Lesson: Circles

Exercise 10.1 (4)

Question: 1

How many tangents can a circle have?

Solution

A circle can have infinite number of tangents.

Question: 2

Fill in the blanks:

(i) A tangent to a circle intersects it in _____ point

(s).

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

a ______.

(iii) A circle can have ______ parallel tangents at the

most.

(iv) The common point of a tangent to a circle and

the circle is called ______.

Solution

(i) One

(ii) Secant

(iii) Two

(iv) Point of contact

Question: 3

A tangent PQ at a point of a circle of radius 5 cm

P

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. cm

119

Solution

(d)

Draw a circle with radius 5 cm as shown in the figure

below.

Fig. Exc_10.1_3

The line from the centre to the tangent is

perpendicular to the tangent.

Thus, .

OP PQ

In , by Pythagoras theorem,

OPQ

2 2 2

OQ OP PQ

2 2

2

12 5 PQ

2

PQ 144 25

2

PQ 119

PQ 119

Thus, the length of PQ is .

119

Question: 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.

Solution

Step 1: Draw a line .

l

Step 2: Set the compass to the desired radius of the

circle.

Step 3: Put the needle point of the compass at the

circle’s centre.

Step 4: Turn the compass through and draw

360

the circle above the line.

Step 5: Draw a line parallel to the given line such

that it intersects the circle at two points. It is a

secant of the circle.

Step 6: Draw a line parallel to the given line such

that it touches the circle at one point only. It is a

tangent of the circle.

Fig. Exc_10.1_4

Exercise 10.2 (13)

In Q.1 to 3, choose the correct option and give

justification.

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

b. 12 cm

c. 15 cm

d. 24.5 cm

Solution

(a)

Let the centre of the circle be O.

Given: OQ 25 cm and PQ 24 cm

In a circle, radius is perpendicular to the tangent.

Fig. Exc_10.2_1

Thus, OP PQ

By Pythagoras theorem in OPQ,

2 2 2

OP PQ OQ

2 2

2

OP 24 25

2

OP 625 576

2

OP 49

OP 7

Hence, the radius of the circle is 7 cm.

Question: 2

If TP and TQ are the two tangents to a circle with

centre O so that , then is equal

POQ 110

PTQ

to

a.

60

b.

70

c.

80

d.

90

Fig. Exc_10.2_2

Solution

(b)

The tangents to the circle are TP and TQ.

Radius of a circle is perpendicular to the tangents.

Therefore, OP TP and OQ TQ

OPT 90

OQT 90

The sum of interior angles in quadrilateral POQT is

360

OPT POQ OQT PTQ 360

90 110 90 PTQ 360

PTQ 70

Hence,

PTQ 70

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 is equal to

POA

a.

50

b.

60

c.

70

d.

80

Solution

(a)

Given: and are tangents.

PA

PB

Fig. Exc_10.2_3

Thus, the radius and are perpendiculars to

OB

OA

the tangents.

Thus, and .

OA PA

OB PB

and

OBP 90

OAP 90

In quadrilateral ,

AOBP

Sum of all interior angles

360

So,

OAP APB PBO BOA 360

90 80 90 BOA 360

BOA 360 260

BOA 100

In ,

OPBand OPA

[Tangents from a point]

AP BP

[Radii of the circle]

OA OB

[Common side]

OP OP

Thus,

[By SSS congruence criterion]

OPB OPA

Thus, [By CPCT]

POB POA

Also,

1

POA AOB

2

100

= 50

2

Hence,

POA 50

Question: 4

Prove that the tangents drawn at the ends of a

diameter of a circle are parallel.

Solution

Fig. Exc_10.2_4

Let the diameter of the circle be .

XY

At point , two tangents are

Xand Y

PQ andRS

drawn.

So, and .

OX RS

OY PQ

OXR 90 OXS

OYP 90 OYQ

Observe that,

[Alternate interior angles]

OXR OYQ

[Alternate interior angles]

OXS OYP

Alternate interior angles are equal. So, lines

are parallel.

PQ andRS

Question: 5

Prove that the perpendicular at the point of contact

of the tangent to a circle passes through the centre.

Solution

Let be the centre of circle and be a tangent

O

XY

that touches the circle at point .

P

We have to prove that the line perpendicular to

XY

at passes through the point .

P

O

We will prove this by a contradiction.

Let the perpendicular to at do not pass

XY

P

through the centre .

O

Let it is passes through another point .

Q

Join and .

OP

QP

Fig. Exc_10.2_5

Perpendicular to at passes through .

XY

P

Q

Thus,

QPY 90 ...... 1

According to the figure, the centre of the circle is

O

and the point of contact is .

P

So,

OPY 90 ...... 2

By equation (1) and (2),

QPY OPY ...... 3

From the figure, observe that

QPY OPY ...... 4

Thus, is not possible.

QPY OPY

It is possible only when the lines and QP

OP

coincide.

Thus, perpendicular to XY through P, passes through

the centre O.

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

Let O as a centre of a circle.

Fig. Exc_10.2_6

From point A, AB tangent is drawn on the circle.

Given: and

OA 5cm

AB 4 cm

In ,

ABO

.

OB AB

Apply, Pythagoras theorem in ,

ABO

2 2 2

AB BO OA

2 2 2

4 BO 5

2

16 BO 25

2

BO 9

BO 3

Thus, radius of the circle is cm.

3

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

Let the point O be the centre of two concentric

circles.

Fig. Exc_10.2_7

Also, chord of the larger circle is AB and touches the

smaller circle at point P.

Thus, AB becomes the tangent to the smaller circle.

[ is radius of smaller circle]

OP AB

OP

Apply Pythagoras theorem in ,

OAP

2 2 2

OP AP OA

2

2 2

3 AP 5

2

AP 25 9

2

AP 16

AP 4

In ,

OAB

OP AB

AP PB

Perpendicular from thecentre

of thecirclebisectsthechord

Now,

AB 2AP

2 4

8

Thus, length of the chord is cm.

8

Question: 8

A quadrilateral ABCD is drawn to circumscribe a

circle. Prove that .

AB CD AD BC

Fig. Exc_10.2_8

Solution

Let ABCD be a quadrilateral.

Observe that,

DR DS

Tangentsfrom point

D on thecircle

...... 1

CR CQ

Tangentsfrom point

C on thecircle

...... 2

BP BQ

Tangents from point

Bon thecircle

...... 3

AP AS

Tangents from point

A on thecircle

...... 4

Add all these equations.

DR CR BP AP DS CQ BQ AS

DR CR BP AP DS AS CQ BQ

CD AB AD BC

Hence proved.

Question: 9

XY and are two parallel tangents to a circle with

XY

centre O and another tangent AB with point of

contact C intersecting XY at A and at B. Prove

XY

that .

AOB 90

Fig. Exc_10.2_9(Ques.)

Solution

Join point O to C.

Fig. Exc_10.2_9(Sol.)

In and ,

OPA

OCA

[radii of same circle]

OP OC

AP AC

Tangents from point

A on thecircle

[Common side]

AO AO

[By SSS congruence criterion]

OPA OCA

POA COA ...... 1

Similarly,

OQB OCB

QOB COB ...... 2

The diameter POQ of the circle is a straight line.

Thus,

POA COA COB QOB 180

From equations (1) and (2),

POA COA COB QOB 180

COA COA COB COB 180

2 COA 2 COB 180

COA COB 90

AOB 90

Hence proved.

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

Let O be the centre of the circle.

Fig. Exc_10.2_10

Let P be the external point and PX and PY are the

two tangents.

XY is the line segment.

Observe that .

OX PX

Thus,

OXP 90

Similarly, .

OY PY

.

OYP 90

In quadrilateral ,

OXPY

OXP XPY PYO YOX 360

90 XPY 90 YOX 360

XPY YOX 360 180

XPY YOX 180

Hence, 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.

Question: 11

Prove that the parallelogram circumscribing a circle

is a rhombus.

Solution

Given: Let ABCD be a parallelogram.

AB CD ...... 1

BC AD ...... 2

Fig. Exc_10.2_11

Observe that,

AW AZ

Tangents from point

A on thecircle

BW BX

Tangents from point

Bon thecircle

CY CX

Tangents from point

C on thecircle

DY DZ

Tangents from point

D on thecircle

Add the above equations.

DY CY BW AW DZ CX BX AZ

DY CY BW AW DZ AZ CX BX

DC AB AD BC

From equations (1) and (2),

DC AB AD BC

2AB 2BC

AB BC ...... 3

Compare equations (1), (2) and (3),

AB BC CD DA

Hence, ABCD is a rhombus.

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. Find the sides

AB and AC.

Fig. Exc_10.2_12(Ques.)

Solution

Sides AB and AC touch the circle at point E and F

respectively.

Let the length of AF be .

x

In ,

ABC

CF CD 6cm

Tangents from point

C on thecircle

BE BD 8cm

Tangents from point

Bon thecircle

AF AE x

Tangents from point

A on thecircle

Fig. Exc_10.2_12(Sol.)

Now,

AB AE EB x 8

BC BD DC

8 6

14

CA CF FA

6 x

By the Hero’s formula,

2s AB BC CA

2s x 8 14 6 x

2s 28 2x

s 14 x

ABC s s a sArea o b sf c

14 x 14 x 14 14 x 6 x 14 x 8 x

14 x 14 x 14 14 x 6 x 14 x 8 x

14 x x 8 6

2

4 3 14x x

1

OBCAre Oa

2

f Do BC

1

4 14

2

28

1

OCAAre Oa

2

f Fo AC

1

4 6 x

2

12 2x

1

OABAr OE AB

2

ea of

1

4 8

2

x

16 2x

Now, find the value of .

x

Area of ABC Area of OBC

Area of OCA

Area of OAB

2

4 3 14 28 12 2 16 2x x x x

2

4 3 14 56 4x x x

2

3 14 14x x x

2

2

3 14 14x x x

2 2

42 3 196 28x x x x

2

2 14 196 0x x

2

7 98 0x x

Solve the quadratic equation.

2

7 98 0x x

2

14 7 98 0x x x

14 7 14 0x x x

14 7 0x x

14,7x

Length is always positive so, is rejected.

14

Thus,

7x

Hence, cm

AB 8 7 8 15x

cm.

CA 6 6 7 13x

Question: 13

Prove that the opposite sides of a quadrilateral

circumscribing a circle subtend supplementary

angles at the centre of the circle.

Solution

Let O be the centre of circle and quadrilateral ABCD

touches the circle at points .

W X Y Z, , ,

Fig. Exc_10.2_13

Join the vertices of the quadrilateral with the centre

of the circle.

In and ,

OAW

OAZ

AW AZ

Tangents from point

A on thecircle

[radii of same circle]

OW OZ

[common side]

OA OA

[SSS congruence criterion]

OAW OAZ

Thus,

WOA AOZ

1 8

Similarly,

2 3

4 5

6 7

Now,

1 2 3 4 5 6 7 8 360

1 8 2 3 4 5 6 7 360

2 1 2 2 2 5 2 6 360

2 1 2 2 5 6 360

1 2 5 6 180

AOB COD 180

AOB COD 180

Similarly,

BOC DOA 180

Thus, the opposite sides of a quadrilateral

circumscribing a circle subtend supplementary

angles at the centre of the circle.