The availability of ICSE Class 9 Maths Solutions S Chand Chapter 13 Circle Ex 13(A) encourages students to tackle difficult exercises.

## S Chand Class 9 ICSE Maths Solutions Chapter 13 Circle Ex 13(A)

Question 1.

Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.

Solution:

In the given circle radius OA = 13 cm and ⊥ OM = 12 cm

Let length of chord AB = 2x

∵ OM ⊥ AB

∴ M is midpoint of AB

∴ AM = \(\frac { 2x }{ 2 }\) = x

Now in right angled ∆OAM,

OA² = OM² + AM² (Pythagoras Theorem)

⇒ (13)² = (12)² + x²

⇒ 169 = 144 + x²

⇒ x² = 169 – 144

⇒ x² = 25 = (5)²

∴ x = 5

Length of chord AB = 2 x 5 = 10 cm

Question 2.

In the figure, the radius of the given circle, with centre C, is 6 cm, if the chord AB is 3 cm away from the centre, calculate its length.

Solution:

In the given circle,

Radius CA = 6 cm and ⊥ CM = 3 cm

Let length of AB = x cm 1 x

Then AMC = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) cm

Now in right angled ∆CAM,

CA² = CM² + AM² (Pythagoras Theorem)

⇒ (6)² = (3)² + (\(\frac { x }{ 2 }\))² ⇒ 36 = 9 + \(\frac { x² }{ 4 }\)

∴\(\frac { x² }{ 4 }\) = 36 – 9 = 27 = (3\(\sqrt{3}\))²

∴ \(\frac { x }{ 2 }\) = 3\(\sqrt{3}\) x = 6\(\sqrt{3}\)

∴ Length of AB = 6\(\sqrt{3}\) cm

Question 3.

In figure, CD is diameter which meets the chord AB in E, such that AE = BE = 4 cm. If CE is 3 cm, find the radius of the circle.

Solution:

In the figure,

CD is the diameter of the circle with centre O

AB is chord which intersects CD at E and EA = EB = 4 cm and CE = 3 cm

Let r be the radius of the circle then

OC = r and OE = (r – 3)

Join OB In right ∆OBC,

OB² = BE² + OE² (Pythagoras Theorem)

⇒ r² = (4)² + (r – 3)²

⇒ r² = 16 + r² – 6r + 9

⇒ r² – r² + 6r = 16 + 9

⇒ 6r = 25 ⇒ r = \(\frac { 25 }{ 6 }\) = 4\(\frac { 1 }{ 6 }\)

∴ Radius of the circle = 4\(\frac { 1 }{ 6 }\) cm

Question 4.

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

(i) the same side of the centre ;

(ii) opposite sides of the centre.

Solution:

(i) When the chords are in the same side of the centre

Radius of the circle = 5 cm

Length of AB = 8 cm and CD = 6 cm

From O, draw a perpendicular on CD which intersects AB at L and meets CD at M

∴ M is the midpoints of CD and L is the midpoint ofAB

Join OA and OC

Now in right ∆OCM,

OC² = CM² + OM² (Pythagoras Theorem)

⇒ (5)² = (3)² + OM² (∵ CM = \(\frac { 1 }{ 2 }\) CD = \(\frac { 1 }{ 2 }\) x 6 = 3 cm)

⇒ 25 = 9 + OM² ⇒ OM² = 25 – 9 = 16 = (4)²

∴ OM = 4 cm

Similarly in right angled ∆OAL,

OA² = AL² + OL²

⇒ (5)² = (4)² + OL² (AL = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 8 = 4 cm)

25 = 16 + OL²

⇒ OL² = 25 – 16 = 9 = (3)²

∴ OL = 3 cm

Now LM = OM – OL = 4 – 3 = 1 cm

∴ Distance between the two chords = 1 cm

(ii) When the chords are in the opposite sides of the centre

Draw OL ⊥ AB and OM ⊥ CD

In right ∆OCM,

OC² = CM² + OM² (Pythagoras Theorem)

⇒ (5)² = (3)² + OM² (∵ CM = \(\frac { 1 }{ 2 }\) CD)

⇒ 25 = 9 + OM²

⇒ OM² = 25 – 9 = 16 = (4)²

OM = 4 cm

Similarly in right ∆OLA,

OA² = OL² + AL²

⇒ (5)² = OL² + (4)² (∵ AL = \(\frac { 1 }{ 2 }\) AB)

⇒ 25 = OL² + 16

OL² = 25 – 16 = 9 = (3)²

∴ OL = 3 cm

∴ LM = OL + OM = 3 + 4 = 7 cm

∴ Distance between the two chords = 7 cm

Question 5.

The radius of a circle is 2.5 cm. AB, CF are two parallel chords 3.9 cm apart. If AB = 1.4 cm, find CF.

Solution:

In right angled ∆OAM,

AM = 0.7 cm

OA = 2.5 cm and OA² = AM² + OM²

(Pythagoras Theorem)

⇒ (2.5)² = (0.7)² + OM²

⇒ 6.25 = 0.49 + OM²

⇒ OM² = 6.25 – 0.49 = 5.76 = (2.4)²

∴ OM = 2.4

But LM = 3.9

∴ OL = LM – OM = 3.9 – 2.4 = 1.5 cm

Now in ∆OLC,

OC² = CL² + OL² (Pythagoras Theorem)

⇒ (2.5)² = CL² + (1.5)²

⇒ 6.25 = CL² + 2.25 CL²

= 6.25 – 2.25 = 4.00 = (2)²

∴ CL = 2

∴ Chord CF = 2 x CL = 2 x 2 = 4 cm

Question 6.

A chord distant 2 cm from the centre of a circle is 18 cm long. Calculate the length of a chord of the same circle which is 6 cm distant from the centre.

Solution:

Let r be the radius of the circle and chord

AB = 18 cm

and distance from the centre O

∴ OL = 2 cm

Let CD be another chord

OM ⊥ CD and OM = 6 cm

Now in right ∆OAL,

OA² = AL² + OL² (Pythagoras Theorem)

= (9)² + (2)² = 81 + 4 = 85 (AL = \(\frac { 1 }{ 2 }\)AB = \(\frac { 1 }{ 2 }\) x 18 = 9 cm)

Similarly in right ∆OCM,

OC² = OM² + CM²

⇒ OA² = OM² + CM² (∵ OA = OC radii of the same circle)

⇒ 85 = (6)² + CM²

⇒ 85 = 36 + CM² ⇒ CM² = 85 – 36 = 49 = (7)²

∴ CM = 7

∴ CD = 2 x CM

= 2 x 7 = 14cm

Question 7.

In figure, circles are concentric with centre O. Find AC.

Solution:

In the figure, two circles are concentric with centre O

Chord AB intersect the small circle at C and D

OB and OC are joined

OB = 17 cm, OC = 10 cm

OL ⊥ AB

Now in right ∆OMB,

OB2 = OM² + MB² (Pythagoras Theorem)

⇒ (17)² = (9)² + MB²

⇒ 289 = 81 + MB²

⇒ MB² = 289 – 81 = 208

MB = \(\sqrt{208}\) = 14.42

or AM = 14.42 cm (∵ M is midpoint of AB)

Similarly in right ∆OCM,

OC² = OM² + CM²

⇒ (10)² = (9)² + CM²

⇒ 100 = 81 + CM²

⇒ CM² = 100 – 81 = 19

∴ CM = \(\sqrt{19}\) = 4.36 cm

∴ AC = AM – CM = 14.42 – 4.36 = 10.06 cm

Question 8.

The length of the common chord of two equal intersecting circles is 10 cm and the distance between the two centres is 6 cm. Find the radius of each circle.

Solution:

Two equal circles with centre O and O’ intersect each other at A and B

OO’, AB, OA, OB, O’A and O’B are joined

AB = 10 cm

OO’ = 6 cm

Let OO’ bisects AB at M i.e. AM = MB = \(\frac { 10 }{ 2 }\) = 5cm

and OM = MO’ = \(\frac { 6 }{ 2 }\) = 3 cm

Now in right ∆OAM,

OA² = OM² + AM² (Pythagoras Theorem)

= (3)² + (5)² = 9 + 25 = 34

∴ OA = \(\sqrt{34}\) = 5.83 cm

Hence radius of each circle = 5.83 cm

Question 9.

In figure, AB = 8 cm, CM = 1 cm, CM is the perpendicular bisector of AB. The radius OA = x cm. Find x.

Solution:

In circle with centre O, OA is its radius

AB is chord, OM ⊥ AB which is produced to meet the circle at C

AB = 8 cm, CM = 1 cm, OA = x

OC = OA = x

∴ OM = OC – CM = (x – 1)

Now in right ∆OAM,

OA² = AM² + OM² (Pythagoras Theorem)

x² = (4)² + (x – 1)² (AM = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 8 = 4 cm)

⇒ x² = 16 + x² – 2x + 1

⇒ x² – x² + 2x = 17

⇒ 2x = 17 ⇒ x = \(\frac { 17 }{ 2 }\) = 8.5

Hence x = 8.5 cm

Question 10.

In figure, CD is the perpendicular bisector of the chord AB. If AB = 2 cm and CD = 4 cm, calculate the radius of the circle.

Solution:

In the circle with centre O, AB is chord CD is perpendicular bisector of AB

AB = 2 cm, CD = 4 cm

Join OA, let r be the radius of the circle

OA = OD = r

∴ OC = CD – OD = 4 – r

AC = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 2 = 1 cm

Now in right ∆OAC,

OA² = OC² + AC² (Pythagoras Theorem)

⇒ r² = (4 – r)² + (1)²

⇒ r² = 16 + r² – 8r + 1

⇒ r² – r² + 8r = 17

⇒ 8r = 17

⇒ r = \(\frac { 17 }{ 8 }\) = 2\(\frac { 1 }{ 8 }\)

∴ Radius of the circle = 2 \(\frac { 1 }{ 8 }\) cm

Question 11.

In figure, OD is perpendicular to the chord AB of a circle, whose centre is O. Prove that CA = 2OD.

Solution:

Given : In the circle AB is chord, OD ⊥ AB

BOC is the diameter of the circle

AC is joined

To prove : CA = 2OD

Proof: OD ⊥ AB

∴ D is midpoint of AB

and O is midpoint of BC

In ∆ABC, DO || AC

∴ In ∆ABC ~ ∆DBO,

∴ \(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{BO}}=\frac{\mathrm{AC}}{\mathrm{OD}}\)

(Sides of similar triangles are proportional)

⇒ \(\frac{\mathrm{AB}}{\frac{1}{2} \mathrm{AB}}=\frac{\mathrm{CA}}{\mathrm{OD}} \Rightarrow \frac{2}{1}=\frac{\mathrm{CA}}{\mathrm{OD}}\) ⇒ CA = 2OD

Hence proved.

Question 12.

In figure, OMNP is a square. A circle drawn with centre O cuts the square in X and Y. Prove that : ∆OXM ≅ ∆OYP. Hence prove that NX = NY.

Solution:

Given : In the figure,

OMNP is a square

A circle with centre O is drawn which intersects the square at X and Y

Join OX and OY

(i) ∆OXM ≅ ∆OYP

(ii) NX = NY

Proof: In right ∆OXM and ∆OYP,

Hyp. OX = OY (radii of the same circle)

Side OM = OP (sides of a square)

∴ ∆OXM ≅ ∆OYP (RHS axiom)

∴ MX = PY

But MN = PN (sides of a square)

∴ MN – MX = PN – PY

⇒ NX = NY

Hence proved.

∴ Least value of x = 2 cm and greatest = 8 cm

Question 13.

Find the centre of given circle.

Solution:

Steps of construction :

(i) Take three points A, B and C on the circle.

(ii) Join AB and AC.

(iii) Draw the perpendicular bisectors of AB and AC. Which intersect each other at O.

Then O is the centre of the circle

(We know that the perpendicular bisectors of chords of a circle passes through the centre of the circle) [Hint, see fig. 15.36]

Question 14.

Draw an arc of a circle to pass through three points A, B and C not in the same st. line.

Solution:

Steps of construction :

(i) Take three points A, B and C.

(ii) Join AB and BC.

(iii) Draw the perpendicular bisector of AB and BC intersecting each other at O.

(iv) With centre O and radius OA or OB or OC, draw an arc ABC.

This is the required arc.

∵ O lies on the perpendicular bisector of AB.

∴ OA = OB

Again O lies on the perpendicular bisector of BC

∴ OB = OC

∴ OA = OB = OC

Question 15.

Answer true or false. Three circles can pass through three given points not in a straight line.

Solution:

False : as through three points which are not in a straight line, one and only one circle can be drawn.

Question 16.

Two chords A B and CD of a circle are parallel and a line l is the perpendicular bisector of AB. Show that l bisects CD.

Solution:

AB and CD are two parallel chords of the circle and a line / is the perpendicular bisector of AB

i. e., AL = LB and ∠ALM = 90°

To prove : Line l is also perpendicular bisector of CD

Proof: ∵ Line l is perpendicular bisector of chord AB

∴ It will pass through the centre of the circle

∵ AB || CD and l is perpendicular to AB

∴ l is also perpendicular to CD

∵ OM ⊥ CD

∴ M is the midpoint of CD

∴ l bisects CD

Hence proved.

Question 17.

In the given figure, O and O’ are the centres of two intersecting circles and APB is parallel to 00′. Prove that AB = 200′.

Solution:

Given : Two circles with centre O and O’

intersect each other at P

APB || OO’ is drawn

To prove : AB = 200′

Construction : From O, draw OM ⊥ AP

and from O’, O’N ⊥ PB

Proof: ∵ OM ⊥ AP and O’N ⊥ PM

and OO’ || AB

∴ MN = OO’

Now ∵ OM ⊥ AP

∴ M is midpoint of AP

AM = MP ⇒ MP = \(\frac { 1 }{ 2 }\) AP

Similarly PN = NB ⇒ PN = \(\frac { 1 }{ 2 }\) PB

Adding we get

MP + PN = \(\frac { 1 }{ 2 }\) AP + \(\frac { 1 }{ 2 }\) PB

⇒ MN = \(\frac { 1 }{ 2 }\) (AP + PB) = \(\frac { 1 }{ 2 }\) AB

⇒ OO’ = \(\frac { 1 }{ 2 }\) AB (∵ OO’ = MN)

⇒ 200′ = AB

or AB = 200′

Hence proved.

Question 18.

Two equal chords AB and CD of a circle intersect at P, show that AP = PD and BP = CP.

Solution:

Given : In a circle with centre O,

Two equal chords AB and CD intersect each other at P

To prove : AP = PD and BP = CP

Construction : Draw OM ⊥ AB and ON ⊥ CD

Join OP

Proof : In right ∆OMP and ∆ONP,

Hyp. OP = OP (common)

Side OM = ON

(Equal chords are equidistant from the centre)

∴ ∆OMP ≅ ∆ONP (R.H.S. axiom)

∴ PM = PN (c.p.c.t.)

∵ OM ⊥ AB and ON ⊥ CD

∴ M and N are the midpoints of AB and CD respectively

or AM = MP and CN = NP

∵ Chords AB = CD

∴ AB – PB = CD – PC

⇒ AP = PD

Similarly BM – PM = CN – PN (∵ PM = PN)

⇒ BP = CP

Hence proved.

Question 19.

In the figure, shows circle with centre O, with equal chords AB and CD; OE ⊥CD at H and OF ⊥ AB at G. Prove that EH = GF.

Solution:

Given : In a circle with centre O,

chord AB = CD

OE ⊥ CD at A and OF ⊥ AB at G

To prove : EH = GF

Proof: ∵ OE ⊥ CD and OF ⊥ AB

∴ G and H are midpoints of AB and CD respectively

∵ and OG = OH

(Equal chords are equidistant from the centre)

But OF = OE (radii of the same circle)

∴ OF – OG = OE – OH

⇒ GF = EH

∴ EH = GF

Hence proved.

Question 20.

In the figure, C is the centre of the circle. CB bisects the ∠DBE, CD ⊥ PQ and CE ⊥ RS. Prove that PQ = RS.

Solution:

Given : In circle with centre C,

CB bisects ∠DBE

CD ⊥ PQ and CE ⊥ RS where PQ and RS are the chords which intersect each other at B

To prove : PQ = RS

Proof: In ∆EBC and ∆DBC,

∠E = ∠D (each 90°)

∠EBC = ∠DBC

(∵ BC is the bisector of ∠DBE)

BC = BC (common)

∴ ∆EBC ≅ ∆DBC (AAS axiom)

∴ and CE = CD (c.p.c.t.)

But these are the perpendicular distance from the centre and equal chords are equidistant from the centre

∴ RS = PQ

or PQ = RS

Hence proved.

Question 21.

In the figure, AB and CD are equal chords of a circle whose centre is O. OM ⊥ AB and ON ⊥ CD. Prove that ∠OMN = ∠ONM.

Solution:

Given : In circle with centre O,

Two equal chords AB and CD intersect each other inside the circle

OM ⊥ AB and ON ⊥ CD

MN arejoined

To prove : ∠OMN = ∠ONM

Proof: ∵ Chord AB = CD

∴ OM = ON

(Equal chords are equidistant from the centre)

In ∆OMN,

∵ OM = ON

∴ ∠OMN = ∠ONM

Hence proved.