQ = PR.

Join OQ, OR, and OP.

In right triangles OQP and ORP:

OQ = OR (radii)

OP = OP (common)

∠OQP = ∠ORP = 90° (tangent perpendicular to radius)

So, ΔOQP ≅ ΔORP (RHS congruence)

Therefore, PQ = PR (CPCT)