# ML Aggarwal Solutions for Class 10 Maths Chapter 15 Circles

ML Aggarwal Solutions for Class 10 Maths Chapter 15 Circles, are provided here in PDF format, which can be downloaded for free. The ML Aggarwal Solutions for the chapter Circles have been designed accurately by mathematics experts at BYJUâ€™S. These serve as a reference tool for the students to do their homework and assignments, as well. These solutions for Class 10 contain the exercise-wise answers for all the chapters, thus being a very useful study material for the students studying in Class 10. Chapter 15 of ML Aggarwal Solutions for Class 10 Maths Circles explains circles, constructions of circles and its applications. A circle is a special kind of ellipse in which the eccentricity is zero and the two foci are coincident. A circle is also termed as the locus of the points drawn at an equidistant from the centre. The distance from the center of the circle to the outer line is its radius. Diameter is the line which divides the circle into two equal parts and is also equal to twice the radius. We, in our aim to help students, have devised detailed chapter wise solutions for them to understand the concepts easily.

## ML Aggarwal Solutions for Class 10 Maths Chapter 15 :-

### Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 15 – Circles

Exercise 15.1

1. Using the given information, find the value of x in each of the following figures:

Solution:

(i)
âˆ ADB and âˆ ACB are in the same segment.

âˆ ADB = âˆ ACB = 50Â°

Now in

âˆ DAB + X + âˆ ADB = 180Â°

= 42o + x + 50o = 180o

= 92o + x = 180o

x = 180o – 92o

x = 88o

(ii) In the given figure we have

= 32o + 45o + x = 180o

= 77o + x = 180o

x = 103o

(iii) From the given number we have

Because angles in the same segment

âˆ BCD = 20o

âˆ CEA = 90o

âˆ CED = 90o

Now in triangle CED,

âˆ CED +
âˆ BCD + âˆ CDE = 180o

90o + 20o + x = 180o

= 110o + x = 180o

x = 180o â€“ 110o

x = 70o

(iv) In
âˆ†ABC

âˆ ABC + âˆ ABC + âˆ BAC = 180o

(Because sum of a triangle)

69o + 31o + âˆ BAC = 180o

âˆ BAC = 180o â€“ 100o

âˆ BAC = 80o

Since âˆ BAC and âˆ BAD are in the same

Segment.

âˆ BAD = xo = 80o

(v) Given âˆ CPB = 120o , âˆ ACP = 70o

To find, xo i,e., âˆ PBD

Reflex âˆ CPB = âˆ BPO + âˆ CPA

1200 = âˆ BPD + âˆ BPD

(BPD = CPA are vertically opposite âˆ s)

2âˆ BPD = 120o âˆ PBD = 1200/2 = 60o

Also âˆ ACP and PBD are in the same segment

âˆ PBD + âˆ ACP = 700

Now, In âˆ†PBD

âˆ PBD + âˆ PDB + âˆ BPD = 180o

(sum of all âˆ s in a triangle)

700 + xo + 600 = 180o

x = 180o – 130o

x = 50o

(vi) âˆ DAB = âˆ BCD

(Angles in the same segment of the circle)

âˆ DAB = 250 (âˆ BCD = 250 given)

In âˆ†DAP,

Ex, âˆ CDA = âˆ DAP + âˆ DPA

xo = âˆ DAB + âˆ DPA

xo = 25o + 35 o

xo = 60o

2. If O is the center of the circle, find the value of x in each of the following figures (using the given information):

Solution:

(i) âˆ ACB = âˆ ADB

(Angles in the same segment of a circle)

âˆ ABC = xo

Now in âˆ†ABC

âˆ CAB + âˆ ABC + âˆ ACB = 180o

40o + 900 + xo = 180o

(AC is the diameter)

130o + xo = 180o

xo = 1800 â€“ 130o = 50o

(ii) âˆ ACD = âˆ ABD

(angles in the same segment)

âˆ ACD = xo

Now in triangle OAC,

OA = OC

âˆ ACO = âˆ AOC

(opposite angles of equal sides)

Therefore, xo = 62o

(iii) âˆ AOB + âˆ AOC + âˆ BOC = 360o

(sum of angles at a point)

âˆ AOB + 80o + 130o = 360o

âˆ AOB + 210o = 360o

âˆ AOB = 360o â€“ 210o = 1500

Now arc AB subtends âˆ AOB at the centre âˆ ACB at the remaining part of the circle

âˆ AOB = 2 âˆ ACB

âˆ ACB = Â½ âˆ AOB = Â½ Ã— 150o = 75o

(iv)
âˆ ACB + âˆ CBD = 180o

âˆ ABC + 75o = 180o

âˆ ABC = 180o â€“ 75oâ€‘ = 105o

Now arc AC Subtends reflex âˆ AOC at the centre and âˆ  ABC at the remaining part of the circle.

Reflex
âˆ AOC = 2 âˆ ABC

= 2Ã— 105o =210o

(v) âˆ AOC + âˆ COB = 180o

135o + âˆ COB = 180o

âˆ COB = 1800 â€“ 135o = 45o

Now arc BC Subtends reflex âˆ COB at the centre and âˆ  CDB at the remaining part of the circle.

âˆ COB = 2 âˆ CDB

âˆ CDB = Â½ âˆ COB

= Â½ Ã— 45o = 45o/2 = 22 1/2o

(vi) Arc AB subtends âˆ AOD at the centre and âˆ ACD at the remaining part of the Circle

âˆ AOD = 2 âˆ ACB

âˆ ACB = Â½ âˆ AOD = Â½ Ã— 70o = 35o

âˆ CMO = 90o

âˆ AMC = 90o

(âˆ AMC + âˆ CMO = 180o)

Now in âˆ†ACM

âˆ ACM + âˆ AMC + âˆ CAM = 180o

35o + 90o + xo = 180o

125o + xo = 180o

Xo = 180 â€“ 125o = 55o

3. (a) In the figure (i) given below, AD || BC. If âˆ ACB = 35Â°. Find the measurement of âˆ DBC.

(b) In the figure (ii) given below, it is given that O is the centre of the circle and âˆ AOC = 130Â°. Find âˆ  ABC

Solution:

(a) Construction: Join AB

âˆ A = âˆ C = 350 (Alt Angles)

âˆ ABC = 35o

(b) âˆ AOC + reflex âˆ AOC = 360o

130o + Reflex âˆ AOC = 360o

Reflex
âˆ AOC = 360o â€“ 130o = 230o

Now arc BC Subtends reflex âˆ AOC at the centre and âˆ  ABC at the remaining part of the circle.

Reflex âˆ AOC = 2 âˆ ABC

âˆ ABC =1/2 reflex âˆ AOC

= Â½ Ã— 230o = 115o

4. a) In the figure (i) given below, calculate the values of x and y.
(b) In the figure (ii) given below, O is the centre of the circle. Calculate the values of x and y.

Solution:

âˆ B + âˆ D = 1800

Y + 400 + 45o = 180o

(y + 85o = 180o)

Y = 180o â€“ 85o = 95o

xo = 40

(a) Arc ADC Subtends âˆ AOC at the centre and
âˆ  ABC at the remaining part of the circle

âˆ AOC = 2 âˆ ABC

xo = 60o

Again ABCD is a Cyclic quadrilateral

âˆ B + âˆ D = 180o

(60o + yo = 180o)

y = 180o â€“ 60o = 120o

5. (a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If âˆ NYB = 50Â° and âˆ YNB = 20Â°, find âˆ MAN and the reflex angle MON.

(b) In the figure (ii) given below, O is the centre of the circle. If âˆ AOB = 140Â° and âˆ OAC = 50Â°, find
(i) âˆ ACB
(ii) âˆ OBC
(iii) âˆ OAB
(iv) âˆ CBA

Solution

(a) âˆ NYB = 50Â°, âˆ YNB = 20Â°.

In
âˆ†YNB,

âˆ NYB + âˆ YNB + âˆ YBN = 180o

50o + 20o + âˆ YBN = 180o

âˆ YBN + 70o = 180o

âˆ YBN = 180o â€“ 70o = 110o

But âˆ MAN = âˆ YBN

(Angles in the same segment)

âˆ MAN = 110o

Major arc MN subtend reflex âˆ MON at the

Centre and âˆ MAN at the remaining part of

the choice.

Reflex âˆ MAN at the remaining part of the circle

Reflex âˆ MON = 2 âˆ MAN = 2 Ã— 110o =220o

(b) (i)
âˆ AOB + reflex âˆ AOB = 360o

(Angles at the point)

140o + reflex âˆ AOB = 360o

Reflex âˆ AOB = 360o â€“ 140o = 220o

Now major arc AB subtends âˆ AOB + âˆ OBC = 360o

50o + 110o + 140o + âˆ OBC = 3600

300o + âˆ OBC = 3600

âˆ 300o + âˆ OBC = 3600

âˆ OBC = 360o â€“ 300o

âˆ OBC = 60o

âˆ OAC + âˆ ACB + âˆ AOB + âˆ OBC = 360o

50o + 110o + 140o + âˆ OBC = 360o

300o + âˆ OBC = 360o

âˆ OBC = 360o â€“ 300o

âˆ OBC =60o

(iii) in âˆ†OAB,

OA = OB

âˆ OAB + âˆ OBA = 180o

2 âˆ OAB = 180o â€“ 140o = 40o

âˆ OAB = 40o/2 = 200

But âˆ OBC = 60o

âˆ CBA = âˆ OBC – âˆ OBA

= 60o â€“ 20o = 40o

6. (a)In the figure (i) given below, O is the centre of the circle and âˆ PBA = 42Â°. Calculate the value of âˆ PQB
(b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that âˆ ECD = âˆ EDC = 32Â°, calculate
(i) âˆ CEF
(ii) âˆ COF.

Solution:

In âˆ†APB,

âˆ APB = 90Â° (Angle in a semi-circle)

But âˆ A + âˆ APB + âˆ ABP = 180Â° (Angles of a triangle)

âˆ A + 90Â° + 42Â°= 180Â°

âˆ A + 132Â° = 180Â°

â‡’ âˆ A = 180Â° â€“ 132Â° = 48Â°

But
âˆ A = âˆ PQB

(Angles in the same segment of a circle)

âˆ PQB = 48o

(b) (i) in âˆ†EDC,

(Ext, angle of a triangle is equal to the sum

of its interior opposite angels)

(ii) arc CF subtends âˆ COF at the centre and

âˆ CDF at the remaining part of the circle

âˆ COF = 2 âˆ CDF = 2 âˆ CDE

=2 Ã— 32o = 2 âˆ CDE

= 2 Ã— 32o = 64o

7. (a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, âˆ A = 35Â°, âˆ Q = 25Â°. Find (i) âˆ PRB (ii) âˆ PBR (iii) âˆ BPR.

(b) In the figure (ii) given below, it is given that âˆ ABC = 40Â° and AD is a diameter of the circle. Calculate âˆ DAC.

Solution

(a) (i) âˆ PRB = âˆ BAP

(Angles in the same segment of the circle)

âˆ´
âˆ PRB = 35Â° (âˆµ âˆ BAP = 35Â° given)

8. (a) In the figure given below, P and Q are centers of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcenter of triangle ABC in which AC = BC. Given that âˆ ACB = 56Â°, calculate

(i)âˆ CAB

(ii)âˆ OAC

Solution:

Given that

(a) Arc AB subtends âˆ APB at the center

and âˆ ACB at the remaining part of the circle

âˆ ACB = Â½ âˆ APB = Â½ Ã— 130o = 65o

But âˆ ACB + âˆ BCD = 180o

(Linear Pair)

65o + âˆ BCD = 180o

âˆ BCD = 180o â€“ 65o = 115o

Major arc BD subtends reflex âˆ BQD at the

Centre and âˆ BCD at the remaining part of

the circle

reflex âˆ BQD = 2
âˆ BCD

=2 Ã— 115o = 2300

But reflex âˆ BQD + x = 360o

(Angles at a point)

230o + x = 360o

x = 360o â€“ 230o = 130o

(b) Join OC

In âˆ†ABC,AC = BC

âˆ A = âˆ B

But âˆ A + âˆ B + âˆ C = 180o

âˆ A + âˆ A + 560 = 1800

2 âˆ A 180o â€“ 560 = 124o

âˆ A = 124/2 = 62o or âˆ CAB = 620

OC is the radius of the circle

OC bisects âˆ ACB

âˆ OCA = Â½ âˆ ACB = Â½ Ã— 56o = 28o

Now in âˆ†OAC

OA = OC

âˆ OAC = âˆ OCA = 28o

Exercise 15.2

1. If O is the center of the circle, find the value of x in each of the following figures (using the given information)

Solution:

From the figure

(i) ABCD is a cyclic quadrilateral

Ext. âˆ DCE = âˆ BAD

Now arc BD subtends âˆ BOD at the center

And âˆ BAD at the remaining part of the circle.

âˆ BOD = 2 âˆ BAD = 2 x

2 x = 150o (x = 750)

(ii) âˆ BCD + âˆ DCE = 180o

(Linear pair)

âˆ BCD + 800 = 180o

âˆ BCD = 1800 â€“ 800 = 100o

Arc BAD subtends reflex âˆ BOD at the

Centre and âˆ BCD at the remaining part of the circle

Reflex âˆ BOD = 2 âˆ BCD

Xo = 2 Ã— 100o = 200o

(iii) In âˆ†ACB,

âˆ CAB + âˆ ABC + âˆ ACB = 180o

(Angles of a triangle)

But
âˆ ACB = 90o

((Angles of a semicircle)

25o + 90o + âˆ ABC = 180o

=115o + âˆ ABC = 180o

âˆ ABC = 180o â€“ 1150=65o

âˆ ABC + âˆ ADC = 180o

(Opposite angles of a cyclic quadrilateral)

65o + xo =180o

xo = 180o -65o = 115o

2. (a) In the figure (i) given below, O is the center of the circle. If âˆ AOC = 150Â°, find (i) âˆ ABC (ii) âˆ ADC

(b) In the figure (i) given below, AC is a diameter of the given circle and âˆ BCD = 75Â°. Calculate the size of (i) âˆ ABC (ii) âˆ EAF.

Solution:

(a) Given,
âˆ AOC = 150Â° and AD = CD

We know that an angle subtends by an arc of a circle

at the center is twice the angle subtended by the same arc

at any point on the remaining part of the circle.

(i) âˆ AOC = 2 Ã— âˆ ABC

âˆ ABC = âˆ AOC/2 = 150o/2 = 75o

(ii) From the figure, ABCD is a cyclic quadrilateral

âˆ ABC + âˆ ADC = 180o

(Sum of opposite angels in a cyclic quadrilateral

Is 180o)

75o + âˆ ADC = 180o

âˆ ADC + 180o â€“ 75o

(b) (i) AC is the diameter of the circle

âˆ ABC = 90o (Angle in a semi-circle)

(ii) ABCD is a cyclic quadrilateral

âˆ BAD + âˆ BCD = 180o

âˆ BAD + 75o = 180o

(âˆ BCD = 75o)

âˆ BAD = 180o -75o = 105o

But âˆ EAF = âˆ BAD

(Vertically opposite angles)

âˆ EAF = 105o

3. (a) In the figure, (i) given below, if âˆ DBC = 58Â° and BD is a diameter of the circle, calculate:

(i) âˆ BDC (ii) âˆ BEC (iii) âˆ BAC

(b) In the figure (if) given below, AB is parallel to DC, âˆ BCE = 80Â° and âˆ BAC = 25Â°. Find:

Solution:

(a) âˆ DBC = 58Â°

BD is diameter

âˆ DCB = 90Â° (Angle in semi-circle)

(i) In âˆ†BDC

âˆ BDC + âˆ DCB + âˆ CBD = 180Â°

âˆ BDC = 180Â°- 90Â° â€“ 58Â° = 32Â°

(ii) BEC = 180o â€“ 32o = 148o

(iii) âˆ BAC = âˆ BDC = 32o

(Angles in same segment)

(b) in the figure, AB âˆ¥DC

âˆ BCE = 80o and âˆ BAC = 25o

ABCD is a cyclic Quadrilateral and DC is

Production to E

(i) Ext, âˆ BCE = interior âˆ A

80o = âˆ BAC + âˆ CAD

80o = 25o + âˆ CAD

âˆ CAD = 80o â€“ 25o = 55o

(ii) But âˆ CAD = âˆ CBD

(Alternate angels)

âˆ CBD = 55o

(iii) âˆ BAC = âˆ BDC

(Angles in the same segments)

âˆ BDC = 25o

(âˆ BAC = 25o)

Now AB âˆ¥ DC and BD is the transversal

âˆ BDC = âˆ ABD

âˆ ABD = 25o

âˆ ABC = âˆ ABD + âˆ CBD = 25o + 55o = 80o

But âˆ ABC + âˆ ADC = 180o

(opposite angles of a cyclic quadrilateral)

80o + âˆ ADC = 180o

âˆ ADC = 180o â€“ 80o = 100o

4. (a) In the figure given below, ABCD is a cyclic quadrilateral. If âˆ ADC = 80Â° and âˆ ACD = 52Â°, find the values of âˆ ABC and âˆ CBD.

(b) In the figure given below, O is the center of the circle. âˆ AOE =150Â°, âˆ DAO = 51Â°. Calculate the sizes of âˆ BEC and âˆ EBC.

Solution:

(a) In the given figure, ABCD is a cyclic quadrilateral

âˆ ADC = 80Â° and âˆ ACD = 52Â°

To find the measure of âˆ ABC and âˆ CBD

âˆ ABC + âˆ ADC = 180o

(Sum of opposite angles = 180o)

âˆ ABC + 80o = 180o

âˆ AOE = 150o, âˆ DAO = 51o

To find âˆ BEC and âˆ EBC

Ext. âˆ BEC = âˆ DAB = 51o

âˆ  AOE = 150o

Ref âˆ AOE = 360o â€“ 150o = 51o

âˆ AOE = 150o

Ref âˆ AOE = 360o â€“ 150o = 210o

Now arc ABE subtends âˆ AOE at the Centre

And âˆ ADE at the remaining part of the circle.

âˆ ADE = Â½ ref âˆ AOE = Â½ Ã— 210o = 105o

But Ext âˆ EBC = âˆ ADE = 105o

Hence âˆ BEC = 51o and âˆ EBC = 105o

5. (a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that âˆ BEF = 80Â°, find âˆ ABC.

(b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and âˆ B = 70Â°, find:

Solution:

Ext. âˆ FEB = âˆ ADF

ABCD is a parallelogram

âˆ B = âˆ D = âˆ ADF = 80Â°

or âˆ ABC = 80Â°

(b)In trapezium ABCD, AD || BC

(i) âˆ B + âˆ A = 180Â°

â‡’ 70Â° + âˆ A = 180Â°

â‡’ âˆ A = 180Â° â€“ 70Â° = 110Â°

(ii) ABCD is a cyclic quadrilateral

âˆ A + âˆ C = 180Â°

â‡’ 110Â° + âˆ C = 180Â°

â‡’ âˆ C = 180Â° â€“ 110Â° = 70Â°

âˆ BCD = 70Â°

6. (a) In the figure given below, O is the center of the circle. If âˆ BAD = 30Â°, find the values of p, q and r.

(a) In the figure given below, two circles intersect at points P and Q. If âˆ A = 80Â° and âˆ D = 84Â°, calculate

(i) âˆ QBC

(ii) âˆ BCP

Solution:

1. (i) ABCD is a cyclic quadrilateral

âˆ A + âˆ C = 180o

30o + p = 180o

P=180o â€“ 30o = 150o

(ii) Arc BD subtends âˆ BOD at the center

And âˆ BAD at the remaining part of the circle

âˆ BOD = 2 âˆ BAD

q = 2 Ã— 30o = 60o

âˆ BAD = âˆ BED are in the same segment of the circle

30o = r

r = 30o

1. Join PQ

âˆ A + âˆ QPD = 180o

7. (a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ.Given âˆ PQR = 58Â°, calculate (i) âˆ RPQ (ii) âˆ STP

(T is a point on the minor arc SP)

(b) In the figure given below, if âˆ ACE = 43Â° and âˆ CAF = 62Â°, find the values of a, b and c (2007)

Solution:

(a) In âˆ†PQR,

âˆ PRQ = 90Â° (Angle in a semi-circle) and âˆ PQR = 58Â°

âˆ RPQ = 90Â° â€“ âˆ PQR = 90Â° â€“ 58Â° = 32Â°

SR || PQ (given)

âˆ SRP = âˆ RPQ = 32o (Alternate angles)

Now PRST is a cyclic quadrilateral,

âˆ STP + âˆ SRP = 180o

âˆ STP = 180o â€“ 32o = 148o

(b) In the given figure,

âˆ ACE 43o and âˆ CAF = 620

Now, in âˆ†AEC

âˆ ACE + âˆ CAE + âˆ AEC = 180o

43o + 62o + âˆ AEC = 180o

105o + âˆ AEC = 180o

âˆ AEC = 180o â€“ 1050 = 75o

But âˆ ABD + âˆ AED = 1800

(sum of opposite angles of acyclic quadrilateral)

and âˆ AED = âˆ AEC

a + 75o = 180o

a = 180o â€“ 75o â€“ 105o

but âˆ EDF = âˆ BAE

(Angles in the alternate segment)

8. (a) In the figure (i) given below, AB is a diameter of the circle. If âˆ ADC = 120Â°, find âˆ CAB.

(b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.

Solution:

(a) Construction: Join BC, and AC then

Now in âˆ†DCF

Ext. âˆ 2 = x + z

and in âˆ†CBE

Ext. âˆ 1 = x + y

x + y + x + z = âˆ 1 + âˆ 2

2 x + y + z = 180o

But x : y : z = 3 : 4 : 5

x/y = Â¾ (y = 4/3 x)

x/z = 3/5 (z = 5/3 x.

Exercise 15.3

1. Find the length of the tangent drawn to a circle of radius 3cm, from a point distnt 5cm from the center.

Solution:

In a circle with center O and radius 3cm and p is at a distance of 5cm.

That is OT = 3 cm, OP = 5 cm

OT is the radius of the circle

OT âŠ¥ PT

Now in right âˆ† OTP, by Pythagoras axiom,

OP2 = OT2 + PT2

(5)2 = (3)2 + PT2

PT2 = (5)2 â€“ (3)2 = 25 â€“ 9 = 16 = (4)2

PT = 4 cm.

2. A point P is at a distance 13 cm from the center C of a circle and PT is a tangent to the given circle. If PT = 12 cm, find the radius of the circle.

Solution:

CP = 13 cm and tangent PT = 12 cm

CT is the radius and TP is the tangent

CT is perpendicular TP

Now in right angled triangle CPT,

CP2 = CT2 + PT2 [using Pythagoras axiom]

(13)2 = (CT)2 + (12)2

169 = (CT)2 + 144

(CT)2 = 169 -144 =25 = (5)2

CT = 5 cm.

Hence the radius of the circle is 5cm

3. The tangent to a circle of radius 6 cm from an external point P, is of length 8 cm. Calculate the distance of P from the nearest point of the circle.

Solution:

Radius of the circle = 6 cm

and length of tangent = 8 cm

Let OP be the distance

i.e. OA = 6 cm, AP = 8 cm .

OA âŠ¥ AP

Now In right âˆ†OAP,

OP2 = OA2 + AP2

(By Pythagoras axiom)

= (6)2 + (8)2

=36 + 64

= 100

= (10)2

OP = 10 cm.

4. Two concentric circles are of the radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

Solution:

Two concentric circles with center O

OP and OB are the radii of the circles respectively, then

OP = 5 cm, OB = 13 cm.

Ab is the chord of outer circle which touches the inner circle at P.

OP is the radius and APB is the tangent to the inner circle.

In the right angled triangle OPB, by Pythagoras axiom,

OB2 = OP2 + PB2

132 = 52 + PB2

169 = 25 + PB2

PB2 = 169 â€“ 25

= 144

PB = 12 cm

But P is the mid-point of AB.

AB = 2PB

= 24 cm

5. Two circles of radii 5 cm and 2-8 cm touch each other. Find the distance between their centers if they touch :

(i) externally

(ii) internally.

Solution:

Radii of the circles are 5 cm and 2.8 cm.

i.e. OP = 5 cm and CP = 2.8 cm.

(i) When the circles touch externally,

then the distance between their centers = OC = 5 + 2.8

= 7.8 cm.

(ii) When the circles touch internally,

then the distance between their centers = OC = 5.0 â€“ 2.8

= 2.2 cm

6. (a) In figure (i) given below, triangle ABC is circumscribed, find x.
(b) In figure (ii) given below, quadrilateral ABCD is circumscribed, find x.

Solution:

(a) From A, AP and AQ are the tangents to the circle

âˆ´ AQ = AP = 4cm

But AC =12 cm

CQ = 12 â€“ 4 = 8 cm.

From B, BP and BR are the tangents to the circle

BR = BP = 6 cm.

Similarly, from C,

CQ and CR the tangents

CR = CQ = 8 cm

x = BC = BR + CR = 6 cm + 8 cm = 14 cm

(b) From C, CR and CS are the tangents to the circle.

CS = CR = 3 cm.

But BC = 7 cm.

BS = BC â€“ CS = 7 â€“ 3 =4 cm.

Now from B,BP and BS are the tangents to the circle.

BP = BS = 4 cm

From A, AP and AQ are the tangents to the circle.

AP = AQ = 5cm

x = AB = AP + BP = 5 + 4

= 9 cm

7. (a) In figure (i) given below, quadrilateral ABCD is circumscribed; find the perimeter of quadrilateral ABCD.

(b) In figure (ii) given below, quadrilateral ABCD is circumscribed and AD âŠ¥ DC ; find x if radius of incircle is 10 cm.

Solution:

(a) From A, AP and AS are the tangents to the circle

âˆ´AS = AP = 6

From B, BP and BQ are the tangents

âˆ´BQ = BP = 5

From C, CQ and CR are the tangents

CR = CQ

From D, DS and DR are the tangents

DS = DR = 4

Therefore, perimeter of the quadrilateral ABCD

= 6 + 5 + 5 + 3 + 3 + 4 + 4 + 6

= 36 cm

(b) in the circle with center O, radius OS = 10 cm

PB = 27 cm, BC = 38 cm

SD = OS = 10 cm.

Now from D,DR and DS are the tangents

To the circle

DR = DS = 10 cm

From B, BP and BQ are tangents to the circle.

BQ = BP = 27 cm.

CQ = CB â€“ BQ = 38 â€“ 27 = 11 cm.

Now from C, CQ and CR are the tangents to the circle

CR = CQ = 11 cm.

DC = x =DR + CR

= 10 + 11 = 21 cm

8. (a) In the figure (i) given below, O is the center of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, find the radius of the circle.

(b) In the figure (ii) given below, from an external point P, tangents PA and PB are drawn to a circle. CE is a tangent to the circle at D. If AP = 15 cm, find the perimeter of the triangle PEC.

Solution:

(i) Join OB

âˆ OBA = 90Â°

(Radius through the point of contact is

perpendicular to the tangent)

OB2 = OA2 â€“ AB2

r2 = (r + 7.5)2 â€“ 152

r2 = r2 + 56.25 + 15r â€“ 225

15r = 168.75

r = 11.25

Hence, radius of the circles = 11.25 cm

(ii) In the figure, PA and PB are the tangents

Drawn from P to the circle.

CE is tangent at D

AP = 15 cm

PA and PB are tangents to the circle

AP = BP = 15 cm

Similarly EA and ED are tangents

EA = ED

Similarly BC = CD

Now perimeter of triangle PEC,

= PE + EC + PC

= PE + ED + CD + PC

PE + EA + CB + PC

(ED = EA and CB = CD)

=AP + PB = 15 + 15

= 30 cm.

9. (a) If a, b, c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by

r = /frac (a + b â€“ c) â€“ (2)

(b) In the given figure, PB is a tangent to a circle with center O at B. AB is a chord of length 24 cm at a distance of 5 cm from the center. If the length of the tangent is 20 cm, find the length of OP.

Solution:

(a) Let the circle touch the sides BC, CA and AB

of the right triangle ABC at points D, E and F respectively,

where BC = a, CA = b

and AB = c (as showing in the given figure).

As the lengths of tangents drawn from an

External point to a circle are equal

AE = AF, BD = BF and CD = DE

OD âŠ¥ BC and OE âŠ¥ CA

OD âŠ¥ BC and OE âŠ¥ CA

ODCE is a square of side r

DC = CE = r

AF = AE = AC â€“ EC = b â€“ r and

BF = BD = BC â€“ DC = a â€“ r

Now, AB = AF + BF

C = (b â€“ r) + (a â€“ r)

2r = a + b â€“ c

r = a + b â€“ c/2

OP2 = 400 + 169

OP = aâˆ’âˆš569cm

10. Three circles of radii 2 cm, 3 cm and 4 cm touch each other externally. Find the perimeter of the triangle obtained on joining the centers of these circles.

Solution:

Three circles with centers A, B and C touch each other externally

at P, Q and R respectively and the radii of these circles are

2 cm, 3 cm and 4 cm.

By joining the centers of triangle ABC formed in which,

AB = 2 + 3 = 5 cm

BC = 3 + 4 = 7 cm

CA = 4 + 2 = 6 cm

Therefore, perimeter of the triangle ABC = AB + BC + CA

= 5 + 7 + 6

= 18 cm

Chapter test

1. (a) In the figure (i) given below, triangle ABC is equilateral. Find âˆ BDC and âˆ BEC.Â (b) In the figure (ii) given below, AB is a diameter of a circle with center O. OD is perpendicular to AB and C is a point on the arc DB. Find âˆ BAD and âˆ ACDÂ

Solution:

(a) triangle ABC is an equilateral triangle

Each angle = 60o

âˆ A = 60o

But âˆ A = âˆ D

(Angles in the same segment)

âˆ D = 600

Now ABEC is a cyclic quadrilateral,

âˆ A = âˆ E = 180o

60o + âˆ E = 180o

600 + âˆ E = 180o (âˆ E = 180o â€“ 60o

âˆ E = 120o

Hence âˆ BDC = 60o and âˆ BEC = 120o

1. AB is diameter of circle with centre O.

OD âŠ¥ AB and C is a point on arc DB.

In âˆ†AOD, âˆ AOD = 900

OA = OD (raddi of the semi â€“ circle)

But âˆ OAD + âˆ ODA = 90o

âˆ OAD + âˆ ODA = 90o

âˆ OAD = 90o/2 = 450

(ii) Arc AD subtends âˆ AOD at the centre and

âˆ ACD at the remaining part of the circle

âˆ AOD = 2 âˆ ACD

90o = 2 âˆ ACD (OD âŠ¥ AB)

âˆ ACD = 90o/2 = 45o

2. (a) In the figure given below, AB is a diameter of the circle. If AE = BE and âˆ ADC = 118Â°, findÂ (i) âˆ BDC (ii)Â âˆ CAE

(B) inthe figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and âˆ BCD = 140Â°, find âˆ AED and âˆ EBD. Also Prove that OE is parallel to BD.Â

Solution:

(a) Join DB, CA and CB.Â âˆ ADC = 118Â° (given)Â and âˆ ADB = 90Â°Â

(Angles in a semi-circle)

= 1180 â€“ 90o = 28o

âˆ ABCD is a cyclic quadrilateral)

âˆ ADC + âˆ ABC = 180o

118o + âˆ ABC = 180o

âˆ ABC = 180o â€“ 118o = 62o

But in âˆ†AEB

âˆ AEB = 90o

(Angles in a semi-circle)

âˆ EAB = âˆ ABE (AE = BE)

âˆ EAB + âˆ ABE = 90o

âˆ EAB = 90o Ã— Â½ = 45o

âˆ CBE = âˆ ABC + âˆ ABE

= 62o + 45o = 107o

But AEBD is a cyclic quadrilateral

âˆ CAE + âˆ CBE = 180o

âˆ CAE + 107o = 180o

âˆ CAE = 180o â€“ 107o = 73o

(b) AB is the diameter of semi-circle ABCDE

With center O.AE = ED and âˆ BCD = 140o

(i) âˆ BCD + âˆ BED = 180o

140o + âˆ BED = 180o

âˆ BED = 180o â€“ 140o = 400

But âˆ AED = 90o

(Angles in a semi circle)

âˆ AED = âˆ AEB + âˆ BED

= 90o + 40o = 130o

(ii) Now in cyclic quadrilateral AEDB

âˆ AED + âˆ DBA = 180o

130o + âˆ DBA =180o

âˆ BDA = 180o â€“ 130o = 50o

Chord AE = ED (given)

âˆ DBE = âˆ EBA

But âˆ DBE + âˆ EBA = 50o

DBE + âˆ DBE = 50o

2âˆ DBE = 50o

âˆ DBE = 25o or âˆ EBD = 25o

In âˆ†OEB,OE = OB

âˆ OEB = âˆ EBO = âˆ DBE

But these are ultimate angles

OE âˆ¥ BD

3. a) In the figure (i) given below, O is the centre of the circle. Prove that âˆ AOC = 2 (âˆ ACB + âˆ BAC).Â (b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z

Solution :

(a) Given: O is the center of the circle.Â To Prove : âˆ AOC = 2 (âˆ ACB + âˆ BAC).Â Proof: In âˆ†ABC,Â âˆ ACB + âˆ BAC + âˆ ABC = 180Â° (Angles of a triangle)Â

âˆ ABC = 180o â€“ (âˆ ACB + âˆ BAC)â€¦.(i)

In the circle, arc AC subtends âˆ AOC at

The center and âˆ ABC at the remaining part of the circle.

Reflex âˆ AOC = 2 âˆ ABC â€¦(ii)

Reflex AOC = 2 { (180o â€“ (ACB + BAC)}

But âˆ AOC = 360o â€“ 2(âˆ ACB + âˆ BAC)

But âˆ AOC = 360o â€“ reflex âˆ AOC

=360 â€“ (360o â€“ 2(âˆ ACB + âˆ BAC)

=360o â€“ 360o + 2 (âˆ ACB + âˆ BAC)

=2 (âˆ ACB + âˆ BAC)

Hence âˆ AOC = 2 (âˆ ACB + âˆ BAC)

(b) Given : in the figure, O is the center of the circle

To Prove : Ã— + y = z.

Proof : Arc BC subtends âˆ AOB at the center and âˆ BEC at the remaining part of the circle.

âˆ BOC = 2 âˆ BEC

But âˆ BEC = âˆ BDC

(Angles in the same segment)

âˆ BOC = âˆ BEC + âˆ BDC â€¦â€¦. (ii)

Similarly in âˆ†ABD

Ext. âˆ BDC = x + âˆ ABD

= x + âˆ EBD â€¦â€¦â€¦â€¦.(iii)

Substituting the value of (ii) and (iii) in (i)

âˆ BOC = y – âˆ EBD + x + âˆ EBD = x + y

Z = x + y

Â