# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 12 – Circles

## Chapter 12 – Circles Exercise Ex. 12A

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of a chord from the centre of the circle.

_{}

_{Prove that the diameter of the circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.}

_{}

_{}

_{Prove that two different circles cannot intersects other at more than two points.}

_{}

_{Two circle of the radii 10 cm and 8 cm intersects each other , and the length of the common chord is 12 cm. Find the distance between their centres.}

_{Two equal circle intersects in P and Q. A straight line through P meets the circles in A and B. Prove that QA= QB.}

_{}

_{If a diameter of the circle bisects each of the two chords of a circle then prove that the chords are parallel.}

_{}

_{In the adjoining figure , two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q . Find the length of PQ.}

_{In the given figure , AB is a chord of a circle with centre O and AB is produced to C such that BC=OB. Also CO is a joined and produced to meet the circle in D. If }_{, prove that x=3y.}

AB and AC are two chords of a

circle of radius r such that AB = 2AC. If p and q are the distances of AB and

AC from the centre then prove that 4q^{2} =

p^{2} + 3r^{2}.

Let

O be the centre of a circle with radius r.

⇒ OB = OC = r

Let

AC = x

Then,

AB = 2x

Let

OM ⊥ AB

⇒ OM = p

Let

ON ⊥ AC

⇒ ON = q

In

ΔOMB, by

Pythagoras theorem,

OB^{2}

= OM^{2} + BM^{2}

In

ΔONC, by

Pythagoras theorem,

OC^{2}

= ON^{2} + CN^{2}

_{In the adjoining figure , O is the centre of a circle . If AB and AC are chordsof a circlesuch thatAB=AC, }_{, prove that PB=QC.}

_{}

_{In the adjoining figure, BC is a diameter of a circle with circle with centre O. If AB and CD are two chords such that AB|| CD, prove that AB=CD.}

Find the length of the chord which is at the distance of 3 cm from the centre of a circle of radius 5 cm.

_{}

_{}

_{An equilateral triangle of side 9 cm is inscribed in a circle . Find the radius of the circle.}

_{}

_{}

_{}

_{}

_{In the adjoining figure , OPQR is a square. A circle drawn with centre O cuts the square in X and Y. prove that QX =QY.}

Two circle 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 = 2OO’.

Draw

OM ⊥ PQ and O’N ⊥ PQ

⇒ OM ⊥ AP

⇒ AM = PM (perpendicular from the centre of a circle bisects the chord)

⇒ AP = 2AM ….(i)

And,

O’N ⊥ PQ

⇒ O’N ⊥ AQ

⇒ AN = QN (perpendicular from the centre of a circle bisects the chord)

⇒ AQ = 2AN ….(ii)

Now,

PQ = AP + PQ

⇒ PQ = 2AM + 2AN

⇒ PQ = 2(AM + AN)

⇒ PQ = 2MN

⇒ PQ = 2OO’ (since MNO’O is a rectangle)

A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find the radius of the circle.

_{}

_{}

_{}

In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6cm respectively. Calculate the distance between the chords if they are

(i) On the same side of the centre

(ii) On the opposite side of the centre.

_{}

_{ }

_{}

_{}

_{}

_{}

_{}

Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of a radius 17 cm. Find the distance between the chords.

_{}

_{}

_{}

In the given figure , the diameter CD of a circle with centre O is perpendicular to chord AB. If AB=12 cm and CE=3cm, calculate the radius of the circle.

_{}

_{ }

_{}

_{}

In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE=ED=8 cm and EB=4 cm. Find the radius of the circle.

_{}

_{ }

_{ }

In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC||DO and AC=2xOD.

_{}

_{}

In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects _{. Prove that AB=CD.}

_{}

_{}

## Chapter 12 – Circles Exercise Ex. 12B

(i) In figure (1) , O is the centreof the circle . If _{(ii) In figure(2), A, B and C are three points on the circlewithcentre O such that .}

_{}

_{}

_{In the given figure , O is the centre of a circle, }_{, find .}

_{}_{ }

_{In the adjoining figure , chords AC and BD of a circle with centre O, intersect at right angles at E. if }_{, calculate .}

_{In the given figure , O is the centre of a circle in which . F}_{ind }_{ }

In the given figure, O is the centre of the circle and ∠BCO

= 30°.

Find x and y.

Given,

∠AOD = 90° and ∠OEC = 90°

⇒ ∠AOD = ∠OEC

But

∠AOD and ∠OEC are corresponding

angles.

⇒ OD || BC and OC is

the transversal.

∴ ∠DOC = ∠OCE (alternate angles)

⇒ ∠DOC = 30° (since ∠OCE = 30°)

We

know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a

point on the circumference.

⇒ ∠DOC = 2∠DBC

Now,

∠ABE = ∠ABC = ∠ABD + ∠DBC = 45° + 15° = 60°

In

ΔABE,

∠BAE + ∠AEB + ∠ABE = 180°

⇒ x + 90° + 60° = 180°

⇒ x + 150° = 180°

⇒ x = 30°

In the given figure, O is the centre of the circle, BD = OD and CD ⊥

AB. Find ∠CAB

Construction:

Join AC

Given,

BD = OD

Now,

OD = OB (radii of same circle)

⇒ BD = OD = OB

⇒ ΔODB is an

equilateral triangle.

⇒ ∠ODB = 60°

We

know that the altitude of an equilateral triangle bisects the vertical angle.

Now,

∠CAB = ∠BDC (angles in the same segment)

⇒ ∠CAB = ∠BDE = 30°

_{In the given figure, PQ is a diameter of a circle with centre O. If }

_{}

_{ }

In the figure given below, P and Q

are centres of two circles, intersecting at B and

C, and ACD is a straight line.

We know that the angle

subtended by an arc of a circle at its centre is

twice the angle subtended by the same arc at a point on the circumference.

⇒ ∠APB = 2∠ACB

Now, ACD is a straight

line.

⇒ ∠ACB + ∠DCB = 180°

⇒ 75° + ∠DCB = 180°

⇒ ∠DCB = 105°

Again,

_{In the given figure , }_{. Show that BC is equal to the radius of the circumcircle of whose centre is O.}

_{ }

_{ }

In the given figure, AB and CD are

two chords of a circle, intersecting each other at a point E. Prove that

Join

AC and BC

We

know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a

point on the circumference.

⇒ ∠AOC = 2∠ABC ….(i)

Similarly,

∠BOD = 2∠BCD ….(ii)

Adding

(i) and (ii),

∠AOC + ∠BOD = 2∠ABC + 2∠BCD

⇒ ∠AOC + ∠BOD = 2(∠ABC + ∠BCD)

⇒ ∠AOC + ∠BOD = 2(∠EBC + ∠BCE)

⇒ ∠AOC + ∠BOD = 2(180° – ∠CEB)

⇒ ∠AOC + ∠BOD = 2(180° – [180° – ∠AEC])

⇒ ∠AOC + ∠BOD = 2∠AEC

_{In the given figure, O is the centre of the circle and .}

_{Calculate the value of .}

_{}_{ }

_{In the given figure , O is the centre of the circle .If , find the value of }

_{In the given figure , O is the centre of the circle. If }

_{}

_{ }

_{In the given figure, O is the centre of the circle .If }_{ find .}

In the given figure , _{, calculate }

In the adjoining figure , DE is a chord parallel to diameter AC of the circle with centre O. If _{, calculate .}

_{In the adjoining figure, O is the centre of the circle. Chord CD is parallel to diameter AB. If }_{ calculate }

_{In the given figure, AB and CD are straight lines through the centre O of a circle. If ,}_{ find }

## Chapter 12 – Circles Exercise Ex. 12C

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that _{ .}

_{Find }

_{}

_{In the given figure, O is the centre of the circle and }_{. Calculate the vales of x and y.}

_{}

_{ }

_{ }

_{In the given figure , sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If , }_{ find the value of x.}

_{}

_{}

_{In the given figure, AB is a diameter of a circle with centre O and DO||CB. If }_{, calculate}

_{Also , show that is an equilateral triangle.}

_{}

_{ }

_{ }

_{In the given figure , O is the centre of a circle. If }_{, calculate}

_{}

_{}

_{In the given figure, is an isosceles triangle in which AB=AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.}

_{}

_{}

_{ }

_{In the given figure, AB and CD are two parallel chords of a circle . If BDE and ACE are straight lines , intersecting at E, prove that is isosceles.}

_{}

_{In the given figure, }_{ Find the values of x and y.}

_{}

_{In the given figure , ABCD is a quadrilateral in which AD=BC and }_{. Show that the pints A, B, C, D lie on a circle.}

_{}

_{}

_{Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.}

_{}

_{ }

_{In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If }

_{}

_{}

_{Prove that the circles described with the four sides of a rhombus . as diameter , pass through the point of intersection of its diagonals.}

_{}

_{ }

_{}

_{}

_{}

_{ABCD is a rectangle . Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.}

_{}

_{ }

_{}

_{Give a geometrical construction for finding the fourth point lying on a circle passing through three given points , without finding the centre of the circle. Justify the construction.}

_{}

_{ }

_{}

_{}

_{In a cyclic quadrilateral ABCD, if }_{, show that the smaller of the two is }

_{}

_{The diagonal s of a cyclic quadrilateral are at right angles . Prove that perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.}

_{}

_{}

On a common hypotenuse AB, two

right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC

= ∠BDC.

AB

is the common hypotenuse of ΔACB and ΔADB.

⇒ ∠ACB = 90° and ∠BDC = 90°

⇒ ∠ACB + ∠BDC =

180°

⇒ The opposite

angles of quadrilateral ACBD are supplementary.

Thus,

ACBD is a cyclic quadrilateral.

This

means that a circle passes through the points A, C, B and D.

⇒ ∠BAC = ∠BDC (angles in the same segment)

ABCD is a quadrilateral such that

A is the centre of the circle passing through B, C

and D. Prove that

Construction:

Take a point E on the circle. Join BE, DE and BD.

know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a

point on the circumference.

⇒ ∠BAD = 2∠BED

Now,

EBCD is a cyclic quadrilateral.

⇒ ∠BED + ∠BCD =

180°

⇒ ∠BCD =

180° – ∠BED

In ΔBCD, by

angle sum property

∠CBD + ∠CDB + ∠BCD =

180°

_{In the given figure , O is the centre of the circle and arc ABC subtends an angle of at the centre . If AB is extended to P, find .}

_{}

_{}

_{}

_{In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced . If }

_{}

_{}

_{In the given figure, BD=DC and }

_{}

_{}

_{In the given figure, O is the centre of the given circle and measure of arc ABC is Determine .}

_{}

_{In the given figure, is equilateral. Find }

_{}

_{In the adjoining figure, ABCD is a cyclic quadrilateral in which }_{.}

_{}

_{}

_{ }

_{}

_{In the given figure , O is the centre of a circle and }_{Find the values of x and y.}

_{}

_{}

_{}

## Chapter 12 – Circles Exercise MCQ

The radius of a circle is 13 cm and

the length of one of its chords is 10 cm. The distance of the chord from the

centre is

- 11.5 cm
- 12 cm
- 23 cm

Correct option: (b)

In the given figure, BOC is a diameter of a circle with

centre O. If AB and CD are two chords such that AB ‖

CD. If AB = 10 cm, then CD =?

- 5 cm
- 12.5 cm
- 15 cm
- 10 cm

In the given figure, AB is a chord of a circle with centre

O and AB is produced to C such that BC = OB. Also, CO is joined and produced

to meet the circle in D. If ∠ACD = 25°,

then ∠AOD

=?

- 50°
- 75°
- 90°
- 100°

In the given figure, AB is a chord

of a circle with centre O and BOC is a diameter. If OD ⟘

AB such that OD = 6 cm, then AC =?

- 9 cm
- 12 cm
- 15 cm
- 7.5 cm

An equilateral triangle of side 9

cm is inscribed in a circle. The radius of the circle is

- 3 cm
- 6 cm

The angle in a semicircle measures

- 45°
- 60°
- 90°
- 36°

Angles in the same segment of a

circle area are

- equal
- complementary
- supplementary
- none of these

In the given figures, ⧍ABC

and ⧍DBC

are inscribed in a circle such that ∠BAC = 60°

and ∠DBC

= 50°.

Then, ∠BCD

=?

- 50°
- 60°
- 70°
- 80°

In the given figure, BOC is a

diameter of a circle with centre O. If ∠BCA = 30°,

then ∠CDA

=?

- 30°
- 45°
- 60°
- 50°

In the given figure, O is the centre of a circle. If

∠OAC

= 50°,

then ∠ODB

=?

- 40°
- 50°
- 60°
- 75°

In the given figure, O is the centre of a circle in which ∠OBA

= 20°

and ∠OCA

= 30°.

Then, ∠BOC

=?

- 50°
- 90°
- 100°
- 130°

A chord is at a distance of 8 cm from the centre of a

circle of radius 17 cm. The length of the chord is

- 25 cm
- 12.5 cm
- 30 cm
- 9 cm

Correct option: (c)

In the given figure, O is the centre of a circle. If ∠AOB

= 100°

and ∠AOC

= 90°,

then ∠BAC

=?

- 85°
- 80°
- 95°
- 75°

In the given figure, O is the

centre of a circle. Then, ∠OAB =?

- 50°
- 60°
- 55°
- 65°

In the given figure, O is the centre of a circle and ∠AOC

= 120°.

Then, ∠BDC

=?

- 60°
- 45°
- 30°
- 15°

In the given figure, O is the

centre of a circle and ∠OAB = 50°.

Then, ∠CDA

=?

- 40°
- 50°
- 75°
- 25°

In the given figure, AB and CD are two intersecting chords

of a circle. If ∠CAB = 40°

and ∠BCD

= 80°,

then ∠CBD

=?

- 80°
- 60°
- 50°
- 70°

In the given figures, O is the centre of a circle and

chords AC and BD intersect at E. If ∠AEB = 110°

and ∠CBE

= 30°,

then ∠ADB

=?

- 70°
- 60°
- 80°
- 90°

In the given figure, O is the

centre of a circle in which ∠OAB =20°

and ∠OCB

= 50°.

Then, ∠AOC

=?

- 50°
- 70°
- 20°
- 60°

In

the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If ∠ADC

= 120°,

then ∠BAC

=?

- 60°
- 30°
- 20°
- 45°

In the given figure ABCD is a cyclic quadrilateral in which

AB ‖

DC and ∠BAD

= 100°.

Then ∠ABC

=?

- 80°
- 100°
- 50°
- 40°

In the given figure, O is the

centre of a circle and ∠AOC =130°.

Then, ∠ABC

=?

- 50°
- 65°
- 115°
- 130°

In the given figure, BOC is a diameter of a circle and AB =

AC. Then ∠ABC =?

- 30°
- 45°
- 60°
- 90°

In the given figure, AOB is a

diameter of a circle and CD AB. If ∠BAD = 30°,

then ∠CAD

=?

- 30°
- 60°
- 45°
- 50°

In the given figure, O is the

centre of a circle in which ∠AOC =100°.

Side AB of quad. OABC has been produced to D. Then, ∠CBD =?

- 50°
- 40°
- 25°
- 80°

In the given figure, O is the

centre of a circle and ∠OAB = 50°.

Then, ∠BOD=?

- 130°
- 50°
- 100°
- 80°

In the given figures, ABCD is a cyclic quadrilateral in

which BC = CD and ∠CBD = 35°.

Then, ∠BAD

=?

- 65°
- 70°
- 110°
- 90°

In the given figure, equilateral ⧍

ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then ∠BDC

=?

- 90°
- 60°
- 120°
- 150°

In the give figure, side Ab and AD of quad. ABCD are

produced to E and F respectively. If ∠CBE = 100°,

then ∠CDF

=?

- 100°
- 80°
- 130°
- 90°

In the given figure, O is the centre of a circle and ∠AOB

= 140°.

The, ∠ACB

=?

- 70°
- 80°
- 110°
- 40°

In the given figure, O is the

centre of a circle and ∠AOB = 130°.

Then, ∠ACB

=?

- 50°
- 65°
- 115°
- 155°

In the given figure, ABCD and ABEF

are two cyclic quadrilaterals. If ∠BCD = 110°,

then ∠BEF

=?

- 55°
- 70°
- 90°
- 110°

In the given figure,

ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn

parallel to AB such that ∠ADC = 95°

and ∠ECF

=20°.

Then, ∠BAD

=?

- 95°
- 85°
- 105°
- 75°

In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB =?

- 30°
- 15°
- 60°
- 90°

Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD =?

- 10.5 cm
- 9.5 cm
- 8.5 cm
- 7.5 cm

In the given figure, A and B are

the centres of two circles having radii 5 cm and 3 cm respectively and

intersecting at points P and Q respectively. If AB = 4 cm, then the length of

common chord PQ is

- 3 cm
- 6 cm
- 7.5 cm
- 9 cm

In the given figure, ∠AOB = 90° and ∠ABC = 30°. Then ∠CAO =?

- 30°
- 45°
- 60°
- 90°

In the given figure, O is the centre of a circle. If ∠OAB

= 40°

and C is a point on the circle, then ∠ACB =?

- 40°
- 50°
- 80°
- 100°

In the given figure, AOB is a diameter of a circle with

centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then, the

distance of CD from AB is

- 8 cm
- 15 cm
- 18 cm
- 6 cm

AB and CD are two equal chords of a circle with centre O

such that ∠AOB = 80°, then ∠COD

=?

- 100°
- 80°
- 120°
- 40°

In the given figure, CD is the diameter of a circle with

centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm,

then radius of the circle is

- 6 cm
- 9 cm
- 7.5 cm
- 8 cm

In the given figure, O is the centre of a circle and

diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB

= 4 cm. The radius of the circle is

- 10 cm
- 12 cm
- 6 cm
- 8 cm