11 / 19

CHAPTER 10

Circles

Exercise 10.3

1. If arcs AXB and CYD of a circle are congruent, find the ratio of AB and chord CD.

Sol. We have

Since if two arcs of a circle are congruent, then their corresponding arcs are equal, so we

have chord AB = chord CD Hence, AB : CD = 1 : 1.

2. If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at

P and Q, prove that arc PXA ≅ Arc PYB.

Sol. As PQ is the perpendicular bisect of AB,

http

s://sc

hoolc

onne

cts.in

22 / 19

So, AM = BM

In and we have

AM = BM [Proved above]

[Each = 90o]

PM = PM [Common side]

[By SAS congruence rule]

So, AP = BP [By C.P.C.T.]

Hence, arc PXA ≅ Arc PYB

[If two chords of a circle are equal, then their corresponding arcs are congruent]

3. A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB,

BC and CA are concurrent.

Sol. Given: Three non-collinear points A, B and C are on a circle.

To prove: Perpendicular bisectors of AB, BC and CA are concurrent.

Construction: Join AB, BC and CA.

Draw perpendicular bisectors ST of AB, PM of BC and QR of CA are respectively. As point A, B

http

s://sc

hoolc

onne

cts.in

33 / 19

and C are not collinear, so ST, PM and QR are not parallel and will intersect.

Proof: O lies on ST, the bisector of AB

OA = OB…(1)

Similarly, O lies on PM, the bisector of BC

OB = OC…(2)

And, O lies on QR, the bisector of CA

OC = OA…(3)

From (1), (2) and (3), OA = OB = OC = r(say)

With O as a centre and r as the radius, draw circle C(O, r) which will pass through A, B and C.

This prove that there is a circle passing through the points A, B and C. Since ST, PM or QR can

cut each other at one and only one point O.

O is the only point equidistance from A, B and C.

Hence, the perpendicular bisectors of AB, BC and CA are concurrent.

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.

Sol. Given: AB and AC are two chords which are equal with centre O. AM is the bisector of

http

s://sc

hoolc

onne

cts.in

44 / 19

To prove: AM passes through O.

Construction: Join BC. Let AM intersect BC at P.

Proof: In and

AB = AC [Given]

[Given]

And AP = BP [Common side]

[By SAS congruency]

[By C.P.C.T.] And

CP = PB But [Linear

pair ]

AP is perpendicular bisector of the chord BC, which will pass through the centre O on

being produced.

Hence, AM passes through O.

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.

http

s://sc

hoolc

onne

cts.in

55 / 19

Sol. Given: AB and CD are two chords of a circle whose centre of O. The mid-points of AB and

CD are L and M respectively.

To prove: AB||CD

Proof: L is the mid-point of chord AB or

[ The line joining the centre of a circle to the mid-point of a chord is perpendicular to the

chord]

Similarly,

But, these are corresponding angles.

So, AB||CD.

Hence, proved

6. ABCD is such a quadrilateral that A is the centre of the circle passing through B, C

and D. Prove that

Sol. ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D.

We have to prove that

http

s://sc

hoolc

onne

cts.in

66 / 19

Join AC.

Since angle subtended by an arc at the centre is double the angle subtended by it at point on

the remaining part of the circle.

Therefore, …(1)

And …(2)

Adding (1) and (2), we get

Hence,

7. O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove

that ∠BOD = ∠A.

http

s://sc

hoolc

onne

cts.in

77 / 19

Sol. Given: O is the circumcentre of and

To prove:

Construction: Join OB and OC.

Proof: In and we have

OB = OC [Each equal to radius of the circumcircle]

[Each of 90o]

OD = OD [Common]

[By SAS congruency]

Therefore, BOD = COD [By C.P.C.T.]

Now, are BC subtends at the centre and A at a point in the remaining part of the

circle.

Hence, proved.http

s://sc

hoolc

onne

cts.in

88 / 19

8. On a common hypotenuse AB, two right triangles ACB and ADB are situated on

opposite sides. Prove that ∠BAC = ∠BDC.

Sol. In right triangles ACB and ADB, we have

and

If the sum of any pair of opposite angles of a quadrilateral is 180o, then the quadrilateral is

cyclic. So, ADBC is a cyclic quadrilateral.

Join CD. Angles and are made by in the same segment BDAC.

Hence,

[ Angles in the same segment of a circle are equal]

9. Two chords AB and AC of a circle subtends angles equal to 90º and 150º, respectively

at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.

Sol. We have

Reflex

Now,ht

tps:/

/scho

olcon

nects

.in

9 / 19

[Since, angle subtended by an arc at the centre is double of the angle subtended by the same

arc on the remaining part of the cirlce]

Hence, BAC = BOC =

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.

Sol. As BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC

Since, if a line segment (here BC) joining two points (here B and C) subtends equal angles

(here and ) at M and N on the same side of the line (here BC) containing the

segment, the four points (here B, C, M and N) are concyclic.http

s://sc

hoolc

onne

cts.in

1010 / 19

Hence, B, C, M and N are concyclic.

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.

Sol. is an isosceles triangle in which AB = AC. DE is drawn parallel to BC. We have to

prove that quadrilateral BCED is a cyclic quadrilateral i.e., point B, C, E and D lie on a circle.

In we have

AB = AC[Given]

[ Angles oppossite to equal sides are equal]

Now, DE||BC and AB cuts them,

[ Sum of interior angles on the same side of the transversal]

Similarly, we can show that

Since if pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is

cyclic.

Hence, BCED is a quadrilateral.

12. If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals

are also equal.

http

s://sc

hoolc

onne

cts.in

1111 / 19

Sol. ABCD is cyclic quadrilateral in which one pair of opposite sides AB = DC. We have to

prove that diagonal AC = diagonal BD.

In and we have

[Angles in the same segment of the circle are equal]

AB = DC[Given]

Also, [Angles in the same segment of the circle are equal]

[By ASA congruence rule]

AO = OD [By C.P.C.T.]…(1)

And OC = BO...(2)

Now, adding (1) and (2),we get

AO + OC = BO + OD

AC = BD

Hence, proved.

13. The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.

Sol. ABC is a triangle and O is the circumcentre.http

s://sc

hoolc

onne

cts.in

1212 / 19

Draw Join OB and OC.

In right and right we have

hyp. OB = hyp. OC

[Radii of the same circle]

OD = OD

[Common side]

[By RHS cong. Rule]

and [By C.P.C.T.]

Now, and

…(1)

[Adding to both sides] [

and ]

Hence, proved.

14. A chord of a circle is equal to its radius. Find the angle subtended by this chord at a

http

s://sc

hoolc

onne

cts.in

1313 / 19

point in major segment.

Sol. Since chord of a circle is equal radius, so we have AB = OA = OB.

Therefore, ABC is an equilateral triangle.

Since each angle of an equilateral triangle is 60o, so we have Since angle

subtended by an arc at the centre is double the angle subtended by it at any point on the

remaining part of the circle, so we have

Hence,

15. In Fig.10.13, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.

Sol. In the given figure, we have and chord BC = chord BE. We have to find

Since ABCD is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral

are supplementary.

http

s://sc

hoolc

onne

cts.in

1414 / 19

…(1)

In and we have

BC = BE[Given]

OC = OE [Radii of same circle]

OB = OB [Common side]

[By SSS cong. Rule]

[By C.P.C.T. and by (1) ]

Hence,

16. In Fig.10.14, ∠ACB = 40º. Find ∠OAB.

Sol. Since the angle subtended by an arc at the centre is double the angle subtended by it at

any point on the remaining part of the circle, so we have

http

s://sc

hoolc

onne

cts.in

1515 / 19

So, in we have

OAB = OBA [Angle opposite to equal sides are equal]

Hence,

17. A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC =

130º. Find ∠BAC.

Sol. Since the opposite angles of a cyclic quadrilateral are supplementary.

Now, in [ Angle in a semi-circle = 90o]

And

http

s://sc

hoolc

onne

cts.in

1616 / 19

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 PQ = 2 OO′.

Sol. Two circles with centre 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.

Draw and

We have to prove that PQ = 2 OO’.

Since perpendicular from the centre to a chord bisect the chord, so

PA = 2CA…(1)

And AQ = 2 AD…(2)

Adding (1) and (2), we get

PA + AQ = 2CA + 2AD

PQ = 2(CA + AD) = 2CD

Hence, PQ = 2 OO’ [ CD and OO’ are opposite sides of a rectangle]

19. In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the

semi-circle. Find the value of ∠ACD + ∠BED.http

s://sc

hoolc

onne

cts.in

1717 / 19

Sol. Join BC.

Since angle in a semicircle is 90o, we have

∠ACB = 90o

As ABCD is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are

supplementary

Now, adding to both sides, we get

Hence,

20. In Fig. 10.16, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.ht

tps:/

/scho

olcon

nects

.in

1818 / 19

Sol. In we have

OB = OC [Radii of the same circle]

Now, in we have

[ (Given)]

Again, in we have

http

s://sc

hoolc

onne

cts.in

Hence, and

19 / 19

http

s://sc

hoolc

onne

cts.in