Lesson: Circles

Exercise 10.1 (10 Multiple Choice Questions and

Answers)

Question: 1

AD is a diameter of a circle and

AB

is a chord. If

34 cm, 30 cmAD AB

, the distance of

AB

from

the centre of the circle is:

(a) 17 cm

(b) 15 cm

(c) 4 cm

(d) 8 cm

Solution:

d

Given,

34 cm, 30 cm,AD AB

Let us draw the figure.

Now,

diameter

cm

34

17

2 2 2

AD

OA

and

1

2

AM AB

(

Perpendicular from the center of circle to the

chord bisects it)

Now, in right

AMO

,

(

2 2 2

) ( ) ( )OM OA AM

(By Pythagoras theorem)

cm

2 2

(17) (15)

289 225

64

64

8

OM

Hence, the distance of

AB

from the center of circle is

8 cm.

Question: 2

In Fig. 10.3, if

5 cm, 8 cmOA AB

and

OD

is

perpendicular to

AB

, then CD is equal to:

a cm

cm

c cm

d cm

b

2

3

4

5

Solution:

a

Given,

cm cm5 , 8OA AB

and

OD AB

,

Let us draw a figure,

OD AB

1

2

AC AB

(

Perpendicular from the centre of

circle to the chord bisects it)

cm

1

8

2

4

AC

AC

Now in right

ACO

,

(

2 2 2

) ( ) ( )OC OA AC

(By Pythagoras theorem)

cm

2 2

(5) (4)

25 16

9

9

3

OC

Then,

cm

5 3

2

CD OD OC

Question: 3

If

12 cm, 16 cmAB BC

and

AB

is perpendicular to

BC

, then the radius of the circle passing through the

points

A

,

B

and

C

is:

(a) 6 cm

(b) 8 cm

(c) 10 cm

(d) 12 cm

Solution:

c

Given,

12 cm, 16 cmAB BC

and

AB BC

,

AB BC

AC

is diameter

Now, in right

CBA

,

2

2 2

( ) ( )AC AB BC

(By Pythagoras theorem)

2 2

(12) (16)

144 256

=400

400AC

= 20 cm

Then, radius

2

AC

cm

20

10

2

Hence, required radius is 10 cm.

Question: 4

In Fig.10.4, if

20ABC

, then

AOC

is equal to:

(a) 20°

(b) 40°

(c) 60°

(d) 10°

Solution:

b

Given,

20ABC

,

Now,

2AOC ABC

(

By inscribed angle

1

2

central

angle)

2

2 20

40

AOC ABC

Question: 5

In Fig.10.5, if

AOB

is a diameter of the circle

and

AC BC

, then

CAB

is equal to:

(a) 30°

(b) 60°

(c) 90°

(d) 45°

Solution:

d

Given,

AC BC

and

AOB

is a diameter,

1 90

(Angle in semicircle)

In right

,ACB AC BC

(Given)

2 3

(By triangle isosceles property)

Now,

1 2 3 180

(By Angle sum property of a

triangle)

[ ]90 3 3 180 1 90 , 2 3

2 3 180 90

2 3 90

45CAB

Question: 6

In Fig. 10.6, if

40OAB

, then

ACB

is equal to:

(a) 50°

(b) 40°

(c) 60°

(d) 70°

Fig. 10.6

Solution:

a

Given,

40OAB

,

In

,AOB OA OB

(

Both are radius of the given

circle)

1 40OAB

(By triangle isosceles

property)

Now,

1 2 180OAB

(By Angle sum property of

)

40 2 40 180

2 180 80

2 100

Then,

1

2

ACB AOB

(

By inscribed angle

1

2

central

angle)

1

100

2

50

Question: 7

In Fig. 10.7, if

60 , 50 DAB ABD

, then

ACB

is equal to:

(a) 60°

(b) 50°

(c) 70°

(d) 80°

Solution:

c

Given,

60DAB

,

50ABD

In

ADB

180ADB DAB DBA

(By Angle sum

property of a

triangle

)

60 50 180

180 110

70

ADB

ADB

ADB

Therefore,

70ACB ADB

(As angles on the

same segment are equal)

Question: 8

ABCD is a cyclic quadrilateral such that AB is a

diameter of the circle circumscribing it and

140 ADC

, then

BAC

is equal to:

(a) 80°

(b) 50°

(c) 40°

(d) 30°

Solution:

b

Given,

140 , ADC

4 90

(Angle in semicircle)

3 180 ADC

(

Opposite angles of cyclic

quadrilateral)

140 3 180

3 180 140

3 40

Now,

In

,ACB

4 3 2 180

90 40 2 180

2 180 130

2 50

Hence,

2 50BAC

Question: 9

In Fig. 10.8,

BC

is a diameter of a circle and

60BAO

. Then

ADC

is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 120°

Solution:

c

Given,

60 BAO

and

BC

is a diameter of a circle.

In

,BAO OA OB

(

Both are radius of the given

circle)

1 60OAB

(By triangle isosceles

property)

2 1 OAB

(By exterior angle property of ∆)

60 60

120

Now,

2 2 3

(

The angle at the centre is twice the

angle at the circumference)

120 2 3

3 60

Therefore,

60ADC

Question 10

In Fig. 10.9,

90AOB

and

30ABC

, then

CAO

is equal to:

Fig 10.9

(a) 30°

(b) 45°

(c) 90°

(d) 60°

Solution:

d

Given,

90AOB

and

30ABC

In the above figure,

2AOB ACB

(

The angle at the centre is twice

the angle at the circumference)

90 2

45

ACB

ACB

In

,ACB

180ACB CAB ABC

(By Angle sum

property of

)

45

30 180

180 75

105

CAB

CAB

Now, in

,AOB OA OB

(

Both are radius of the

given circle)

... )OAB OBA (i

(By triangle isosceles

property)

(By ( ))

2

180

90 2 180

180 90 90

90

45

2

AOB OAB OBA

OAB i

OAB

OAB

Now,

105 45

60

CAO CAB OAB

Exercise 10.2

Write True or False and justify your answer in each

of the following:

Question 1

Two chords

AB

and

CD

of a circle are each at

distances 4 cm from the centre. Then

AB CD

Solution:

True

In

and ,OEB OFD

cm4OE OF

(Given)

OB OD

(

Both are radius of the circle)

And

OEB OFD

(

Each 90°)

OEB OFD

(

By RHS congruence rule

)

BE DF

(By CPCT)

2 2BE DF

(By multiplying 2 on both the sides)

AB CD

(

The perpendicular from the centre of

a circle to a chord bisects the chord

)

Question 2

Two chords

AB

and

AC

of a circle with centre

O

are

on the opposite sides of

OA

. Then

OAB OAC

Solution:

False

Given:

AB

and

AC

are two chords of a circle.

Joining

OB

and

OC

,

In

and ,AOB AOC

OC OB

(

Both are radius of the circle)

OA OA

(Common)

Here,

OAB OAC

if

AOB AOC

or if

AB AC

Question 3

The congruent circles with centres

O

and

O

intersect

at two points

A

and

B.

Then

AOB AO B

.

Solution:

True,

Because equal chords of congruent circles subtend

equal angles at the respective centres.

Question 4

Through three collinear points, a circle can be drawn.

Solution:

False,

A circle can pass through two collinear points but

not through three collinear points.

Question 5

A circle of radius 3 cm can be drawn through two

points

A

,

B

such that

cm6AB

.

Solution:

True

If AB is diameter, then the radius

diameter

2

cm

cm

2

6

2

3

AB

Question 6

If

AOB

is a diameter of a circle and

C

is a point on

the circle, then

2 2 2

AC BC AB

.

Solution:

True

90ACB

(

Angle in a semicircle is a right angle

)

In right

ACB

,

2 2 2

AC BC AB

(

By Pythagoras theorem

)

Question 7

ABCD

is a cyclic quadrilateral such that

and 90 , 70 , 95 105A B C D

Solution:

False

In a cyclic quadrilateral,

A C

180B D

But,

175 18070 105 B D

∴

ABCD

is not a cyclic quadrilateral.

Question 8

If

A

,

B

,

C

,

D

are four points such

that

and 30 60BAC BDC

, then

D

is the

centre of the circle through

A

,

B

and

C

.

Solution:

True

We know, that angle subtended by an arc at the

centre of the circle is double that of the angle

subtended by the same arc at the circumference.

So,

BDC

can be considered as angle at centre and

BAC

as the angle at circumference. Thus

D

is the

centre of the circle through

A

,

B

and

C

.

Question 9

If

A

,

B

,

C

and

D

are four points such that

45 BAC

and

45 BDC

, then

A

,

B

,

C

,

D

are

concyclic.

Solution:

True

As angles in the same segment are equal, therefore

A

,

B

,

C

,

D

are concyclic.

Question10

In Fig. 10.10, if

AOB

is a diameter and

120ADC

,

then

30CAB

.

Solution:

True

Let us join

AC

.

90 ACB

… (i)

(

Angle in the semi-circle is a right

angle

)

Since

ABCD

is cyclic,

180 ABC ADC

120 180 ABC

(

Opposite angle of cyclic

)

...( )

180 120

60 ii

ABC

ABC

Now, in

ABC

,

Using angle sum property we have,

(Using (i) and (ii))

180

60 90 180

180 150

30

CAB ABC ACB

CAB

CAB

CAB

Exercise 10.3

Question: 1

If arcs

AXB

and

CYD

of a circle are congruent, find

the ratio of

AB

and

CD

.

Solution:

Given,

arc arcAXB CYD

AB CD

(Equal arcs have equal corresponding

chords.)

1

1

AB

CD

: 1:1. AB CD

Question: 2

If the perpendicular bisector of a chord

AB

of a circle

PXAQB

Y intersects the circle at

P

and

Q

, prove that

arc

arcPXA PYB

.

Solution:

Let

AB

be a chord of the circle

PXAQB

Y and

PQ

be

the perpendicular bisector of the chord

AB

, which

intersect it at

M.

In

APM

and

BPM

,

PM PM

(Common side)

PMA PMB

(

PM AB

)

AM MB

(

PM

bisects

AB

)

Therefore,

APM BPM

(By SAS congruence

rule)

PA PB

(By CPCT)

arc arc PXA PYB

Question: 3

A

,

B

and

C

are three points on a circle. Prove that the

perpendicular bisectors of

AB

,

BC

and

CA

are

concurrent.

Solution:

Consider the perpendicular bisectors OD and OF of

lines AB and AC intersecting at the point O.

Now, draw perpendicular OE from the point O to the

line BC.

We need to show that

BE EC

In

AOD

and

BOD

,

AD BD

(OD is perpendicular bisector of AB)

ADO BDO

(Each

90

)

OD OD

(

Common)

Therefore,

AOD BOD

(By SAS congruence

rule)

OA OB

(By CPCT)

Similarly,

OFA OFC

(By SAS congruence rule)

OA OC

(By CPCT)

say( )OA OB OC r

Now, in

OEB

and

OEC

,

OB OC

(Proved above)

OE OE

(Common side)

And,

OEB OEC

(Each

90

)

Therefore,

OEB OEC

(By RHS congruence rule)

BE EC

(By CPCT)

Hence,

OE

is the perpendicular bisector of

BC

.

Hence, the perpendicular bisectors of

AB

,

BC

and

CA

are concurrent.

Question: 4

AB

and

AC

are two equal chords of a circle. Prove

that the bisector of the angle

BAC

passes through

the centre of the circle.

Solution:

Let us draw bisector of

BAC

and join

BC

and let

the intersecting point is

D

.

In

BAD

and

CAD

AB AC

(Given)

1 2

(AD is the bisector of

BAC

)

AD AD

(Common)

Therefore,

BAD CAD

(By SAS congruence

rule)

BD CD

(By CPCT)

And

BDA CDA

(By CPCT)

But,

180 BDA CDA

(Linear pair axiom)

180

2 180

90

90

BDA BDA

BDA

BDA

BDA CDA

Also,

BD CD

(As proved above)

So,

AD

is the perpendicular bisector of the

chord

.BC

Hence, bisector of

BAC

(

i.e.

AD

)

passes through

the centre

O

.

Question: 5

If a line segment joining mid-points of two chords of

a circle passes through the centre of the circle, prove

that the two chords are parallel.

Solution:

Let

P

and

Q

are the midpoints of two chords

AB

and

CD

respectively and the line segment joining

PQ

passes through centre.

Now, we need to prove that

AB CD

.

Since

P

is the midpoint of

AB

,

Therefore,

OP AB

.

(Since the line joining the centre of a circle to the

midpoint a chord is perpendicular to the chord)

90 ...( )APO i

Similarly,

OQ CD

...( )90OQD ii

From eq. (

i

) and (

ii

),

90 APO OQD

Thus, alternate interior angles are equal.

So,

AB CD

Question: 6

ABCD

is such a quadrilateral that

A

is the centre of

the circle passing through

B

,

C

and

D

. Prove that

1

2

CBD CDB BAD

Solution:

In a circle,

...( )1 2 2 i

(The angle subtended by an arc at

the centre is twice the angle subtended by it in the

remaining part of the circle.)

Similarly,

...( )3 2 4 ii

(The angle subtended by an arc at

the centre is twice the angle subtended by it in the

remaining part of the circle.)

On adding eq. (

i

) and (

ii

), we get

1 3 2 2 2 4

2 2 4

2 2 4

1

2

BAD

BAD

CDB CBD BAD

Question: 7

O

is the circumcenter of the triangle

ABC

and

D

is

the mid-point of the base

BC

. Prove that

BOD A

.

Solution:

Let us join OB, OC and OD.

In

BOD

and

COD

,

OD OD

(Common side)

BD DC

(D is the midpoint of BC)

OB OC

(Both are radius of circle)

Therefore,

BOD COD

(By SSS congruence

rule)

BOD COD

(By CPCT)

(i)

1

...

2

BOD COD BOC

In a circle, the angle subtended by an arc at the

centre is twice the angle subtended by it in the

remaining part of the circle.

2 BAC BOC

1

2

BAC BOC

BAC BOD

(Using (i))

( )A BOD BAC A

Question: 8

On a common hypotenuse

AB

, two right triangles

ACB

and

ADB

are situated on opposite sides. Prove

that

BAC BDC

.

Solution:

Let us join

CD.

90 90 180 C D

ADBC

is a cyclic quadrilateral.

BAC BDC

(Angles in the same segment

BC

of a circle are equal.)

Question: 9

Two chords

AB

and

AC

of a circle subtend angles

equal to 90° and 150°, respectively at the centre.

Find

BAC

, if

AB

and

AC

lie on the opposite sides

of the centre.

Solution:

In

Δ , AOB OA OB

(Both are radius of the circle.)

2 1

(Angles opposite to equal sides are equal.)

1 2 3 180

(By angle sum property of

Δ

)

2 2 90° 180° ( 2 1)

2 45°

In

Δ , AOC OA OC

(Both are radius of the circle.)

5 4

(Angles opposite to equal sides are equal.)

4 5 180AOC

(By angle sum property of

Δ

)

4 4 150 180 ( 5 4)

4 15

2 4

45 15

60

BAC

Question: 10

If

BM

and

CN

are the perpendiculars drawn on the

sides

AC

and

AB

of the triangle

ABC

, prove that the

points

B

,

C

, M and

N

are concyclic.

Solution:

We need to prove that

B, C, M, N

are concyclic.

Let us suppose BC is a diameter of the circle.

Then,

BC

subtends

1

and

2

on the same side and

Each1 2 90

So, points

M

and

N

lie on circle.

∴

Points

B, C, M, N

are concyclic.

Question: 11

If a line is drawn parallel to the base of an isosceles

triangle to intersect its equal sides, prove that the

quadrilateral so formed is cyclic.

Solution:

Given,

PBCQ

is cyclic

Now, in

ΔABC

AB AC

(Given)

C B

(Isosceles

property)

1 180

1 180

PQ BC

B

C B C

PBCQ

is cyclic

Question: 12

If a pair of opposite sides of a cyclic quadrilateral is

equal, prove that its diagonals are also equal.

Solution:

Let us suppose

ABCD

is a cyclic quadrilateral in

which

AD BC

.

We need to prove that

AC BD

.

Let us join

AC

and

BD

and draw

,DP AB CQ AB

Since

AD BC

,

arc arc AD BC

1 2

(Equal chords of a circle subtends equal

angles at the centre and also, the angle subtended by

a chord at the centre is twice the angle subtended by

it at the remaining part of the circle)

But these are alternate interior angles.

DC AB

DP CQ

…(i) (Distance between parallel lines)

Now, in

DAP

and

CBQ

3 4

(Each

90

)

DP CQ

(Using (i))

AD BC

(Given)

DAP CBQ

(By R.H.S. property)

A B

(CPCT)

In

DAB

and

CBA

,

DA CB

(Given)

A B

(Proved above)

AB BA

(Common)

DAB CBA

by SAS property

BD AC

Question: 13

The circumcenter of the triangle

ABC

is

O

. Prove

that

90 OBC BAC

.

Solution:

In

OBC

,

OB OC

(Radii of a triangle)

2 1

(Isosceles triangle property)

Now,

1 2 3 180

(Angle sum property of

triangle)

2 1 3 180 1 2

But

3 2 A

2 1 2 180 A

1 90A

90OBC BAC

Question: 14

A chord of a circle is equal to its radius. Find the

angle subtended by this chord at a point in major

segment.

Solution:

Let

AB

is the chord of a circle with centre

O

such

that

AB OA OB

We need to find

D

in the figure.

Now, in

OAB

,

OA OB AB

(Given)

∴

OAB

is an equilateral triangle

.

Hence,

60 AOB

(Each angle of equilateral triangle)

But,

2AOB D

60

2

30

D

D

Question: 15

In Fig. 10. 13,

130ADC

and chord

BC

chord

BE

. Find

CBE

.

Solution:

Let

M

be the centre of the circle.

Let us join

CM

,

EM

and

CE

.

ADCB

is cyclic

1 180

1 130 180

1 180 130

1 50

ADC

Now,

CM EM

(Radius of a circle) and

BC BE

(Given)

BCME

is a kite

MB CE

(Diagonals of a kite are perpendicular

to each other)

BCE

is an isosceles triangle (

BC BE

)

And

BO CE

Thus,

BO

bisects

CBE

2 1 50

Now,

1 2

50 50

100

CBE

CBE

CBE

Question: 16

In Fig.10.14,Find

OAB

.

Solution:

It is given that

40 ACB

Now,

2 2 1

(∵ Angle subtended by a segment to

the circle is half the angle subtended to the centre)

2 2 40

2 80

In

OAB

,

OA OB

(Radius of a circle)

4 3

(Isosceles triangle property)

Now,

2 3 4 180

(Angle sum property of∆)

80 3 3 180

2 3 180 80

2 3 100

3 50

50OAB

Question: 17

A quadrilateral

ABCD

is inscribed in a circle such

that

AB

is a diameter and

130ADC

. Find

BAC

.

Solution:

Since

ABCD

is inscribed in a circle,

ABCD

is a cyclic

quadrilateral.

1 2 180

(The sum of opposite angles of a

cyclic quadrilateral is supplementary)

130 2 180

2 180 130

2 50

Now,

3=90

(Angle in the semi-circle is 90

o

)

In

Δ ACB

2 3 4 180

(Angle sum property of a triangle)

50 90 4 180

4 180 140

40 BAC

Question: 18

Two circles with centres

O

and

O

intersect at two

points

A

and

B

. A line

PQ

is drawn parallel to

OO

through

A

(or

B

) intersecting the circles at

P

and

Q

.

Prove that

2 .PQ OO

Solution:

First, let us draw

OM PQ

and

'O N PQ

Since

OM

chord

PA

(∵

OM PQ

),

(i)

1

2

PM AM PA....

(Perpendicular drawn from

the centre of a circle to a chord, bisects it)

Similarly,

' chordO N AQ

(ii)

1

2

A N QN QA...

2 2PQ PA AQ AM AN

(Using (i) and (ii))

( )

2( )

2 ..... iii

PQ AM AN

PQ MN

Now,

'PQ OO

(Given)

'OM O N

(Both are the perpendicular distance

between the parallel lines

PQ

and

'OO

)

Therefore

'OMNO

is a parallelogram.

Hence,

(iv)' .....OO MN

(Opposite sides of a

parallelogram)

Using (iii) and (iv), we get

2 'PQ OO

Question: 19

In Fig.10.15,

AOB

is the diameter of a circle and

C

,

D

,

E

are any three points on the semi-circle. Find the

value of

.ACD BED

Solution:

Let us join

AE

and

BC

.

Now,

Angle 2 = 90

o

... (i) (Angle subtended by a diameter

of a circle)

Since

AEDC

is inscribed in a circle,

AEDC

is a cyclic

quadrilateral.

4 180 ACD

.... (ii) (

The sum of opposite angles

of a cyclic quadrilateral)

Adding (i) and (ii)

4 2 180 90

180 90

ACD

ACD BED

270

Question: 20

In Fig. 10.16,

30OAB

and

57OCB

. Find

BOC

and

.AOC

Fig. 10.16

Solution:

In ∆

OCB,

OB OC

(Both are the radius of a circle)

1 3 57

(Isosceles ∆ Property)

Also,

1 3 4 180

(Angle sum property of a ∆)

57 57 4 180

4 180 114

( )66 .... iBOC

In

OAB

OA OB

(Both are the radius of a circle)

5 2 30

(Isosceles ∆ Property)

Also,

2 5 6 180

(Angle sum property of ∆)

30 30 6 180

6 180 60

( )6 120 .... ii

Now,

6 4

120 66 54

AOC AOB BOC

AOC

Exercise 10.4

Question: 1

If two equal chords of a circle intersect, prove that

the parts of one chord are separately equal to the

parts of the other chord.

Solution:

Let us consider

AB

and

CD

are two equal chords of a

circle, meet at point

P

.

We need to show that

AP CP

and

BP DP

.

Let us draw

OE AB

and

OF CD

.

Then,

2

AB

AE EB

and

2

CD

CF FD

(Perpendicular drawn from the centre of a circle to a

chord bisects the chord)

Now,

AB CD

(Given)

∴

..(i)

2 2

..

AB CD

EB FD

(Using above result)

In

OPE

and

OPF

,

OEP OFP

(Both are right angles)

OE OF

(Equal chords are equidistant from the

centre)

OP OP

(Common)

OPE POF

by RHS property

Hence,

PE PF

.... (ii)

(By CPCT)

Adding (i) and (ii), we get

(iii)

EB+PE DF+PF

PB=PD....

Now,

AP AB PB CD PD

(

AB CD

(Given)

and using (iii))

( )AP CP CD PD CP

Hence, it is proved.

Question: 2

If non-parallel sides of a trapezium are equal, prove

that it is cyclic.

Solution:

Let ABCD be a trapezium in which

AB DC

and

.AD BC

Let us draw

DP AB

and

CQ AB

Now, in

DPA

and

CQB

3 1 90

DA CB

(Given)

DP CQ

(Distance between parallel lines)

DPA CQB

(By R.H.S property))

Thus,

A B

(By CPCT)

But,

180B C

(

AB CD

and BC is a transversal)

180A C

ABCD

is cyclic

Question: 3

If

P

,

Q

and

R

are the mid-points of the sides

BC

,

CA

and

AB

of a triangle and

AD

is the perpendicular

from

A

on

BC

, prove that

P

,

Q, R

and

D

are concyclic.

Solution:

Let us join

RQ, QP

and

RD

.

R

is the midpoint of hypotenuse

AB

of right angle

ADB

(

AD BC

)

RB RD

(

Midpoint of the hypotenuse is

equidistant from all 3 vertices in a right angles

triangle)

Therefore, in

DBR

........ i2 1

(Isosceles triangle property)

Now,

RQ

joins midpoint of sides

AB

and

AC

respectively in

ABC

.

RQ BC

(Midpoint theorem)

RQ BP

Similarly,

QP RB

is BPQR gm

So,

...(ii)1 3

(Opposite angles of a

gm

)

From (i) and (ii), we have

2 3

But

2 4 180

(linear pair axiom)

3 4 180

PQRD

is cyclic

Thus, the points

P

,

Q

,

R

,

D

are concyclic.

Question: 4

ABCD

is a parallelogram. A circle through

A

,

B

is so

drawn that it intersects

AD

at

P

and

BC

at

Q

. Prove

that

P, Q

,

C

and

D

are concyclic.

Solution:

Let us join

PQ

.

1 A

(Exterior angle property of cyclic

quadrilateral)

But

A C

(Opposite angles of a

gm

)

1 C

But,

180 C D

(Adjacent angles of a parallelogram)

1 180D

QCDP

is cyclic

Thus, points

P, Q, C, D

are concylic.

Question: 5

Prove that if the angle bisector of any angle of a

triangle and perpendicular bisector of the opposite

side intersect, they will intersect on the circumcircle

of the triangle.

Solution:

Let ABC be a triangle inscribed in a circle.

Let angle bisector of

A

and perpendicular bisector

of

BC

intersect at point

P

.

We need to prove that

A, B, P

and

C

are concyclic.

Let us join

BP

and

CP

.

We know that any 3 points are concyclic.

Therefore,

A, B, C

are concyclic.

Let us assume that P lies outside the circle.

Now, in

BMP

and

CMP

,

MP=MP

(Common)

BMP CMP

(Each right angle)

BM=CM (MP

is perpendicular bisector of

BC

)

Thus,

BMP CMP

(By SAS Rule)

So,

BP=CP

(By CPCT)…..(i)

Also,

BAP CAP

(Given).... (ii)

From (i) and (ii), we can say that P lies on the circle

as equal chords of a circle subtend equal angle at the

at the centre and also, the angle subtended by a

chord at the centre is twice the angle subtended by it

at the remaining part of the circle. Thus, equal

chords subtend equal angle at the circle.)

Hence,

A, B, C

and

P

are concyclic.

Question: 6

If two chords

AB

and

CD

of a circle

AYDZBWCX

intersect at right angles (See Fig.10.18), prove that

arc

CXA

arc

DZB

arc

AYD

arc

BWC

semicircle.

Fig.10.18

Solution:

Let the chords

AB

and

CD

of a circle

AYDZBWCX

intersect at

T

.

Let us draw a diameter

MN CD

.

Let the centre be

O

.

Since

MN CD

, arc

CM

=arc

DN.

....(i)

Also, arc

MCXA

= arc

MWB

(Symmetrical about

diameter of a circle)

And arc

AN

= are

NB

.... (ii)

Now,

MN i

s a diameter.

So, arc

MCXAYDN

= arc

MWBZN

= semicircle..... (iii)

Now,

arc arc arc MCXAYDN MC CXA arc AN

arc arc arc DN CXA NB

arc arc CXA DNB

arc arc ....CXA DZB iv

arc = arc arc MWBZN MWB NB

arc arc MWB AN

(Using (i) and (ii))

arc arc arc arc ( ) ( )CWB CM AYD DN

arc arc arc arc CWB CM AYD CM

arc arc ....CWB AYD v

Using (iii), (iv) and (v), we get

arc

CXA

arc

DZB

arc

AYD

arc

BWC

semicircle

Hence, it is proved.

Question: 7

If

ABC

is an equilateral triangle inscribed in a circle

and

P

be any point on the minor arc

BC

which does

not coincide with

B

or

C

, prove that

PA

is an angle

bisector of

BPC

.

Solution:

Since ABC is an equilateral triangle.

So,

( ).... i ABC ACB

Now,

ABC

and

APC

are the angles in the same

segment

AC

( ).... ii ABC APC

Similarly,

( ).... iii ACB APB

Using (i), (ii) and (iii), we get

APC APB

Thus,

PA

is an angle bisector of

BPC

.

Question: 8

In Fig. 10.19,

AB

and

CD

are two chords of a circle

intersecting each other at point

E

.

Prove that

1

2

AEC

(Angle subtended by arc

CXA

at centre + angle subtended by arc

DYB

at the

centre).

Solution:

We need to show that

1

( )

2

AEC AOC DOB

Let us join

AC, BC

and

BD

.

In

CEB

,

AEC ECB CBE

(Exterior angle property)

AEC DCB ABC

.....(i)

Since AC is a chord,

2 AOC ABC

..... (ii) (Angle subtended by the

chord AC )

Similarly,

2 DOB DCB

....(iii) (Angles subtended

by the chord BD)

Adding (ii) and (iii)

2( ) AOC DOB ABC DCB

.... (iv)

Now, using (i) in (iv),

)

2

1

(

2

AOC DOB AEC

AOC DOB AEC

Question: 9

If bisectors of opposite angles of a cyclic

quadrilateral

ABCD

intersect the circle,

Circumscribing it at the points

P

and

Q

, prove that

PQ

is a diameter of the circle.

Solution:

To prove that

PQ

is the diameter of the circle, we

need to prove that

90 PAQ

.

Let us join

AQ

and

QD

.

Since

ABCD

is a cyclic quadrilateral,

180A C

(Opposite angles of a cyclic

quadrilateral are supplementary)

90

2 2

A C

...(i)

It is given that

AP

bisects

A

and

CQ

bisects

C

.

So,

90PAD DCQ

....(i)

Also,

DCQ DAQ

...(ii) (Angles in the same segment

DF)

On comparing (i) and (ii), we have

90

90

PAD DAQ

PAQ

PAQ

is the angle in a semi-circle.

PQ

is the diameter of the circle.

Question: 10

A circle has radius

2

cm. It is divided into two

segments by a chord of length 2cm.

Prove that the angle subtended by the chord at a

point in major segment is 45°.

Solution:

Let us consider a circle with centre O and

radius

2

cm.

Let

2BC

cm be the chord and let

A

be the point in major segment.

We need to show that

45BAC

In

OBC

2 2

2 2

2 2 OB OC

2

2 2 4 (2)

2 2 2

OB OC BC

Thus, by reverse of Pythagoras theorem,

OBC

is a right angle triangle,

So,

90BOC

Now,

1

2

BAC BOC

(Angle subtended by a chord on

the centre is double of the angle subtended by same

chord on circumference)

90

45

2

BAC

Question: 11

Two equal chords

AB

and

CD

of a circle when

produced intersect at a point

P

.

Prove that

.PB PD

Solution:

Let us draw

OM AB

and

ON CD

Since

OM AB

,

2

AB

AM MB

…. (i) (Perpendicular from the

centre to a chord bisects it.)

Similarly,

2

CD

CN ND

… (ii)

Now,

AB CD

(Given)

2 2

AB CD

...(iii) MB ND

(Using (

i

) and (

ii

))

In

OMP

and

,ONP

1 2 90

OP OP

(Common)

OM ON

(Equal chords are equidistant from the centre)

OMP ONP

(By RHS property)

...(iv) MP NP

(By CPCT)

Subtracting (iii) from (iv), we get

MP MB NP ND

PB PD

Question: 12

AB

and

AC

are two chords of a circle of radius

r

such

that

2AB AC

. If

p

and q are the distances of

AB

and

AC from the centre, prove that

2 2 2

4 3 .q p r

Solution:

Let

2AC a

then

4AB a

.

Since

,ON AC

2

2

a

AN a

(Perpendicular from the centre to a

chord bisects it)

Similarly,

,OM AB

4

2

22

a

a

AB

AM

In

ANO

,

90ANO

2 2 2

r q a

(By Pythagoras theorem)

...(i)

2 2 2

a r q

In

AMO

,

90AMO

(ii)

2 2 2

(2 ) .....r p a

(By Pythagoras theorem)

2 2 2

4 r p a

2 2 2 2

4( ) r p r q

(Using (i))

2 2 2 2

4 4 r p r q

2 2 2 2

4 4 q r p r

2 2 2

4 3q r p

Hence proved.

Question: 13

In Fig. 10.20, O is the centre of the circle,

30 .BCO

Find

x

and

y

.

Solution:

In

OPC

,

180OPC PCO POC

(Angle sum property)

90 30 180

180 120 60

POC

POC

Now,

1 POD POC

90 60

30

…(i)

CD

subtends

1

at the centre and

y

on

circumference.

1 2y

30 2 y

(Using (i))

15 y

arc

AD

subtends

2

at the centre and

3

on

circumference.

2 2 3

But

2 90

3 45

…(ii)

In the right angle

APB

,

4 180 x ABP

(Angle sum property of ∆)

( 3 ) 90 180 x y

45 15 90 180 x

(Using (ii))

180 150 x

30 x

Question: 14

In Fig. 10.21, O is the centre of the circle,

BD OD

and

CD AB

. Find

.CAB

Solution:

It is given that

BD OD

But

OD OB

(Both are radius)

So, in

OBD

BD OD OB

is an equilateral triangle. OBD

1 60

(Angle of an equilateral triangle)

But

2 1

(Angle in same segment)

2 1 60

In

CAM

,

2 3 4 180

(Angle sum property of a

triangle)

60 90 4 180

1504 180

30 CAB