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MATHEMATICS - I         

 

 

 

 

 

 

 

 

 

 

Written as per the latest textbook prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune.

Std. XI Sci. & Arts

Printed at: Quarterfold Printabilities, Navi Mumbai 

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

P.O. No. 192310TEID: 13664

 Balbharati Registration No.: 2018MH0022

PERFECT

Written as per the new textbook Exhaustive coverage of entire syllabus Topic-wise distribution of textual questions and practice problems at the start of

every chapter. Precise theory for every topic Covers answers to all exercises and miscellaneous exercises given in the textbook. All derivations and theorems covered Includes additional problems for practice and MCQs Illustrative examples for selective problems Recap of important formulae at the end of the book Activity Based Questions covered in every chapter Smart Check to enable easy rechecking of solutions ‘Competitive Corner’ presents questions from prominent Competitive Examinations

Salient Features

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“The only way to learn Mathematics is to do Mathematics” – Paul Halmos “Mathematics – I : Std. XI” forms a part of ‘Target Perfect Notes’ prepared as per the New Textbook. It is a complete and thorough guide critically analysed and extensively drafted to boost the students’ confidence. The book provides answers to all textbook questions included in exercises as well as miscellaneous exercises. Apart from these questions, we have provided ample questions for additional practice to students based on every exercise of the textbook. Only the final answer has been provided for such additional practice questions. At the start of the chapter, we have provided a table to birfucate the textbook questions and additional practice questions as per the different type of problems/concepts in the chapter. This will help in systematic study of the entire chapter. Precise theory has been provided at the required places for better understanding of concepts. Further, all derivations and theorems have been covered wherever required. A recap of all important formulae has been provided at the end of the book for quick revision. We have also included activity based questions in every chapter. We have newly introduced ‘competitive corner’ in this book wherein we have included questions from prominent competitive exams. It will help students to get an idea about the type of questions that are asked in Competitive Exams. We all know that there are certain sums that can be solved by multiple methods. Besides, there are also other ways to check your answer in Maths. ‘Smart Check’ has been included to help you understand how you can check the correctness of your answer. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Pls write to us on: mail@targetpublications.org A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants!

From, Publisher Edition: First

This reference book is transformative work based on textbook Mathematics - I; First edition: 2019 published by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

Disclaimer

PREFACE 

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Chapter No. Chapter Name Page No.

1 Angle and its Measurement 1

2 Trigonometry – I 24

3 Trigonometry – II 65

4 Determinants and Matrices 107

5 Straight Line 181

6 Circle 230

7 Conic Sections 263

8 Measures of Dispersion 328

9 Probability 366

Important formulae 416

 

CONTENTS

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230  230

Type of Problems Exercise Q. Nos.

Finding the equation of the circle when centre and radius are given

6.1 Q.1 (i, ii, iii)

Practice Problems (Based on Exercise 6.1)

Q.1, 2 (i, ii, iii)

Finding the centre and radius of the given circle

6.1 Q.2

Practice Problems

(Based on Exercise 6.1) Q.3

6.2 Q.1

Practice Problems

(Based on Exercise 6.2) Q.1, 3

Miscellaneous Exercise 6 Q.II (1, 2)

(Based on Miscellaneous Exercise 6)

Q.1

Finding the equation of the circle for the given conditions by using centre-radius form

6.1 Q.1 (iv), 3, 4, 6, 8

Practice Problems

(Based on Exercise 6.1) Q.2 (iv), 4, 5, 6, 7, 9, 11, 13, 14

Miscellaneous Exercise 6 Q.II (3, 10)

(Based on Miscellaneous Exercise 6)

Q.2

Finding the equation of the circle when endpoints of a diameter are given

Practice Problems

(Based on Exercise 6.1)

Q.8

Finding the equation of the circle for the given conditions by using diameter form

6.1 Q.5, 7

Practice Problems

(Based on Exercise 6.1) Q.10, 12

Miscellaneous Exercise 6 Q.II (4, 6)

(Based on Miscellaneous Exercise 6)

Q.3

Show that the given equation represents a circle

6.2 Q.2

Practice Problems

(Based on Exercise 6.2) Q.2,3

Finding the equation of the circle passing through three points

6.2 Q.3

Practice Problems

(Based on Exercise 6.2) Q.4

Show that the given points are concyclic

6.2 Q.4

Practice Problems

(Based on Exercise 6.2) Q.5, 6

Miscellaneous Exercise 6 Q.II (5)

(Based on Miscellaneous Exercise 6)

Q.4

Circle6

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Chapter 06: Circle

Parametric equation of a circle 6.3 Q.1, 2

Practice Problems (Based on Exercise 6.3)

Q.1, 2, 3

Equation of a tangent to the given circle

6.3 Q.3, 5

Practice Problems (Based on Exercise 6.3)

Q.4, 5

Miscellaneous Exercise 6 Q.II (8, 16, 17, 18, 20)

(Based on Miscellaneous Exercise 6)

Q.5

Show that the given line is a tangent to the given circle and find their point of contact

6.3 Q.4

Practice Problems (Based on Exercise 6.3)

Q.6, 7

Miscellaneous Exercise 6 Q.II (7, 19)

Director circle Miscellaneous Exercise 6 Q.II (9, 25)

Finding the lengths of the intercepts made by the circle on the coordinate axes

Miscellaneous Exercise 6 Q.II (11)

(Based on Miscellaneous Exercise 6)

Q.12

Circles touching internally/externally, their point of contact and equations of their common tangents

Miscellaneous Exercise 6 Q.II (12, 13)

(Based on Miscellaneous Exercise 6)

Q.13, 14

Length of tangent segment Miscellaneous Exercise 6 Q.II (14, 15)

(Based on Miscellaneous Exercise 6)

Q. 10, 11

Condition of tangency Miscellaneous Exercise 6 Q.II (21, 22, 23, 24)

(Based on Miscellaneous Exercise 6)

Q. 6, 7, 9

Tangents from a point to the circle Miscellaneous Exercise 6 Q.II (25, 26)

(Based on Miscellaneous Exercise 6)

Q. 8

A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point remains constant. The fixed point is called centre of the circle and the fixed distance is called radius of the circle.

Basic Properties: 1. If C is the centre, AB is a

chord and CD AB at D, then AD = DB

2. If O is the centre, AB = CD OP AB at P and OQ CD at Q, then OP = OQ [ Congruent chords are equidistant from the

centre]

O

C D Q

A B P

C

A B D

Let’s Recall

Syllabus

Equation of a circle and its different forms. Equation of Tangent to a circle.

Condition for tangency. Director circle.

Plane

P

O

(Moving point)

Fixed point

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Std. XI : Perfect Maths ‐ I

3. If AB is a diameter, P is a point on circle except A, B, then APB = 90.

4. If ABCD is a quadrilateral

inscribed in a circle, then A + C = 180 and B + D = 180

5. Tangent is always to radius of the circle. Standard form: If P(x, y) is any point on the circle then the equation of a circle with centre at the origin and radius ‘r’ is given by x2 + y2 = r2. The above equation is the standard form of a circle. Centre radius form: If P(x, y) is any point on the circle, then the equation of a circle with its centre at (h, k) and radius ‘r’ is given by distance formula (x h)2 + (y k)2 = r2

The above equation is the centreradius form of equation of a circle. Diameter form: If P(x, y) is any point on the circle, then the equation of a circle described on the line segment joining the points A(x1, y1) and B (x2, y2) as a diameter is given as follows: From the figure, seg. AP seg. BP (Slope of seg. AP).(Slope of seg. BP) = 1

1 2

1 2

. 1

y y y y

x x x x

(y y1) (y y2) = (x x1) (x x2) (x x1) (x x2) + (y y1) (y y2) = 0 The above equation is the diameter form of

equation of a circle, where (x1, y1) and (x2, y2) are the endpoints of diameter of the circle.

1. Construct a circle in fourth quadrant having

radius 3 and touching Y-axis. How many such circles can be drawn? (Textbook page no. 129)

Solution: Since, the position of the circle with respect to

X-axis is not mentioned, we can draw infinite circles in the fourth quadrant.

C(h, k)

P(x, y) r

A B

P (x, y)

(x1, y1) (x2, y2) C

D

A B

C

B A

P

X

3

Y

X

Y

O

C

M l

O(0, 0)

P(x, y) r

Let’s Study

Try This

i. When a circle with centre (h, k) touchesX-axis, its radius = | y-coordinate of centre|

= | k | ii. When a circle with centre (h, k) touches

Y-axis, its radius = | x-coordinate of centre| = | h | iii. Circles touching both the axes: iv. The point of intersection of diameters of a

circle is the centre of the circle.

Remember This

Y

O X

(h, k)

A (h, 0)

k

h

Y

O X

(h, k)B(0, k)

( r, r) (r, r)

(r, r) ( r, r)

r

X X

Y

Y

Equation of a circle and its different forms

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Chapter 06: Circle

1. Find the equation of a circle with i. centre at origin and radius 4. ii. centre at ( 3, 2) and radius 6. iii. centre at (2, − 3) and radius 5. iv. centre at (− 3, − 3) passing through point

(− 3, − 6). Solution: i. The equation of a circle with centre at origin

and radius ‘r’ is given by x2 + y2 = r2 Here, r = 4 the required equation of the circle is x2 + y2 = 42 i.e., x2 + y2 = 16. ii. The equation of a circle with centre at (h, k)

and radius ‘r’ is given by (x h)2 + (y k)2 = r2 Here, h = 3, k = 2 and r = 6 the required equation of the circle is [x (3)]2 + [y (2)]2 = 62 (x + 3)2 + (y + 2)2 = 36 x2 + 6x + 9 + y2 + 4y + 4 36 = 0 x2 + y2 + 6x + 4y 23 = 0 iii. The equation of a circle with centre at (h, k)

and radius ‘r’ is given by (x – h)2 + (y – k)2 = r2 Here, h = 2, k = –3 and r = 5 the required equation of the circle is (x – 2)2 + [y – (–3)]2 = 52 (x – 2)2 + (y + 3)2 = 25 x2 – 4x + 4 + y2 + 6y + 9 – 25 = 0 x2 + y2 – 4x + 6y – 12 = 0 iv. Centre of the circle is C (– 3, – 3) and it passes

through the point P (– 3, – 6). By distance formula,

Radius (r) = CP = 2 2[ 3 ( 3)] + [ 6 ( 3)]

= 2 2( 3 + 3) + ( 6 + 3)

= 2 20 + ( 3)

= 9 = 3 The equation of a circle with centre at (h, k)

and radius ‘r’ is given by (x – h)2 + (y – k)2 = r2

Here, h = – 3, k = –3, r = 3 the required equation of the circle is [x – (– 3)]2 + [y – (– 3)]2 = 32 (x + 3)2 + (y + 3)2 = 9 x2 + 6x + 9 + y2 + 6y + 9 – 9 = 0 x2 + y2 + 6x + 6y + 9 = 0 2. Find the centre and radius of the following

circles: i. x2 + y2 = 25 ii. (x 5)2 + (y 3)2 = 20

iii. 2

1

2

x + 2

1

3

y = 1

36

Solution: i. Given equation of the circle is x2 + y2 = 25 x2 + y2 = (5)2 Comparing this equation with x2 + y2 = r2, we get r = 5 Centre of the circle is (0, 0) and radius of the

circle is 5. ii. Given equation of the circle is (x 5)2 + (y 3)2 = 20

(x 5)2 + (y 3)2 = 2

20

Comparing this equation with (x h)2 + (y k)2 = r2, we get h = 5, k = 3 and r = 20 = 2 5 Centre of the circle = (h, k) = (5, 3) and radius of the circle = 2 5 . iii. Given equation of the circle is

2

1

2

x + 2

1

3

y = 1

36

2

1

2

x + 2

1

3

y =2

1

6

Comparing this equation with (x h)2 + (y k)2 = r2, we get

h = 1

2, k =

1

3

and r =

1

6

Centre of the circle = (h, k ) = 1 1,

2 3

and radius of the circle = 1

6.

Exercise 6.1

C (– 3, – 3)

P (– 3, – 6)r

If the point (–3, –6) satisfies x2 + y2 + 6x + 6y + 9 = 0, then our answer is correct. L.H.S. = x2 + y2 + 6x + 6y + 9 = (–3)2 + (–6)2 + 6(3) + 6(–6) + 9 = 9 + 36 18 – 36 + 9 = 0 = R.H.S. Thus, our answer is correct.

Smart Check

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Std. XI : Perfect Maths ‐ I

3. Find the equation of the circle with centre i. at (a, b) and touching the Y-axis. ii. at ( 2, 3) and touching the X-axis. iii. on the X-axis and passing through the

origin having radius 4. iv. at (3,1) and touching the line

8x − 15y + 25 = 0 Solution: i. Since circle is touching the Y-axis, radius of circle is X-co-ordinate of the centre. r = a The equation of a circle with centre at (h, k)

and radius r is given by (x h)2 + (y k)2 = r2 Here h = a, k = b the required equation of the circle is (x a)2 + (y b)2 = a2 x2 2ax + a2 + y2 2by + b2 = a2 x2 + y2 2ax 2by + b2 = 0 ii. Since circle is touching the X-axis, radius of

circle is Y co-ordinate of the centre. r = 3 The equation of a circle with centre at (h, k)

and radius r is given by (x h)2 + (y k)2 = r2 Here h = 2, k = 3 the required equation of the circle is (x + 2)2 + (y 3)2 = 32 x2 + 4x + 4 + y2 6y + 9 = 9 x2 + y2 + 4x 6y + 4 = 0

iii. Let the co-ordinates of the centre of the required circle be C (h, 0).

Since, the circle passes through the origin i.e., O (0,0).

OC = radius

2 2(h 0) + (0 0) = 4

h2 = 16 h = 4 the co-ordinates of the centre are (4, 0) or

(– 4, 0). The equation of a circle with centre at (h, k)

and radius r is given by (x – h)2 + (y – k)2 = r2 Here, h = 4, k = 0, r = 4 The required equation of the circle is (x – 4 )2 + (y – 0)2 = 42 or (x + 4)2 + (y – 0)2 = 42 x2 – 8x + 16 + y2 = 16 or x2 + 8x + 16 + y2 = 16 x2 + y2 – 8x = 0 or x2 + y2 + 8x = 0 iv. Centre of the circle is C (3, 1). Let the circle touch the line

8x – 15y + 25 = 0 at point M. CM = radius (r) CM = Length of perpendicular from centre

C (3, 1) on the line 8x – 15y + 25 = 0

= 2 2

8(3) 15(1) + 25

8 + ( 15)

= 24 15 + 25

64 + 225

= 34

289

r = 34

17 = 2

The equation of a circle with centre at (h, k) and radius r is given by

(x – h)2 + (y – k)2 = r2 Here, h = 3, k = 1 and r = 2 the required equation of the circle is (x – 3)2 + (y – 1)2 = 22 x2 – 6x + 9 + y2 – 2y + 1 = 4 x2 + y2 – 6x – 2y + 10 – 4 = 0 x2 + y2 – 6x – 2y + 6 = 0

C(a, b)

r = a

Y

O

Y

X X

C(2, 3)

r = 3

X XO

Y

Y

4 4 (4, 0) (4, 0)

X

Y

C(3, 1)

M 8x – 15y + 25 = 0

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Chapter 06: Circle

4. Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.

Solution: Given equations of diameters are 2x + y = 6

and 3x +2y = 4. Let C (h, k) be the centre of the required circle Since, point of intersection of diameters is

centre of the circle. x = h, y = k Equations of diameters become 2h + k = 6 …(i) and 3h + 2k = 4 …(ii) By (ii) 2 (i), we get h = 8 h = 8 Substituting h = 8 in (i), we get 2(8) + k = 6 k = 6 16 k = 10 Centre of the circle is C (8, 10) and radius,

r = 9 The equation of a circle with centre at (h, k)

and radius r is given by (x h)2 + (y k)2 = r2

Here h = 8, k = 10 the required equation of the circle is (x 8)2 + (y +10)2 = 92 x2 16x + 64 + y2 + 20y + 100 = 81 x2 + y2 16x + 20y + 100 + 64 81 = 0 x2 + y2 16x + 20y + 83 = 0 5. If y = 2x is a chord of the circle

x2 + y2 10x = 0, find the equation of the circle with this chord as diameter.

Solution: y = 2x is the chord of the given circle. It satisfies the equation of given circle. Substituting y = 2x in x2 + y2 10x = 0, we get x2 + (2x)2 10x = 0 x2 + 4x2 10x = 0 5x2 10x = 0 5x (x 2) = 0 x = 0 or x = 2 When x = 0, y = 2x = 2 (0) = 0

A (0, 0) When x = 2, y = 2x = 2 (2) = 4 B (2, 4) End points of chord AB are A(0, 0) and

B(2, 4). Since, chord AB is diameter of the required

circle. The equation of a circle having (x1, y1) and

(x2, y2) as end points of diameter is given by (x x1) (x x2) + (y y1) (y y2) = 0 Here, x1 = 0, y1 = 0, x2 = 2, y2 = 4 the required equation of the circle is (x 0) (x 2) + (y 0) (y 4 ) = 0 x2 2x + y2 4y = 0 x2 + y2 2x 4y = 0 6. Find the equation of a circle with radius

4 units and touching both the co-ordinate axes having centre in third quadrant.

Solution: Radius of the circle = 4 units Since, the circle touches both the coordinate

axes and its centre is in third quadrant. the centre of the circle is C (– 4, – 4).

The equation of a circle with centre at (h, k)

and radius r is given by (x – h)2 + (y – k)2 = r2 Here, h = – 4, k = – 4, r = 4 the required equation of the circle is [x – (– 4)]2 + [y – (– 4)]2 = 42 (x + 4)2 + (y + 4)2 = 16 x2 + 8x + 16 + y2 + 8y + 16 – 16 = 0 x2 + y2 + 8x + 8y + 16 = 0 7. Find the equation of circle passing through

the origin and having intercepts 4 and − 5 on the co-ordinate axes.

Solution: Let the circle intersect X-axis at point A and

intersect Y-axis at point B. the co-ordinates of point A are (4, 0) and the

co-ordinates of point B are (0, – 5).

C(h, k)

2x + y = 6

3x + 2y = 4

B

y = 2x

A

O

Y

4

C(– 4, 4)

Y

X X

4

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Std. XI : Perfect Maths ‐ I

Since, AOB is a right angle. AB represents the diameter of the circle. The equation of a circle having (x1, y1) and

(x2, y2) as end points of diameter is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Here, x1 = 4, y1 = 0, x2 = 0, y2 = – 5 the required equation of the circle is (x – 4) (x – 0) + (y – 0) [y – (– 5)] = 0 x(x – 4) + y(y + 5) = 0 x2 – 4x + y2 + 5y = 0 x2 + y2 – 4x + 5y = 0 8. Find the equation of a circle passing

through the points (1,− 4), (5, 2) and having its centre on the line x − 2y + 9 = 0 .

Solution:

Let C (h, k) be the centre of the required circle

which lies on the line x – 2y + 9 = 0. Equation of line becomes h – 2k + 9 = 0 …(i) Also, the required circle passes through the

points A(1, – 4) and B(5, 2). CA = CB = radius CA = CB By distance formula,

2 2 2 2(h 1) + [k ( 4)] = (h 5) + (k 2)

Squaring both the sides, we get (h – 1)2 + (k + 4)2 = (h – 5)2 + (k – 2)2 h2 – 2h + 1 + k2 + 8k + 16 = h2 – 10h + 25

+ k2 – 4k + 4 – 2h + 8k + 17 = – 10h – 4k + 29 8h + 12k – 12 = 0 2h + 3k – 3 = 0 …(ii) By (ii) – (i) 2, we get 7k = 21

k = 3

Substituting k = 3 in (i), we get

h – 2(3) + 9 = 0

h – 6 + 9 = 0

h = – 3

Centre of the circle is C (– 3, 3).

radius (r) = CA

= 2 2[1 ( 3)] + ( 4 3)

= 2 24 + ( 7)

= 16 + 49

= 65

The equation of a circle with centre at (h, k) and radius r is given by

(x – h)2 + (y – k)2 = r2

Here, h = – 3, k = 3, r = 65

the required equation of the circle is

[x – (–3)]2 + (y – 3)2 = 2

65

(x + 3)2 + (y – 3)2 = 65

x2 + 6x + 9 + y2 – 6y + 9 – 65 = 0

x2 + y2 + 6x – 6y – 47 = 0

The equation of a circle with centre at (h, k) and radius r is

(x h)2 + (y k)2 = r2

x2 2hx + h2 + y2 2ky + k2 r2 = 0

x2 + y2 2hx 2ky + (h2 + k2 r2) = 0

Replacing h by ( g), k by ( f) and

h2 + k2 – r2 by c, we get

x2 + y2 + 2gx + 2fy + c = 0

The above equation is the general equation of circle.

Now by adding 2 2g f to both sides of the

general equation, we get

(x2 + 2gx + g2) + (y2 + 2fy + f 2) + c = 2 2g f

(x + g)2 + (y + f)2 = g2 + f 2 c

[x (g)]2 + [y (f)]2 = 22 2g f c

The above equation is of the form

(x h)2 + (y k)2 = r2

Centre of the circle is C (g, f )

and radius of the circle is r = 2 2g + f c

A(4, 0)

Y

X O (0, 0)

5

Y

X 4

B(0, – 5)

C(h, k)

x – 2y + 9 = 0

B(5, 2)

A(1, 4)

Let’s Study

General equation of a circle

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Chapter 06: Circle

(Textbook page no. 130)

Consider the general equation of a circle x2 + y2 + 2gx + 2fy + c = 0

x2 + 2gx + 2g + y2 + 2fy + 2f = 2g + 2f c

…[Adding g2 + f 2 on both the sides]

2 2

x y g f = g2 + f2 c

2 2 x yg f = 2

2 2g f c

(use centre radius form of equation of the circle.) Therefore, centre of the circle is ( g, f ) and

radius is 2 2g f c .

1. Construct a circle whose equation is

x2 + y2 − 4x + 6y − 12 = 0. Find the area of the circle. (Textbook page no. 129)

Solution: Given equation of circle is x2 + y2 – 4x + 6y – 12 = 0

Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get

2g = –4, 2f = 6, c = –12 g = –2, f = 3, c = –12 Centre of the circle = (–g, –f) = (2, – 3)

Radius (r) = 2 2g + f c

= 2 22 3 12

= 4 9 12

= 25 = 5

Area of circle = r2 = 22

7 (5)2

= 550

7 sq. units

2. Check whether the following equations

represent a circle. If, so then find its centre and radius.

i. x2 + y2 – 6x – 4y + 9 = 0 ii. x2 + y2 − 8x + 6y + 29 = 0 iii. x2 + y2 + 7x – 5y + 15 = 0

(Textbook page no. 130) Solution: i. Given equation of the circle is x2 + y2 – 6x – 4y + 9 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 6, 2f = – 4 and c = 9 g = 3, f = – 2 and c = 9 Now, g2 + f 2 – c = (– 3)2 + (– 2)2 – 9 = 9 + 4 – 9 = 4 > 0 the given equation represents a circle. Centre of the circle = ( g, f) = (3, 2)

and radius of the circle = 2 2g f c

= 2 2( 3) ( 2) 9

= 9 4 9

= 4 = 2 ii. Given equation of the circle is x2 + y2 − 8x + 6y + 29 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 8, 2f = 6 and c = 29 g = 4, f = 3 and c = 29 Now, g2 + f 2 – c = (– 4)2 + 32 – 29 = 16 + 9 – 29 = – 4 < 0 the given equation does not represent a circle. iii. Given equation of the circle is x2 + y2 + 7x – 5y + 15 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 7, 2f = – 5 and c = 15

g = 7

2 , f = –

5

2 and c = 15

Now, g2 + f 2 – c = 2

7

2

+ 2

5

2

– 15

= 49

4 +

25

4 – 15

= 74

4 –15

= 74 60

4

=

7

2 > 0

O XX

Y

Y

C(2, –3) 5

Try This

Textual Activity

i. If 2 2g f c 0 , then the equation

x2 + y2 + 2gx + 2fy + c = 0 represents a circle in the xy plane.

ii. If 2 2g f c 0 , then the equation

x2 + y2 + 2gx + 2fy + c = 0 represents a point which is a true degenerate conic andis the limiting position (radius is 0).

iii. If 2 2g f c 0 , then the equation

x2 + y2 + 2gx + 2fy + c = 0 does notrepresent any point in the xy plane.

iv. The general equation of a circle, a. is a second degree equation in x and y b. has coefficient of xy as 0 c. coefficient of x2 = coefficient of y2

Remember This

SAMPLE C

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Std. XI : Perfect Maths ‐ I

the given equation represents a circle.

Centre of the circle = ( g, f ) = 7 5

,2 2

and radius of the circle = 2 2g f c

= 2 2

7 515

2 2

=

49 2515

4 4

= 7

2

1. Find the centre and radius of each of the

following circles: i. x2 + y2 2x + 4y 4 = 0 ii. x2 + y2 6x 8y 24 = 0 iii. 4x2 + 4y2 24x 8y 24 = 0 Solution: i. Given equation of the circle is x2 + y2 2x + 4y 4 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 2, 2f = 4 and c = 4 g = 1, f = 2 and c = 4 Centre of the circle = ( g, f) = (1, 2)

and radius of the circle = 2 2g f c

= 2 2( 1) (2) ( 4)

= 1 4 4

= 9 = 3 ii. Given equation of the circle is x2 + y2 6x 8y 24 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 6, 2f = 8 and c = 24 g = 3, f = 4 and c = 24 Centre of the circle = (g, f) = (3, 4)

and radius of the circle = 2 2g f c

= 2 2( 3) ( 4) ( 24)

= 9 16 24

= 49 = 7 iii. Given equation of the circle is 4x2 + 4y2 24x 8y 24 = 0 Dividing throughout by 4, we get x2 + y2 6x 2y 6 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 6, 2f = 2 and c = 6

g = 3, f = 1 and c = 6 Centre of the circle = (– g,– f) = (3, 1)

and radius of the circle = 2 2g f c

= 2 2( 3) ( 1) ( 6)

= 9 1 6

= 16 = 4 2. Show that the equation 3x2 + 3y2 + 12x + 18y − 11 = 0 represents a

circle. Solution: Given equation is 3x2 + 3y2 + 12x + 18y – 11 = 0 Dividing throughout by 3, we get

x2 + y2 + 4x + 6y – 11

3 = 0

Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get

2g = 4, 2f = 6, c = 11

3

g = 2, f = 3, c = 11

3

Now, g2 + f 2 – c = (2)2 + (3)2 – 11

3

= 4 + 9 + 11

3

= 50

3 > 0

the given equation represents a circle. 3. Find the equation of the circle passing

through the points (5, 7) (6, 6) and (2, 2). Solution: Let C(h, k) be the centre of the required circle. Since, the required circle passes through the

points A(5, 7), B(6, 6) and D(2, 2). CA = CB = CD = radius Consider, CA = CD By distance formula,

2 2(h 5) (k 7) = 22(h 2) k 2

Squaring both the sides, we get (h 5)2 + (k 7)2 = (h 2)2 + (k + 2)2 h2 10h + 25 + k2 14k + 49

= h2 4h + 4 + k2 + 4k + 4

Exercise 6.2

C(h, k)

A(5, 7)

r B(6, 6)

D(2, 2)

r

r

SAMPLE C

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Chapter 06: Circle

10h 14k + 74 = 4h + 4k + 8 6h + 18k 66 = 0 h + 3k – 11 = 0 …(i) Consider, CB = CD By distance formula,

2 2(h 6) (k 6) = 22(h 2) k 2

Squaring both the sides, we get (h 6)2 + (k 6)2 = (h 2)2 + (k + 2)2 h2 12h + 36 + k2 12k + 36

= h2 4h + 4 + k2 + 4k + 4 12h 12k + 72 = 4h + 4k + 8 8h + 16k 64 = 0 h + 2k – 8 = 0 …(ii) By (i) (ii), we get k = 3 Substituting k = 3 in (i), we get h + 3 (3) – 11 = 0 h + 9 – 11 = 0 h = 2 Centre of the circle is C(2, 3).

radius (r) = CD = 2 2(2 2) (3 2) = 20 + 5

= 25 = 5 The equation of a circle with centre at (h, k)

and radius r is given by (x h)2 + (y k)2 = r2 Here, h = 2, k = 3 the required equation of the circle is (x 2)2 + (y 3)2 = 52 x2 4x + 4 + y2 6y + 9 = 25 x2 + y2 4x 6y + 4 + 9 25 = 0 x2 + y2 4x 6y 12 = 0 4. Show that the points (3, − 2), (1, 0),

(− 1, − 2) and (1, − 4) are concyclic. Solution: Let the equation of the circle passing through

the points (3, – 2), (1, 0) and (– 1, – 2) be x2 + y2 + 2gx + 2fy + c = 0 …(i) For point (3, – 2), Substituting x = 3 and y = – 2 in (i), we get 9 + 4 + 6g – 4f + c = 0 6g – 4f + c = –13 …(ii) For point (1, 0), Substituting x = 1 and y = 0 in (i), we get 1 + 0 + 2g + 0 + c = 0 2g + c = – 1 …(iii) For point (1, 2), Substituting x = – 1 and y = – 2, we get 1 + 4 – 2g – 4f + c = 0 2g + 4f – c = 5 …(iv) Adding (ii) and (iv), we get 8g = – 8 g = – 1

Substituting g = – 1 in (iii), we get – 2 + c = – 1 c = 1 Substituting g = – 1 and c = 1 in (iv), we get – 2 + 4f – 1 = 5 4f = 8 f = 2 Substituting g = – 1, f = 2 and c = 1 in (i), we

get x2 + y2 – 2x + 4y + 1 = 0 …(v) If (1, – 4) satisfies equation (v), the four points

are concyclic. Substituting x = 1, y = – 4 in L.H.S of (v), we

get L.H.S. = (1)2 + (– 4)2 – 2(1) + 4(– 4) + 1 = 1 + 16 – 2 – 16 + 1 = 0 = R.H.S. Point (1, – 4) satisfies equation (v) the given points are concyclic.

Let P(x, y) be any point on the circle x2 + y2 = r2. Then l(OP) = r and MOP =

Draw PM X-axis from P. Δ OMP is a right angled triangle.

cos = r

x and sin =

r

y

x = r cos , y = r sin The above equations are parametric form of equation of a circle x2 + y2 = r2 , where is parameter. A point P with parametric co-ordinates (r cos , r sin ) is denoted as P() = (r cos , r sin ) Note: i. The parametric equations of the circle (x h)2 + (y k)2 = r2 are given by x = h + r cos , y = k + r sin ii. Parametric form contains only one variable

and therefore it is more convenient for calculation.

O

X X

Y

Y

P(x, y)

M

r

x

y

Let’s Study

Parametric form of a circle

SAMPLE C

ONTENT

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240

Std. XI : Perfect Maths ‐ I

A tangent is a line that touches the circle at only one point and that point is known as the point of contact. Equation of a tangent at a point P(x1, y1) to the circle x2 + y2 = r2: The centre of the circle is at the origin O(0, 0).

Slope of OP = 1

1

0

0

y

x

= 1

1

y

x, x1 0, y1 0

Since, OP is perpendicular to tangent at point P.

Slope of tangent at P i.e. m = 1

1

x

y

Equation of the tangent at P(x1, y1) is y y1 = m(x x1)

y y1 = 1

1

x

y (x x1)

(y y1)y1 = x1(x x1) yy1 y1

2 = xx1 + x12

xx1+ yy1 = x12 + y1

2 ...(i) Since P(x1, y1) lies on the circle x2 + y2 = r2. x1

2 + y12 = r2

Equation (i) becomes xx1 + yy1 = r2 Thus, the equation of a tangent to the circle

x2 + y2 = r2 at P(x1, y1) is xx1 + yy1 = r2. Verify that the equation of the tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0.

(Textbook page no. 133)

Solution: Given equation of circle is x2 + y2 + 2gx + 2fy + c = 0 The centre of the circle is C( g, f ).

Slope of CP = 1

1

f

g

y

x

Since, CP is perpendicular to tangent at point P.

Slope of tangent at P i.e. m =

1

1

g

f

x

y

Equation of the tangent at P(x1, y1) is y y1 = m(x x1)

y y1 = 1

1

g

f

x

y(x x1)

(y y1)(y1 + f) = (x1 + g)(x x1) (x x1)(x1 + g) + (y y1)(y1 + f) = 0 x(x1 + g) + y(y1 + f) x1

2 gx1 y12 fy1 = 0

x(x1 + g) + y(y1 + f) = x12 + gx1 + y1

2 + fy1 Adding (gx1 + fy1 + c) to both sides, we get xx1 + yy1 + gx + fy + gx1 + fy1 + c = x1

2+ y12 + 2gx1 + 2fy1+ c …(i)

Since, P(x1, y1) lies on the given circle, x1

2+ y12 + 2gx1 + 2fy1 + c = 0

Thus, equation (i) becomes xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 Note: To find equation of a tangent to a circle (standard or general) replace x2 by xx1, y2 by yy1, 2x by (x + x1), 2y by (y + y1) and keep the constant term same in the equation of a circle. Equation of tangent in parametric form: The equation of a tangent to the circle x2 + y2 = r2 at P(x1, y1) is xx1 + yy1 = r2 …(i) Substituting x1 = r cos 1 and y1 = r sin 1 in (i), we get x.r cos 1 + y.r sin 1 = r2 i.e., x cos 1 + y sin 1 = r Thus, equation of a tangent to the circle x2 + y2 = r2 in parametric form at P(r cos , r sin ) is x cos + y sin = r To find the condition that the line y = mx + c is a tangent to the circle x2 + y2 = a2 and also to find the point of contact. Let the given line touch the circle at P(x1, y1). Equation of this tangent line is xx1 + yy1 = a2 yy1 = xx1 + a2 …(i) But its equation is given to be y = mx + c …(ii) Comparing the coefficients of (i) and (ii), we get

1

1

y= 1

m

x-=

2a

c

x1 = 2a m

cand y1 =

2a

c

O (0, 0)

P(x1, y1)

C( g, f )

P(x1, y1)

Try This

Condition of tangency

Equation of Tangent to a circle

SAMPLE C

ONTENT

241

Chapter 06: Circle

Since (x1, y1) lies on the circle, x12 + y1

2 = a2

22a m

c

æ ö- ÷ç ÷ç ÷ç ÷çè ø+

22a

c

æ ö÷ç ÷ç ÷ç ÷çè ø= a2

a2m2 + a2 = c2 i.e. c2 = a2m2 + a2 This is the required condition of tangency. When this condition is satisfied, line (ii) touches the

circle at 2 2a m a

,c c

i.e. the point of contact is 2 2a m a

,c c

Thus, a line y = mx + c is a tangent to the circle

x2 + y2 = a2, if c2 = a2m2 + a2 i.e. c = 2 2 2a m a

and the point of contact is 2 2a m a

,c c

.

There are two tangents with the same slope m,

y = mx + 2 2a 1 m , y = mx 2 2a 1 m .

From any point outside the circle and in the same plane, two tangents can be drawn to the circle. Let P(x1, y1) be a point from which tangents are drawn. Equation of a tangent with slope ‘m’ to the circle

x2 + y2 = a2 is y = mx 2 2a 1 m

This tangent passes through P(x1, y1).

y1 = mx1 2 2a 1 m

(y1 – mx1)2 = a2(1 + m2) (x1

2 a2)m2 2x1y1m + (y12 a2) = 0 …(i)

This is a quadratic equation in ‘m’. It has two roots say m1 and m2, which are the slopes of two tangents drawn from P.

Thus, two tangents can be drawn to a circle from a given point in its plane.

Sum of the roots (m1 + m2) = 1 1

2 21

2

a

x y

x

= 1 12 2

1

2

ax y

x

Product of roots (m1m2) =2 2

12 2

1

a

a

y

x

Director circle of a given circle is the locus of the point P, such that the tangents from P to the given circle are perpendicular to each other. The two tangents are mutually perpendicular, m1m2 = –1

2 2

12 2

1

a

a

y

x= 1

y12 a2 = x1

2 + a2 x1

2 + y12 = 2a2

Equation of the locus of the point P is x2 + y2 = 2a2, which is the director circle of the circle x2 + y2 = a2.

(Textbook page no. 135)

Equation of a circle is x2 + y2 = 9.

Its centre is at ,0 0 and radius is 3

Equation of a line is 3x – 4y + 15 = 0

y = 3

4 x +

15

4

Comparing it with y = mx + c,

m = 3

4 and c =

15

4

P(x, y)

m

B

A

C

The tangency of a straight line to the circle can bechecked by showing that the perpendicular fromthe centre to the line is equal to the radius.

Remember This

Tangents from a point to the circle

Director circle

Radius of the director circle = 2 (radius of the given circle)

Remember This

Textual Activity

director circle circle x2 + y2 = a2

x2 + y2 = 2a2 Y

O X

P(x1, y1)

X

Y

SAMPLE C

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Std. XI : Perfect Maths ‐ I

We know that, if the line y = mx + c is a tangent to x2 + y2 = a2, then c2 = a2m2 + a2

c2 = 2

15

4

= 225

16 …(i)

Also, c2 = a2m2 + a2 = 9 9

16 + 9

= 99

116

= 9(25)

16

= 225

16 …(ii)

From equations (i) and (ii), we conclude that the line 3x 4y + 15 = 0 is a tangent to the circle x2 + y2 = 9.

Common tangents to two circles:

Case Number of tangents Condition 1. Circles touch each other externally.

Exactly three common tangents can be drawn.

d(C1C2) = r1 + r2

2. Circles touch each other internally.

Exactly one common tangent can be drawn.

d(C1C2) = | r1 – r2 |

3. Disjoint circles.

Exactly four common tangents can be drawn.

d(C1C2) > r1 + r2

4. Circles intersecting each other. The chord joining the point of intersection of two given circles is called their common chord and also it is called as the radical axis. seg AB is the chord.

Exactly two common tangents can be drawn.

d(C1C2) < r1 + r2

C1 C2

r2r1

C1 C2

A

B

r1r2

C1 C2 r1 r2

C1 C2 r1 r2

SAMPLE C

ONTENT

243

Chapter 06: Circle

5. Concentric circles.

No common tangent can be drawn.

d(C1C2) = 0

1. Write the parametric equations of the circles: i. x2 + y2 = 9 ii. x2 + y2 + 2x 4y 4 = 0 iii. (x 3)2 + ( y + 4)2 = 25 Solution: i. Given equation of the circle is x2 + y2 = 9 x2 + y2 = 32 Comparing this equation with x2 + y2 = r2, we get r = 3 The parametric equations of the circle in terms

of are x = r cos and y = r sin x = 3 cos and y = 3 sin ii. Given equation of the circle is x2 + y2 + 2x 4y 4 = 0 x2 + 2x + y2 – 4y – 4 = 0 x2 + 2x +1 – 1 + y2 – 4y + 4 – 4 – 4 = 0 (x2 + 2x + 1 ) + (y2 – 4y + 4) – 9 = 0 (x + 1)2 + (y – 2)2 = 9 (x + 1)2 + (y – 2)2 = 32 Comparing this equation with (x – h)2 + (y – k)2 = r2, we get h = – 1, k = 2 and r = 3 The parametric equations of the circle in terms

of are x = h + r cos and y = k + r sin x = – 1 + 3cos and y = 2 + 3sin iii. Given equation of the circle is (x 3)2 + (y + 4)2 = 25 (x – 3)2 + (y + 4)2 = 52 Comparing this equation with (x h)2 + (y k)2 = r2 , we get h = 3, k = 4 and r = 5 The parametric equations of the circle in terms

of are x = h + r cos and y = k + r sin x = 3 + 5 cos and y = 4 + 5 sin

2. Find the parametric representation of the circle 3x2 + 3y2 – 4x + 6y 4 = 0.

Solution: Given equation of the circle is 3x2 + 3y2 – 4x + 6y – 4 = 0 Dividing throughout by 3, we get

x2 + y2 – 4

3x + 2y

4

3= 0

x2 – 4

3x + y2 + 2y –

4

3 = 0

x2 – 4

3x +

4

9 –

4

9+ y2 + 2y + 1 – 1 –

4

3 = 0

2 4 4

3 9

x x + 2 2 1y y – 25

9 = 0

2

2

3

x + (y + 1)2 = 25

9

2

2

3

x + [y (1)]2 = 2

5

3

Comparing this equation with (x – h)2 + (y – k)2 = r2, we get

h = 2

3, k = – 1 and r =

5

3

The parametric representation of the circle in terms of are

x = h + r cos and y = k + r sin

x = 2

3+

5

3 cos and y = – 1 +

5

3sin

3. Find the equation of a tangent to the circle

x2 + y2 − 3x + 2y = 0 at the origin. Solution: Given equation of the circle is x2 + y2 – 3x + 2y = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = – 3, 2f = 2, c = 0

g = 3

2 , f = 1, c = 0

The equation of a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0

Exercise 6.3

C1 C2

r1

r2

SAMPLE C

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Std. XI : Perfect Maths ‐ I

the equation of the tangent at (0, 0) is

x(0) + y(0) + 3

2

(x + 0) + 1(y + 0) + 0 = 0

3

2

x + y = 0

3x – 2y = 0 4. Show that the line 7x − 3y − 1 = 0 touches

the circle x2 + y2 + 5x − 7y + 4 = 0 at point (1, 2).

Solution: Given equation of the circle is x2 + y2 + 5x – 7y + 4 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 5, 2f = – 7, c = 4

g = 5

2, f =

7

2 , c = 4

The equation of a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 the equation of the tangent at (1, 2) is

x(1) + y(2) + 5

2 (x + 1) –

7

2 (y + 2) + 4 = 0

x + 2y + 5

2 x +

5

2 –

7

2y – 7 + 4 = 0

7

2x –

3

2y –

1

2 = 0

7x – 3y – 1 = 0, which is same as the given line. The line 7x – 3y – 1 = 0 touches the given

circle at (1, 2). 5. Find the equation of tangent to the circle

x2 + y2 − 4x + 3y + 2 = 0 at the point (4, −2). Solution: Given equation of the circle is x2 + y2 – 4x + 3y + 2 = 0 Comparing this equation with x2 + y2 +2gx + 2fy + c = 0, we get 2g = –4, 2f = 3, c = 2

g = – 2, f = 3

2, c = 2

The equation of a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 the equation of the tangent at (4, – 2) is

x(4) + y(– 2) – 2(x + 4) + 3

2(y – 2) + 2 = 0

4x – 2y – 2x – 8 + 3

2y – 3 + 2 = 0

2x – 1

2y – 9 = 0

4x – y – 18 = 0

I. Choose the correct alternative. 1. Equation of a circle which passes through

(3, 6) and touches the axes is (A) x2 + y2 + 6x + 6y + 3 = 0 (B) x2 + y2 − 6x − 6y − 9 = 0 (C) x2 + y2 − 6x − 6y + 9 = 0 (D) x2 + y2 − 6x + 6y − 3 = 0 2. If the lines 2x − 3y = 5 and 3x − 4y = 7 are the

diameters of a circle of area 154 sq. units, then find the equation of the circle.

(A) x2 + y2 − 2x + 2y = 40 (B) x2 + y2 − 2x − 2y = 47 (C) x2 + y2 − 2x + 2y = 47 (D) x2 + y2 − 2x − 2y = 40 3. Find the equation of the circle which passes

through the points (2, 3) and (4, 5) and the centre lies on the straight line y − 4x + 3 = 0. (A) x2 + y2 − 4x − 10y + 25 = 0

(B) x2 + y2 − 4x − 10y − 25 = 0 (C) x2 + y2 − 4x + 10y − 25 = 0 (D) x2 + y2 + 4x − 10y + 25 = 0 4. The equation(s) of the tangent(s) to the circle

x2 + y2 = 4 which are parallel to x + 2y + 3 = 0 are

(A) x − 2y = 2 (B) x + 2y = ± 2 3

(C) x + 2y = ± 2 5 (D) x − 2y = ± 2 5 5. If the lines 3x − 4y + 4 = 0 and

6x − 8y − 7 = 0 are tangents to a circle, then find the radius of the circle.

(A) 3

4 (B)

4

3 (C)

1

4 (D)

7

4

6. Area of the circle having centre at (1, 2) and

passing through (4, 6) is (A) 5 (B) 10 (C) 25 (D) 100 If a circle passes through the points (0, 0),

(a, 0) and (0, b), then find the co-ordinates of its centre.

(A) a b

,2 2

(B) a b

,2 2

(C) a b

,2 2

(D) a b

,2 2

8. The equation of a circle with origin as centre and

passing through the vertices of an equilateral triangle whose median is of length 3a is

(A) x2 + y2 = 9a2 (B) x2 + y2 = 16a2 (C) x2 + y2 = 4a2 (D) x2 + y2 = a2

Miscellaneous Exercise – 6

SAMPLE C

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Chapter 06: Circle

9. A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is

(A) 2 π

63 (B)

π3

3

(C) π

3–

3

6 (D)

π3 1

6

10. The parametric equations of the circle

x2 + y2 + mx + my = 0 are

(A) x = m m

cos θ2 2

, y =

m msin θ

2 2

(B) x = m m

cos θ2 2

, y =

+m msin θ

2 2

(C) x = 0, y = 0 (D) x = m cos , y = m sin Answers: 1. (C) 2. (C) 3. (A) 4. (C) 5. (A) 6. (C) 7. (D) 8. (C) 9. (B) 10. (A) Hints: 2. Centre of circle = Point of intersection of

diameters. Solving 2x – 3y = 5 and 3x – 4y = 7, we get x = 1, y = –1 Centre of the circle C (h, k) = C (1, –1) Area = 154 r2 = 154

22

7 r2 = 154

r2 = 154 7

22 = 49

r = 7 equation of the circle is (x 1)2 + (y + 1)2 = 72 x2 + y2 2x + 2y = 47 5. Tangents are parallel to each other. Perpendicular distance between tangents = diameter

22

74

2

3 4

= 2r

15

25

= 2r

r = 3

4

6. r = CA

= 2 24 1 6 2

= 9 16

= 25 = 5 area = r2 = 52 = 25 8. Since the triangle is equilateral. The centroid of the triangle is same as the

circumcentre;

and radius of the circumcircle = 2

(median)3

= 2

(3a)3

= 2a

Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is 2 2 24a x y

9. In OAP,

sin 30 = 1

OP

OP = 2

cos 30 = AP

OP

3

2 =

AP

2

AP = 3 A( AOBP) = 2A( OAP)

= 2 1

2 1 3

= 3

A(sector AOB) = 1

2 (1)2

2

3

= 3

Required area = A( AOBP) A(sector AOB)

= 3 3

II. Answer the following:

1 Find the centre and radius of the circle x2 + y2 x + 2y 3 = 0.

Solution:

Given equation of the circle is

x2 + y2 x + 2y 3 = 0

Comparing this equation with

x2 + y2 + 2gx + 2fy + c = 0, we get

2g = 1, 2f =2 and c = 3

g = 1

2

, f = 1 and c = 3

Centre of the circle = (g, –f) = 1

, 12

A(4, 6)

C(1, 2)

r

3x 4y + 4 = 0

3x 4y 7

2 = 0

A

B

P(x1, y1) 3030

O (0, 0)

1

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Std. XI : Perfect Maths ‐ I

and radius of the circle = 2 2g f c

= 2

21(1) 3

2

= 1

1 34

=

17

4

=

17

2

2. Find the centre and radius of the circle

x = 3 – 4 sin , y = 2 – 4cos . Solution: Given, x = 3 4 sin, y = 2 4 cos x 3 = 4 sin, y 2 = 4 cos On squaring and adding, we get (x 3)2 + (y 2)2 = (4 sin)2 + (4 cos)2 (x 3)2 + (y 2)2 = 16sin2 + 16cos2 (x 3)2 + (y 2)2 = 16(sin2 + cos2) (x 3)2 + (y 2)2 = 16(1) (x 3)2 + (y 2)2 = 16 (x 3)2 + (y 2)2 = 42 Comparing this equation with (x h)2 + (y k)2 = r2, we get h = 3, k = 2, r = 4 Centre of the circle is (3, 2) and radius is 4. 3. Find the equation of circle passing through

the point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x − 2y + 4 = 0.

Solution: Required circle passes through the point of

intersection of the lines x + 3y = 0 and 2x – 7y = 0.

x + 3y = 0 x = 3y …(i) 2x 7y = 0 …(ii) Substituting x = 3y in (ii), we get 2(3y) 7y = 0 6y 7y = 0 13y = 0 y = 0 Substituting y = 0 in (i), we get x = 3(0) = 0 Point of intersection is O(0, 0). This point O(0,0) lies on the circle. Let C(h, k) be the centre of the required circle. Since, point of intersection of lines x + y = –1

and x – 2y = – 4 is the centre of circle. x = h, y = k

Equations of lines become h + k = – 1 …(iii) h – 2 k = – 4 …(iv) By (iii) – (iv), we get 3k = 3 k = 1 Substituting k = 1 in (iii), we get h + 1 = – 1 h = – 2 Centre of the circle is C(–2, 1) and it passes

through point O(0, 0).

Radius(r) = OC = 2 20 2 0 1

= 4 1 = 5 The equation of a circle with centre at (h, k)

and radius r is given by (x – h)2 + (y – k)2 = r2 Here, h = – 2, k = 1 the required equation of the circle is

(x + 2)2 + (y – 1)2 = 2

5

x2 + 4x + 4 + y2 – 2y + 1 = 5 x2 + y2 + 4x – 2y = 0 4. Find the equation of the circle which passes

through the origin and cuts off chords of lengths 4 and 6 on the positive side of X - axis and Y-axis respectively.

Solution: Let the circle cut the chord of length 4 on

X-axis at point A and the chord of length 6 on Y-axis at point B.

the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).

Since, BOA is a right angle. AB represents the diameter of the circle The equation of a circle having (x1, y1) and

(x2, y2) as end points of diameter is given by (x – x1) (x – x2) + (y – y1) (y – y2) = 0 Here, x1 = 4, y1 = 0, x2 = 0, y2 = 6 the required equation of the circle is (x 4) (x 0) + ( y 0) (y 6) = 0 x2 4x + y2 6y = 0 x2 + y2 4x 6y = 0

Y

X

B

A O(0, 0)

6

4

(0, 6)

(4, 0)X

Y

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5. Show that the points (9, 1), (7, 9), (−2, 12) and (6, 10) are concyclic.

Solution: Let the equation of circle passing through the

points (9, 1), (7, 9), (2, 12) be x2 + y2 + 2gx + 2fy + c = 0 …(i) For point (9, 1), Substituting x = 9 and y = 1 in (i), we get 81 + 1 + 18g + 2f + c = 0 18g + 2f + c = 82 …(ii) For point (7, 9), Substituting x = 7 and y = 9 in (i), we get 49 + 81 + 14g + 18f + c = 0 14g + 18f + c = 130 …(iii) For point (2, 12), Substituting x = – 2 and y = 12 in (i), we get 4 + 144 4g + 24f + c = 0 4g + 24f + c = 148 …(iv) By (ii) – (iii), we get 4g 16f = 48 g 4f = 12 …(v) By (iii) – (iv), we get 18g 6f = 18 3g f = 3 …(vi) By 3 (v) – (vi), we get 11f = 33 f = 3 Substituting f = 3 in (vi), we get 3g (3) = 3 3g + 3 = 3 g = 0 Substituting g = 0 and f = 3 in (ii), we get 18 (0) + 2(3) + c = 82 6 + c = 82 c = 76 Equation of the circle becomes x2 + y2 + 2(0)x + 2(–3)y + (–76) = 0 x2 + y2 6y 76 = 0 …(vii) Now for the point (6, 10), Substituting x = 6 and y = 10 in L.H.S. of (vii),

we get L.H.S = 62 + 102 6(10) 76 = 36 + 100 60 76 = 0 = R.H.S. Point (6,10) satisfies equation (vii). the given points are concyclic. 6. The line 2x − y + 6 = 0 meets the circle

x2 + y2 + 10x + 9 = 0 at A and B. Find the equation of circle with AB as diameter.

Solution: 2x y + 6 = 0

y = 2x + 6 Substituting y = 2x + 6 in x2 + y2 + 10x + 9 = 0,

we get x2 + (2x + 6)2 + 10x + 9 = 0

x2 + 4x2 + 24x + 36 + 10x + 9 = 0 5x2 + 34x + 45 = 0 5x2 + 25x + 9x + 45 = 0 (5x + 9) (x + 5) = 0 5x = 9 or x = 5

x = 9

5

or x = 5

When x = 9

5

,

y = 9

25

+ 6 =

18

5

+ 6 =

18 +30

5

=

12

5

Point of intersection is A 9 12

,5 5

.

When x = 5, y = 10 + 6 = 4 Point of intersection in B ( 5, 4). By diameter form, equation of circle with AB

as diameter is

9 125 4

5 5x x y y

= 0

(5x + 9) (x + 5) + (5y 12) ( y + 4) = 0 5x2 + 25x + 9x + 45 + 5y2 + 20y 12y 48 = 0 5x2 + 5y2 + 34x + 8y 3 = 0 7. Show that x = −1 is a tangent to circle

x2 + y2 – 4x 2y 4 = 0 at (−1, 1). Solution: Given equation of circle is x2 + y2 4x 2y 4 = 0. Comparing this equation with

x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 4, 2f = 2, c = 4 g = 2, f = 1, c = 4 The equation of a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 the equation of the tangent at ( 1, 1) is x( 1)+ y(1) 2(x 1) 1(y + 1) 4 = 0 3x 3 = 0 x 1 = 0 x = 1 x = 1 is the tangent to the given circle at

(1, 1). [Note: The question has been modified.] 8. Find the equation of tangent to the circle

x2 + y2 = 64 at the point P 2π

3

.

Solution: Given equation of circle is x2 + y2 = 64 Comparing this equation with x2 + y2 = r2, we get r = 8

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The equation of a tangent to the circle x2 + y2 = r2 at P() is x cos + y sin = r

the equation of the tangent at P2π

3

is

x cos 2π

3 + y sin

2π

3 = 8

x1

2

+ y3

2

= 8

x + 3y = 16

x 3y + 16 = 0 9. Find the equation of locus of the point of

intersection of perpendicular tangents drawn to the circle x = 5cos and y = 5 sin .

Solution: The locus of the point of intersection of

perpendicular tangents is the director circle of the given circle.

x = 5cos and y = 5 sin x2 + y2 = 25 cos2 + 25 sin2 x2 + y2 = 25 (cos2 + sin2 ) x2 + y2 = 25(1) = 25 The equation of the director circle of the circle

x2 + y2 = a2 is x2 + y2 = 2a2. Here, a = 5 the required equation is x2 + y2 = 2(5)2 = 2(25) x2 + y2 = 50 10. Find the equation of the circle concentric

with x2 + y2 – 4x + 6y = 1 and having radius 4 units.

Solution: Given equation of circle is x2 + y2 4x + 6y = 1 i.e., x2 + y2 4x + 6y 1 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 4, 2f = 6 g = 2, f = 3 Centre of the circle = (g, f) = (2, 3) Given circle is concentric with the required

circle. They have same centre. Centre of the required circle = (2, 3) The equation of a circle with centre at (h, k)

and radius r is (x h)2 + (y k)2 = r2 Here, h = 2, k = 3 and r = 4 the required equation of the circle is (x 2)2 + [y (3)]2 = 42

(x 2)2 + (y + 3)2 = 16 x2 4x + 4 + y2 + 6y + 9 16 = 0 x2 + y2 4x + 6y 3 = 0 11. Find the lengths of the intercepts made on

the co-ordinate axes, by the circles. i. x2 + y2 – 8x + y – 20 = 0 ii. x2 + y2 – 5x + 13y – 14 = 0 Solution: To find x-intercept made by the circle

x2 + y2 + 2gx + 2fy + c = 0, substitute y = 0 and get a quadratic equation in x, whose roots are, say, x1 and x2.

These values represent the abscissae of ends A and B of x intercept. Length of x intercept = | AB | = | x2 x1 |

Similarly, substituting x = 0, we get a quadratic equation in y whose roots, say, y1 and y2 are ordinates of the ends C and D of y-intercept.

Length of y intercept = | CD | = | y2 y1 | i. Given equation of the circle is

x2 + y2 8x + y 20 = 0 …(i) Substituting y = 0 in (i), we get x2 8x 20 = 0 …(ii) Let AB represent the x-intercept, where

A = (x1, 0), B = (x2, 0) Then from (ii), x1 + x2 = 8 and x1x2 = 20 (x1 x2)2 = (x1 + x2)2 4 x1x2 = (8)2 4( 20) = 64 + 80 = 144

| x1 x2 | = 2

1 2x x = 144 = 12

Length of x intercept = 12 units Substituting x = 0 in (i), we get y2 + y 20 = 0 …(iii) Let CD represent the y intercept, where

C = (0, y1) and D = (0, y2) Then from (iii), y1 + y2 = 1 and y1 y2 = 20 (y1 y2)2 = (y1 + y2)2 4 y1 y2 = ( 1)2 4( 20) = 1 + 80 = 81

| y1 y2 | = 2

1 2y y = 81 = 9

Length of y intercept = 9 units.

Y

D(0, y2)

C(0, y1)

B(x2, 0)A(x1, 0)X

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Chapter 06: Circle

Alternate method: Given equation of the circle is

x2 + y2 8x + y 20 = 0 …(i) x-intercept: Substituting y = 0 in (i), we get x2 8x 20 = 0 (x 10)(x + 2) = 0 x = 10 or x = 2 length of x-intercept = | 10 (2) | = 12 units y-intercept: Substituting x = 0 in (i), we get y2 + y 20 = 0 (y + 5)(y 4) = 0 y = 5 or y = 4 length of y-intercept = | 5 4 | = 9 units ii. Given equation of the circle is

x2 + y2 5x + 13y 14 = 0 …(i) Substituting y = 0 in (i), we get

x2 5x 14 = 0 …(ii) Let AB represent the x-intercept, where

A = (x1, 0), B = (x2, 0). Then from (ii), x1 + x2 = 5 and x1x2 = 14 (x1 x2)2 = (x1 + x2)2 4 x1x2

= (5)2 4( 14) = 25 + 56 = 81

| x1 x2 | = 2

1 2x x = 81 = 9

Length of x-intercept = 9 units Substituting x = 0 in (i), we get y2 + 13y 14 = 0 …(iii) Let CD represent the y-intercept, where

C = (0, y1), D = (0, y2). Then from (iii), y1 + y2 = 13 and y1 y2 = 14 (y1 y2)2 = (y1 + y2)2 4 y1 y2 = ( 13)2 4( 14) = 169 + 56 = 225

| y1 y2 | = 2

1 2y y = 225 = 15

Length of y-intercept = 15 units 12. Show that the circles touch each other

externally. Find their point of contact and the equation of their common tangent.

i. x2 + y2 – 4x + 10y +20 = 0 x2 + y2 + 8x – 6y – 24 = 0 ii. x2 + y2 – 4x – 10y + 19 = 0 x2 + y2 + 2x + 8y – 23 = 0 Solution: i. Given equation of the first circle is x2 + y2 4x + 10y + 20 = 0

Here, g = 2, f = 5, c = 20 Centre of the first circle is C1 = (2, 5)

Radius of the first circle is r1 = 2 22 5 20

= 4 25 20

= 9 = 3 Given equation of the second circle is x2 + y2 + 8x 6y 24 = 0 Here, g = 4, f = 3, c = 24 Centre of the second circle is C2 = (4, 3) Radius of the second circle is

r2 = 224 3 24

= 16 9 24

= 49 = 7 By distance formula,

C1C2 = 224 2 3 5

= 36 64

= 100 = 10 r1 + r2 = 3 + 7 = 10 Since, C1C2 = r1 + r2 the given circles touch each other externally. Let P(x, y) be the point of contact. P divides C1 C2 internally in the ratio r1 : r2 i.e. 3:7 By internal division,

x = 3 4 7 2

3 7

= 12 14

10

=

1

5

and y = 3 3 7 5

3 7

= 9 35

10

=

13

5

Point of contact = 1 13

,5 5

Equation of common tangent is (x2 + y2 4x + 10y + 20)

(x2 + y2 + 8x 6y 24) = 0 4x + 10y + 20 8x + 6y + 24 = 0 12x + 16y + 44 = 0 3x 4y 11 = 0 Note: i. The equation of the common tangent to two

circles S1 = 0 and S2 = 0 touching externally is S1 S2 = 0.

ii. Point of contact divides C1C2 internally in the ratio r1:r2.

ii. Given equation of the first circle is x2 + y2 4x 10y + 19 = 0 Here, g = 2, f = 5, c = 19 Centre of the first circle is C1 = (2, 5)

3 7 C1 (2, 5)

P(x, y) C2 (– 4, 3)

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Std. XI : Perfect Maths ‐ I

Radius of the first circle is

r1 = 2 22 5 19

= 4 25 19

= 10 Given equation of the second circle is x2 + y2 + 2x + 8y 23 = 0 Here, g = 1, f = 4, c = 23 Centre of the second circle is C2 = ( 1, 4) Radius of the second circle is

r2 = 2 21 4 23

= 1 16 23

= 40 = 2 10 By distance formula,

C1C2 = 2 21 2 4 5

= 9 81

= 90 = 3 10

r1 + r2 = 10 2 10 = 3 10

Since, C1C2 = r1 + r2 the given circles touch each other externally. r1 : r2 = 10 : 2 10 = 1:2

Let P(x, y) be the point of contact. P divides C1 C2 internally in the ratio r1 : r2 i.e. 1:2 By internal division,

x = 1 1 2 2

1 2

= 1 4

3

= 1

and y = 1 4 2 5

1 2

= 4 10

3

= 2

Point of contact = (1, 2) Equation of common tangent is (x2 + y2 – 4x – 10y + 19)

– (x2 + y2 + 2x + 8y – 23) = 0 – 4x – 10y + 19 – 2x – 8y + 23 = 0 – 6x – 18y + 42 = 0 x + 3y – 7 = 0 13. Show that the circles touch each other

internally. Find their point of contact and the equation of their common tangent.

i. x2 + y2 – 4x – 4y – 28 = 0, x2 + y2 – 4x – 12 = 0 ii. x2 + y2 + 4x – 12y + 4 = 0, x2 + y2 – 2x – 4y + 4 = 0 Solution: i. Given equation of the first circle is x2 + y2 4x 4y 28 = 0

Here, g = 2, f = 2, c = 28 Centre of the first circle is C1 = (2, 2) Radius of the first circle is

r1 = 2 22 2 28

= 4 4 28

= 36 = 6 Given equation of the second circle is x2 + y2 4x 12 = 0 Here, g = 2, f = 0, c = 12 Centre of the second circle is C2 = (2, 0) Radius of the second circle is

r2 = 2 22 0 12

= 4 12

= 16 = 4 By distance formula,

C1C2 = 2 22 2 0 2

= 4 = 2

|r1 r2| = 6 4 = 2 Since, C1C2 = |r1 r2| the given circles touch each other internally. Equation of common tangent is (x2 + y2 – 4x – 4y – 28) – (x2 + y2 – 4x – 12) = 0 – 4x – 4y – 28 + 4x + 12 = 0 – 4y – 16 = 0 y + 4 = 0 y = 4 Substituting y = – 4 in x2 + y2 – 4x – 12 = 0,

we get x2 + (– 4)2 – 4x – 12 = 0 x2 + 16 – 4x – 12 = 0 x2 – 4x + 4 = 0 (x – 2)2 = 0 x = 2 Point of contact is (2, – 4) and equation of

common tangent is y + 4 = 0. Note: i. The equation of the common tangent to two

circles S1 = 0 and S2 = 0 touching internally is S1 S2 = 0.

ii. Point of contact divides C1C2 externally in the ratio r1:r2.

ii. Given equation of the first circle is x2 + y2 + 4x 12y + 4 = 0 Here, g = 2, f = 6, c = 4 Centre of the first circle is C1 = ( 2, 6) Radius of the first circle is

r1 = 222 6 4

= 4 36 4

= 36 = 6

1 2 C1 (2, 5) P(x, y)

C2 (– 1, – 4)

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Chapter 06: Circle

Given equation of the second circle is x2 + y2 2x 4y + 4 = 0 Here, g = 1, f = 2, c = 4 Centre of the second circle is C2 = (1, 2) Radius of the second circle is

r2 = 2 21 2 4

= 1 4 4

= 1 = 1 By distance formula,

C1C2 = 2 21 2 2 6

= 9 16

= 25 = 5 |r1 r2| = 6 1 = 5 Since, C1C2 = |r1 r2| the given circles touch each other internally. Equation of common tangent is (x2 + y2 + 4x – 12y + 4)

– (x2 + y2 – 2x – 4y + 4) = 0 4x – 12y + 4 + 2x + 4y – 4 = 0 6x – 8y = 0 3x – 4y = 0

y = 3

4

x

Substituting y = 3

4

x in x2 + y2 – 2x – 4y + 4 = 0,

we get

x2 + 2

3

4

x

– 2x – 4 3

4

x

+ 4 = 0

x2 + 29

16

x – 2x – 3x + 4 = 0

225

16

x – 5x + 4 = 0

25x2 80x + 64 = 0 (5x 8)2 = 0 5x 8 = 0

x = 8

5

Substituting x = 8

5 in y =

3

4

x, we get

y = 3 8

4 5

= 6

5

Point of contact is 8 6

,5 5

and equation of

common tangent is 3x 4y = 0. 14. Find the length of the tangent segment

drawn from the point (5, 3) to the circle x2 + y2 + 10x – 6y – 17 = 0. Solution: Given equation of circle is x2 + y2 + 10x – 6y – 17 = 0

Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 10, 2f = –6, c = –17 g = 5, f = –3, c = –17 Centre of circle = ( g, f ) = C( 5, 3)

Radius of circle = 2 2g + f c

= 225 3 17

= 25 9 17

= 51

BC = 2 25 5 3 3

= 100 0 = 10

In right angled ABC, BC2 = AB2 + AC2 …[Pythagoras theorem]

(10)2 = AB2 + 2

51

AB2 = 100 – 51 = 49 AB = 7 Length of the tangent segment from (5, 3) is

7 units. Alternate Method: Given equation of circle is

x2 + y2 + 10x – 6y – 17 = 0 Here, g = 5, f = 3, c = –17 Length of the tangent segment to the circle x2 + y2 + 2gx + 2fy + c = 0 from the point

(x1, y1) is 2 21 1 1 12g 2f cx y x y .

Length of the tangent segment from (5, 3)

= 2 25 3 10 5 6 3 17

= 25 9 50 18 17

= 49 = 7 units 15. Find the value of k, if the length of the

tangent segment from the point (8, –3) to the circle x2 + y2 – 2x + ky – 23 = 0 is 10 .

Solution: Given equation of the circle is x2 + y2 – 2x + ky – 23 = 0

Here, g = 1, f = k

2, c = 23

Length of the tangent segment to the circle x2 + y2 + 2gx + 2fy + c = 0 from the point

(x1, y1) is 2 21 1 1 12g 2f cx y x y

B(5, 3)

C(–5, 3)

51

A

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Std. XI : Perfect Maths ‐ I

Length of the tangent segment from (8, 3)

= 10

228 3 2 8 k 3 23 = 10

64 + 9 16 3k 23 = 10 …[Squaring both the sides]

34 3k = 10 3k = 24 k = 8 16. Find the equation of tangent to circle

x2 + y2 – 6x – 4y = 0, at the point (6, 4) on it. Solution: Given equation of the circle is x2 + y2 6x 4y = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 6, 2f = 4, c = 0 g = 3, f = 2, c = 0 The equation of a tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1, y1) is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0 the equation of the tangent at (6, 4) is x(6) + y(4) 3(x + 6) 2(y + 4) + 0 = 0 6x + 4y 3x 18 2y 8 = 0 3x + 2y 26 = 0 Alternate method: Given equation of the circle is x2 + y2 6x 4y = 0 x(x 6) + y(y 4) = 0, which is in diameter

form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of OP = 4 0

6 0

= 2

3

Since, OP is perpendicular to the required tangent.

Slope of the required tangent = 3

2

the equation of the tangent at (6, 4) is

y 4 = 3

2

(x 6)

2(y 4) = 3(x 6) 2y 8 = 3x + 18 3x + 2y 26 = 0

17. Find the equation of tangent to circle x2 + y2 = 5, at the point (1, –2) on it.

Solution: Given equation of the circle is x2 + y2 = 5 Comparing this equation with x2 + y2 = r2,

we get r2 = 5 The equation of a tangent to the circle x2 + y2 = r2 at (x1, y1) is xx1 + yy1 = r2 the equation of the tangent at (1, – 2) is x(1) + y(2) = 5 x 2y = 5 18. Find the equation of tangent to circle

x = 5 cos , y = 5 sin , at the point = 3

on

it. Solution: The equation of a tangent to the circle x2 + y2 = r2 at P() is x cos + y sin = r

Here, r = 5, = π

3

the equation of the tangent at Pπ

3

is

x cos π

3 + y sin

π

3 = 5

x1

2

+ y 3

2

= 5

x + 3y = 10 19. Show that 2x + y + 6 = 0 is a tangent to

x2 + y2 + 2x – 2y – 3 = 0. Find its point of contact.

Solution: Given equation of circle is x2 + y2 + 2x – 2y – 3 = 0 …(i) Given equation of line is 2x + y + 6 = 0 y = 6 2x …(ii) Substituting y = 6 2x in (i), we get x2 + ( 6 2x)2 + 2x 2( 6 2x) 3 = 0 x2 + 36 +24x + 4x2 + 2x + 12 + 4x 3 = 0 5x2 + 30x + 45 = 0 x2 + 6x + 9 = 0 (x + 3)2 = 0 x = 3 Since, the roots are equal. 2x + y + 6 = 0 is a tangent to x2 + y2 + 2x 2y 3 = 0 Substituting x = 3 in (ii), we get y = 6 2( 3) = 6 + 6 = 0 Point of contact = ( 3, 0)

Y

O X

P(6, 4)

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Chapter 06: Circle

20. If the tangent at (3, – 4) to the circle x2 + y2 = 25 touches the circle

x2 + y2 + 8x – 4y + c = 0, find c. Solution: The equation of a tangent to the circle x2 + y2 = r2 at (x1, y1) is xx1 + yy1 = r2 Equation of the tangent at (3, 4) is x(3) + y(4) = 25 3x 4y 25 = 0 …(i) Given equation of circle is x2 + y2 + 8x 4y + c = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 8, 2f = – 4 g = 4, f = – 2

C ( 4, 2) and r = 224 2 c = 20 c

Since line (i) is a tangent to this circle also, the perpendicular distance from C( 4, 2) to line (i) is equal to radius r.

2 2

3 4 4 2 25

3 4

= 20 c

45

25

= 20 c

45

5

= 20 c

| 9| = 20 c

81 = 20 c …[Squaring both the sides] c = 61 21. Find the equations of the tangents to the

circle x2 + y2 = 16 with slope –2. Solution: Given equation of the circle is x2 + y2 = 16 Comparing this equation with x2 + y2 = a2,

we get a2 = 16 Equations of the tangents to the circle x2 + y2 = a2 with slope m are

y = mx 2 2a 1 m

Here, m = 2, a2 = 16 the required equations of the tangents are

y = 2x 216 1 2

= 2x 16 5

y = 2x 4 5

2x + y 4 5 = 0

22. Find the equations of the tangents to the circle x2 + y2 = 4 which are parallel to

3x + 2y + 1 = 0. Solution: Given equation of the circle is x2 + y2 = 4 Comparing this equation with x2 + y2 = a2,

we get a2 = 4 Given equation of the line is 3x + 2y + 1 = 0

Slope of this line = 3

2

Since, the required tangents are parallel to the given line.

Slope of required tangents (m) = 3

2

Equations of the tangents to the circle x2 + y2 = a2 with slope m are

y = mx 2 2a 1 m

the required equations of the tangents are

y = 3

2

x

23

4 12

= 3

2

x

94 1

4

y = 3

2

x 13

2y = 3x 2 13

3x + 2y 2 13 = 0

23. Find the equations of the tangents to the

circle x2 + y2 = 36 which are perpendicular to the line 5x + y = 2.

Solution: Given equation of the circle is x2 + y2 = 36 Comparing this equaiton with x2 + y2 = a2,

we get a2 = 36 Given equation of line is 5x + y = 2 Slope of this line = 5 Since, the required tangents are perpendicular

to the given line.

Slope of required tangents (m) = 1

5

Equations of the tangents to the circle x2 + y2 = a2 with slope m are

y = mx 2 2a 1 m

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Std. XI : Perfect Maths ‐ I

the required equations of the tangents are

y = 1

5x

21

36 15

= 1

5x

136 1

25

y = 1

5x

626

5

5y = x 6 26

x 5y 6 26 = 0 24. Find the equations of the tangents to the circle

x2 + y2 – 2x + 8y – 23 = 0 having slope 3. Solution: Let the equation of the tangent with slope 3 be

y = 3x + c. 3x y + c = 0 …(i) Given equation of circle is x2 + y2 2x + 8y 23 = 0 Comparing this equation with x2 + y2 + 2gx + 2fy + c = 0, we get 2g = 2, 2f = 8, c = – 23 g = – 1, f = 4, c = – 23 The centre of the circle is C(1, 4) and its

radius = 1 16 23 = 40 = 2 10 Since line (i) is a tangent to this circle the

perpendicular distance from C(1, 4) to line (i) is equal to radius r.

3 1 4 c

9 1

= 2 10

7 c

10

= 2 10

(7 + c) = 20 7 + c = 20 or 7 + c = 20 c = 13 or c = 27

Equations of the tangents are 3x y + 13 = 0 and 3x y 27 = 0 25. Find the equation of the locus of a point, the

tangents from which to the circle x2 + y2 = 9 are at right angles. Solution: Given equation of the circle is x2 + y2 = 9 Comparing this equation with x2 + y2 = a2, we

get a2 = 9 The locus of the point of intersection of

perpendicular tangents is the director circle of the given circle.

The equation of the director circle of the circle x2 + y2 = a2 is x2 + y2 = 2a2.

the required equation is x2 + y2 = 2(9) x2 + y2 = 18 Alternate method: Given equation of the circle is x2 + y2 = 9 Comparing this equation with x2 + y2 = a2, we

get a2 = 9 Let P(x1, y1) be a point on the required locus. Equations of the tangents to the circle x2 + y2 = a2 with slope m are

y = mx 2 2a 1 m

Equations of the tangents are

y = mx 29 m 1

y = mx 23 1 m

Since, these tangents pass through (x1, y1).

y1 = mx1 23 1 m

y1 mx1 = 23 1 m

(y1 mx1)2 = 9(1 + m2) …[Squaring both the sides]

21y 2mx1y1 + 2 2

1m x = 9 + 9m2

21 9x m2 2mx1y1 + 2

1 9y = 0

This is a quadratic equation which has two roots m1 and m2.

m1m2 = 2121

9

9

y

x

Since, the tangents are at right angles. m1m2 = 1

2121

9

9

y

x

= 1

21y 9 = 9 2

1x

21x + 2

1y = 18

Equation of the locus of point P is x2 + y2 = 18. 26. Tangents to the circle x2 + y2 = a2 with

inclinations, 1 and 2 intersect in P. Find the locus of P such that

i. tan 1 + tan 2 = 0 ii. cot 1 + cot 2 = 5 iii. cot 1. cot 2 = c. Solution: Let P(x1, y1) be a point on the required locus. Equations of the tangents to the circle x2 + y2 = a2 with slope m are

y = mx 2 2a 1 m

Since, these tangents pass through (x1, y1).

y1 = mx1 ± 2 2a 1 m

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Chapter 06: Circle

y1 mx1 = ± 2 2a 1 m

y12 2mx1y1 + m2x1

2 = a2 + a2m2

(x12 a2)m2 2mxly1 + (y1

2 a2) = 0

This is a quadratic equation which has two roots m1 and m2.

m1 + m2 = 1 12 21

2

a

x y

x - and m1 m2 =

2 212 21

a

a

--

y

x

i. Let m1 = tan 1 and m2 = tan 2

Given, tan 1 + tan 2 = 0

m1 + m2 = 0

1 12 21

2

a

x y

x - = 0

2x1y1 = 0

x1y1 = 0

Equation of the locus of point P is xy = 0. ii. Given, cot 1 + cot 2 = 5

1 2

1 1

tanθ tanθ = 5

1

1

m +

2

1

m= 5

1 2

1 2

m +m

m m = 5

1 12 212 212 21

2

a

a

a

x yxy

x

= 5

1 12 2

1

2

ax y

y = 5

2x1y1 = 5y12 5a2

5 y12 2 x1y1 = 5a2

Equation of the locus of point P is 5y2 2xy = 5a2

iii. cot 1cot 2 = c

1 2

1 1

tanθ tanθ = c

1 2

1

m m = c

2 212 21

1

a

a

y

x

= c

2 2

12 2

1

a

a

x

y = c

x12 a2 = c(y1

2 a2)

Equation of the locus of point P is x2 a2 = c(y2 a2).

1. The point P(x, y) where x = 5cos,

y = –3 + 5 sin moves on a circle. Complete the following activity to find its centre, the length of intercept made by the circle on the X-axis.

Solution: x = 5cos, y = –3 + 5sin x = 5cos, y + 3 = 5sin Squaring and adding, we get x2 + (y + 3)2 = 25(sin2 + cos2) = 25 …(i) which is equation of the circle with

centre = 0,

Substituting y = 0 (to get x-intercept) in (i), we get

x2 + 9 = 25 x2 = 16

The circle meets X-axis at , 0 and , 0

The length of the intercept = 2. By completing the following activity, find the

parametric form of the equation of the circle whose Cartesian equation is x2 + y2 + 6x + 2y – 6 = 0.

Solution: x2 + y2 + 6x + 2y – 6 = 0 (x2 + 6x) + (y2 + 2y) = 6

2

x + (y + 1)2 = 16,

which is of the form (x – h)2 + (y – k)2 = r2

r =

The parametric form is x = h + r cos, y = k + r sin

x = + r cos, y = + r sin 3. By completing the following activity, find the

radius of the circle C2, concentric to the circle C1 : x2 + y2 – 4x – 6y + 9 = 0 and having area four times the area of circle C1.

Solution: The circle C1 is given by x2 + y2 – 4x – 6y + 9 = 0

(x – 2)2 + (y – 3)2 =

The area of circle C1 =

The area of circle C2 =

The radius of circle C2 =

Activities for Practice

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Std. XI : Perfect Maths ‐ I

4. Find the equation of the circle whose centre is (4, 3) and cuts the intercept on the line y = 0 of length 8 units. Fill in the boxes.

Solution: The equation y = 0 represents the X-axis. Centre of the circle is C (4, 3). Let M be the mid point of x-intercept AB. |AB| = 8 and abscissa of the centre = 4 M = (4, 0) Circle passes through the origin and point A

lies on the origin. Radius = |CA|

=

The required equation of the circle is

x2 + y2 + x + y + = 0 5. Complete the following activity, to find the

equation of director circle of the circle C1 : x2 + y2 – 8x – 8y + 16 = 0

Solution: The circle C1 is given by x2 + y2 – 8x – 8y + 16 = 0 i.e. (x – 4)2 + (y – 4)2 = (4)2 The equation of the director circle is

(x – 4)2 + (y – 4)2 =

x2 + y2 + x + y + = 0 6. If the lines 5x + 12y – 28 = 0 and

10x + 24y – 4 = 0 are tangents to a circle, then find the radius of the circle by completing the following activity.

Solution: 10x + 24y – 4 = 0 can be written as 5x + 12y – 2 = 0 …(i) Other line is given by 5x + 12y – 28 = 0 …(ii) (i) and (ii) represent parallel lines with

slope = 5

12 , which are tangents to the circle.

The distance between the lines is equal to the length of diameter.

The distance between two parallel lines ax + by + C1 = 0 and ax + by + C2 = 0

is given by 1 2

2 2

C C

a + b

Length of the diameter = 2 2

28

+

The radius of the circle = 7. A pair of tangents are drawn to a unit circle

with centre at the origin. These tangents intersect at a point enclosing an angle of 120. Find the area enclosed by these tangents and the circle by completing the following activity.

Solution: PAQ = 120 OAQ = 60, OQ = 1

QA = OQ cot 60

=

A( OPAQ) = 2 A( OAQ)

= 21

QA OQ2

=

POQ =

area of sector OPQ = 21(1)

2 3

=

6

Required area = A( OPAQ) area of sector OPQ

= 6

8. Find the equation of the tangent to the circle

x2 + y2 + 4x + 6y 12 = 0 at (1, 1) by completing the following activity.

Solution: x2 + y2 + 4x + 6y 12 = 0 Here, g = , f = , c =

Let the given point be (x1, y1) = (1, 1) Equation of the tangent at (x1, y1) is xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0

The required equation of the tangent is 3x + 4y + = 0

1. i. –3 ii. – 4 iii. 4 iv. 8 2. i. –3 ii. 4 iii. –3 iv. –1 3. i. 4 ii. 4 iii. 16 iv. 4

O

Q

A

P

1

A M B4

C(4, 3)

X

Y

y = 0

Answers

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Chapter 06: Circle

4. i. 5 ii. –8 iii. –6 iv. 0 5. i. 32 ii. –8 iii. –8 iv. 0 6. i. –2 ii. 5 iii. 12 iv. 1

7. i. 1

3 ii.

1

3

iii. 60 iv. 1

3

8. i. 2 ii. 3 iii. 12 iv. 7 Based on Exercise 6.1  +1. Find the equation of a circle with centre at

origin and radius 3. 2. Find the equation of the circle with i. centre at (8, – 10) and radius 9

ii. centre at 1 3

,2 2

and radius 3

iii. centre at (– 5, – 1) and radius 3 iv. centre at (– 2, 3) passing through (1, 7) 3. Find the centre and radius of the following

circles. i. x2 + y2 = 36 ii. (x + 2)2 + (y 9)2 = 18

iii. 2

3

2

x + (y – 2)2 = 25

4

+4. Find the equation of a circle whose centre is

(− 3, 1) and which passes through the point (5, 2).

5. Find the equation of the circle with centre at

( 5, 2) and touching the Y-axis. +6. Find the equation of circle touching the Y-axis

at point (0, 3) and whose centre is at (– 3, 3). 7. Find the equation of the circle with centre on

the X-axis and passing through the origin having radius 5.

+8. Find the equation of the circle with A(2, − 3)

and B(− 3, 5) as end points of its diameter. 9. Find the equation of the circle, if the equations

of two diameters are x 3y = 2 and 5x + y = –6 and radius is 4.

10. If x = 3y is a chord of the circle x2 + y2 + 20y = 0, find the equation of the circle with this chord as diameter.

11. Find the equation of a circle with radius 3

units and touching both the coordinate axes having centre in second quadrant.

12. Find the equation of the circle which passes

through the origin and cuts off chords of lengths 3 and 5 on the positive side of the X-axis and Y-axis respectively.

+13. Find the equation of the circle whose centre is

at (3,− 4) and the line 3x − 4y − 5 = 0 cuts the circle at A and B; where l(AB) = 6.

14. Find the equation of circle passing through the

points (1, 3), (2, 1) and whose centre lies on the line x + 5y = 2.

Based on Exercise 6.2  1. Find the centre and radius of each of the

following circles. i. x2 + y2 – 6x + 4y – 12 = 0 ii. x2 + y2 – x + 2y – 3 = 0 iii. 2x2 + 2y2 – 3x + 5y – 7 = 0 2. Show that the equation 2x2 + 2y2 + 8x + 10y – 39 = 0 represents a

circle. +3. Prove that 3x2 + 3y2 − 6x + 4y − 1 = 0,

represents a circle. Find its centre and radius. +4. Find the equation of the circle passing through

the points (5,− 6), (1, 2) and (3, − 4). +5. Show that the points (5, 5), (6, 4), (− 2, 4) and

(7, 1) are on the same circle; i.e. these points are concyclic.

6. Show that the points (– 2, 1), (0, 0), (3, 11)

and (– 3, 2) are concyclic. Based on Exercise 6.3  1. Write the parametric equations of the circles: i. x2 + y2 = 4 ii. 16x2 + 16y2 = 9 iii. (x + 2)2 + (y – 3)2 = 16 +2. Find the parametric equation of the circle

x2 + y2 − 6x + 4y − 3 = 0. 3. Find the parametric representation of the circle

5x2 + 5y2 10x + 30y + 49 = 0. 4. Find the equation of the tangent to the circle x2 + y2 – 3x – 5y = 0 at the origin.

Additional Problems for Practice

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Std. XI : Perfect Maths ‐ I

+5. Find the equation of the tangent to the circle x2 + y2 − 4x − 6y − 12 = 0 at (−1, −1).

6. Show that the line x 4y + 9 = 0 touches the

circle 2x2 + 2y2 + 3x 4y + 1 = 0 at the point (1, 2).

+7. Show that the line 3x − 4y + 15 = 0 is a

tangent to the circle x2 + y2 = 9. Find the point of contact.

Based on Miscellaneous Exercise – 6  1. Find the centre and radius of the circle x = 2 + 3 cos, y = 3 3 sin 2. Find the equation of the circle passing through

the point of intersection of the lines x + 2y = 4 and 2x + 3y 5 = 0 and whose centre is the point of intersection of the lines x y = 1 and 2x + y = 2.

3. Find the equation of the circle which passes

through the origin and cuts off chords of lengths 3 and 5 on the positive side of the X-axis and Y-axis respectively.

4. Show that the points (3, 7), (4, 6), (4, 6) and

(5, 3) are concyclic. 5. Find the equation of tangent to the circle

x2 + y2 = 16 at Pπ

3

.

6. Find the equations of the tangents to the circle

x2 + y2 4x 6y 12 = 0 which are parallel to 3x + 4y = 0.

7. Find the equations of the tangents to the circle x2 + y2 6x 2y 10 = 0 which are

perpendicular to the line 2x y + 5 = 0. 8. Find the equation of the locus of a point, the

tangents from which to the circle 2x2 + 2y2 = 11 are at right angles.

9. Find the equations of the tangents to the circle

x2 + y2 + 4x 2y 8 = 0 with slope 3

2

.

10. Find the length of the tangent segment drawn

from the point (0, 0) to the circle x2 + y2 8x + 12 = 0. 11. Find the value of k, if the length of tangent

segment from the point (3, 4) to the circle

x2 + y2 = k2 is 21 units. 12. Find the lengths of the intercepts made by the

circle x2 + y2 4x 6y 5 = 0 on the co-ordinate axes.

13. Show that the given circles touch each other externally. Find their point of contact and the equation of their common tangent.

x2 + y2 − 8x + 2y + 9 = 0 x2 + y2 + 2x + 12y + 19 = 0 14. Show that the given circles touch each other

internally. Find their point of contact and the equation of their common tangent.

x2 + y2 + 2x 8 = 0 x2 + y2 6x + 6y 46 = 0 Multiple Choice Questions 1. If the line x + 2by + 7 = 0 is a diameter of the

circle x2 + y2 6x + 2y = 0, then b = (A) 3 (B) – 5 (C) –1 (D) 5 2. If a circle whose centre is (1, –3) touches the

line 3x 4y 5 = 0, then the radius of the circle is

(A) 2 (B) 4

(C) 2

5 (D) 2

7

3. The equation of the circle which touches both

the axes and whose radius is a, is (A) x2 + y2 2ax 2ay + a2 = 0 (B) x2 + y2 + ax + ay a2 = 0 (C) x2 + y2 + 2ax + 2ay a2 = 0 (D) x2 + y2 ax ay + a2 = 0 4. The area of the circle whose centre is at (1, 2)

and which passes through the point (4, 6) is (A) 5 (B) 10 (C) 25 (D) None of these 5. The centre and radius of the circle

2 22 2 0 x y x are

(A) 1

, 04

and 1

4

(B) 1

, 02

and 1

2

(C) 1

, 02

and 1

2

(D) 1

0,4

and 1

4

6. The circle 2 2 3 4 2 0 x y x y cuts X-axis

at (A) (2,0),( 3,0) (B) (3,0),(4,0)

(C) (1,0),( 1,0) (D) (1,0),(2,0)

Multiple Choice Questions

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Chapter 06: Circle

7. A circle touches the Y-axis at the point (0, 4) and cuts the X-axis in a chord of length 6 units. The radius of the circle is

(A) 3 (B) 4 (C) 5 (D) 6 8. The equation of the circle which touches

x-axis and whose centre is (1, 2), is (A) x2 + y2 2x + 4y + 1 = 0 (B) x2 + y2 2x 4y + 1 = 0 (C) x2 + y2 + 2x + 4y + 1 = 0 (D) x2 + y2 + 4x + 2y + 4 = 0 9. If the radius of the circle x2 + y2 18x + 12y + k = 0 be 11, then k = (A) 347 (B) 4 (C) 4 (D) 49 10. Centre of circle (x x1)(x x2) + (y y1)(y y2) = 0 is

(A) 1 1 2 2,2 2

x y x y

(B) 1 1 2 2,2 2

x y x y

(C) 1 2 1 2,2 2

x x y y

(D) 1 2 1 2,2 2

x x y y

11. If the equation px2 + (2 q)xy + 3y2 6qx + 30y + 6q = 0

represents a circle, then the values of p and q are (A) 3, 1 (B) 2, 2 (C) 3, 2 (D) 3, 4 12. The equation of circle passing through (4, 5)

and having the centre at (2, 2), is (A) x2 + y2 + 4x + 4y 5 = 0 (B) x2 + y2 4x 4y 5 = 0 (C) x2 + y2 4x = 13 (D) x2 + y2 4x 4y + 5 = 0 13. Radius of circle (x 5)(x 1) + (y 7)(y 4) = 0 is (A) 3 (B) 4 (C) 5/2 (D) 7/2 14. The equation of the circle which passes

through the points (2, 3) and (4, 5) and the centre lies on the straight line y 4x + 3 = 0, is

(A) x2 + y2 + 4x 10y + 25 = 0 (B) x2 + y2 4x 10y + 25 = 0 (C) x2 + y2 4x 10y + 16 = 0 (D) x2 + y2 14y + 18 = 0 15. The equation of the circle with centre on X-axis,

radius 5 and passing through the point (2, 3), is (A) x2 + y2 + 4x 21 = 0 (B) x2 + y2 + 4x + 21 = 0

(C) x2 + y2 4x 21 = 0 (D) x2 + y2 + 5x 21 = 0 16. If the lines x + y = 6 and x + 2y = 4 be

diameters of the circle whose diameter is 20, then the equation of the circle is

(A) x2 + y2 16x + 4y 32 = 0 (B) x2 + y2 + 16x + 4y 32 = 0 (C) x2 + y2 + 16x + 4y + 32 = 0 (D) x2 + y2 + 16x 4y + 32 = 0 17. The equation of the circle whose diameters

have the end points (a, 0) (0, b) is given by (A) x2 + y2 ax by = 0 (B) x2 + y2 + ax by = 0 (C) x2 + y2 ax + by = 0 (D) x2 + y2 + ax + by = 0 18. A circle touches the axes at the points (3, 0)

and (0, –3). The centre of the circle is (A) (3, –3) (B) (0, 0) (C) (–3, 0) (D) (6, –6) 19. Radius of the circle x2 + y2 + 2x cos + 2ysin 8 = 0, is (A) 1 (B) 3 (C) 2 3 (D) 10 20. If the coordinates of one end of the diameter

of the circle x2 + y2 8x 4y + c = 0 are (–3, 2), then the coordinates of other end are

(A) (5, 3) (B) (6, 2) (C) (1, –8) (D) (11, 2) 21. The centre of the circle x = 1 + 2cos, y = 3 + 2sin, is (A) (1, –3) (B) (–1, 3) (C) (1, 3) (D) None of these 22. If (, ) is the centre of a circle passing

through the origin, then its equation is (A) x2 + y2 x y = 0 (B) x2 + y2 + 2x + 2y = 0 (C) x2 + y2 2x 2y = 0 (D) x2 + y2 + x + y = 0 23. The centre of a circle is (2, –3) and the

circumference is 10. Then the equation of the circle is

(A) x2 + y2 + 4x + 6y + 12 = 0 (B) x2 + y2 4x + 6y + 12 = 0 (C) x2 + y2 4x + 6y 12 = 0 (D) x2 + y2 4x 6y 12 = 0 24. The radius of the circle x2 + y2 + 4x + 6y + 13 = 0 is (A) 26 (B) 13

(C) 23 (D) 0

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Std. XI : Perfect Maths ‐ I

25. The equation of the tangent to the circle 2 2 2r x y at (a,b) is a b 0 x y , where

is (A) 2a (B) 2b (C) 2r (D) None of these 26. The equations of the tangents to the circle

2 2 13 x y at the points whose abscissa is 2 are

(A) 2 3 13,2 3 13 x y x y

(B) 3 2 13,2 3 13 x y x y

(C) 2 3 13, 3 2 13 x y x y

(D) None of these 27. The number of common tangents to two

circles 2 2 4 x y and 2 2 8 12 0 x y x is

(A) 1 (B) 2 (C) 3 (D) 4 28. The two circles 2 2 2 6 6 0 x y x y and

2 2 5 6 15 0 x y x y touch each other. The

equation of their common tangent is

(A) 3x

(B) 6y (C) 7 12 21 0 x y

(D) 7 12 21 0 x y Based on Exercise 6.1  1. x2 + y2 = 9 2. i. x2 + y2 – 16x + 20y + 83 = 0 ii. 2x2 + 2y2 – 2x – 6y – 13 = 0 iii. x2 + y2 + 10x + 2y + 17 = 0 iv. x2 + y2 + 4x – 6y – 12 = 0 3. i. (0, 0); 6

ii. (– 2, 9); 3 2

iii. 3

, 22

; 5

2

4. x2 + y2 + 6x – 2y – 55 = 0 5. x2 + y2 + 10x – 4y + 4 = 0 6. x2 + y2 + 6x – 6y + 9 = 0 7. x2 + y2 10x = 0 8. x2 + y2 + x – 2y – 21 = 0 9. x2 + y2 + 2x + 2y – 14 = 0 10. x2 + y2 + 6x + 2y = 0 11. x2 + y2 + 6x – 6y + 9 = 0 12. x2 + y2 – 3x – 5y = 0 13. x2 + y2 – 6x + 8y = 0 14. x2 + y2 + x – y 8 = 0

Based on Exercise 6.2 

1. i. (3, – 2); 5 ii. 1

, 12

; 17

2

iii. 3 5

,4 4

; 3 10

4

3. 2

1,3

; 4

3

4. x2 + y2 – 22x – 4y + 25 = 0 Based on Exercise 6.3  1. i. x = 2cos , y = 2sin

ii. x = 3

4cos , y =

3

4sin

iii. x = – 2 + 4cos , y = 3 + 4sin 2. x = 3 + 4cos , y = – 2 + 4sin

3. x = 1 + 1

5cos , y = – 3 +

1

5sin

4. 3x + 5y = 0 5. 3x + 4y + 7 = 0

7. 9 12

,5 5

Based on Miscellaneous Exercise – 6  1. (2, 3); 3 2. x2 + y2 2x 17 = 0 3. x2 + y2 3x 5y = 0 5. x + 3y 8 = 0

6. 3x + 4y + 7 = 0 3x + 4y 43 = 0 7. x + 2y + 5 = 0 x + 2y 15 = 0 8. x2 + y2 = 11 9. 3x + 2y + 17 = 0 3x + 2y 9 = 0 10. 2 3 units 11. 2 12. x-intercept = 6, y-intercept = 2 14 13. (2, 3), x + y + 1 = 0

14. 17 9

,5 5

, 4x 3y + 19 = 0

1. (D) 2. (A) 3. (A) 4. (C) 5. (A) 6. (D) 7. (C) 8. (B) 9. (C) 10. (C) 11. (C) 12. (B) 13. (C) 14. (B) 15. (A) 16. (A) 17. (A) 18. (A) 19. (B) 20. (D) 21. (B) 22. (C) 23. (C) 24. (D) 25. (C) 26. (A) 27. (C) 28. (A)

Answers to Multiple Choice Questions

Answers to Additional Practice Problems

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Chapter 06: Circle

1. The number of common tangents to the circles

x2 + y2 4x 6y 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is

[JEE (Main) 2015] (A) 1 (B) 2 (C) 3 (D) 4 2. The centres of those circles which touch the

circle, x2 + y2 8x 8y 4 = 0, externally and also touch the X - axis, lie on:

[JEE (Main) 2016] (A) a circle. (B) an ellipse which is not a circle. (C) a hyperbola. (D) a parabola. 3. If one of the diameters of the circle, given by

the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (– 3, 2), then the radius of S is [JEE (Main) 2016]

(A) 5 3 (B) 5

(C) 10 (D) 5 2 4. The radius of a circle, having minimum area,

which touches the curve y = 4 – x2 and the lines y = | x | is [JEE (Main) 2017]

(A) 4 2 1 (B) 2 2 1

(C) 2 2 1 (D) 4 2 1 5. If the tangent at (1, 7) to the curve x2 = y – 6

touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is [JEE (Main) 2018]

(A) 185 (B) 85 (C) 95 (D) 195 6. Let the orthocentre and centroid of a triangle

be A(– 3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter,is [JEE (Main) 2018]

(A) 2 10 (B) 5

32

(C) 3 5

2 (D) 10

7. The sides of a rectangle are given by

x = a and y = b. The equation of the circle passing through the vertices of the rectangle is

[MHT CET 2018] (A) x2 + y2 = a2 (B) x2 + y2 = a2 + b2 (C) x2 + y2 = a2 – b2 (D) (x – a)2 + (y –b)2 = a2 + b2

Answers: 1. (C) 2. (D) 3. (A) 4. (D) 5. (C) 6. (B) 7. (B) Hints: 1. x2 + y2 4x 6y 12 = 0

C1 = (2, 3), r = 2 22 3 12 = 5 x2 + y2 + 6x + 18y + 26 = 0 C2 = (3, 9),

r = 9 81 26

= 64 = 8 l(C1C2) = r1 + r2 The circles touch externally. Number of common tangents is 3. 2. Given, x2 + y2 8x 8y 4 = 0 (x2 8x + 16) + (y2 8y + 16) = 16 + 16 + 4 (x 4)2 + (y 4)2 = 36 Equation of circle touching X – axis (x h)2 + (y k)2 = k2 Since, both circle touches externally distance between their centre = r1 + r2

2 24 h 4 k = 6 + k

(4 h)2 + (4 – k)2 = (6 + k)2 (4 h)2 = 36 + 12k + k2 16 + 8k k2 (4 h)2 = 20k + 20 which is equation of a parabola. 3. The centre of the given circle is C1(2, 3) and

radius

= 2 22 3 12

= 5

C1C2 = 2 23 2 2 3

= 25 25

= 50 Radius of S is C2A

= 2250 5

= 75

= 5 3

Competitive Corner

C1 (4, 4)C2 (h, k)

C1 (2,3)5

A

C2(3, 2)

SAMPLE C

ONTENT

262 

262

Std. XI : Perfect Maths ‐ I

4. Let the radius of circle be r.Then, centre = (0, 4 – r)

Length of perpendicular from (0, 4 – r) to the

line y = x is r = 0 (4 r)

2

r – 4 = r 2

r = 4

1 2

r = 4

1 2 …

4r (rejected)

1 2

r = 4 2 1

5. Equation of the tangent at (1, 7) to x2 = y – 6 is2x – y + 5 = 0Centre of the given circle is (– 8, – 6).

Length of perpendicular from the centre(– 8, – 6) to the line 2x – y + 5 = 0 is equal tothe radius of the circle.

2 2

2 8 6 5

2 1

= 2 28 6 c

5

5

= 100 c

c = 95

6. AB = 2 23 3 3 5

= 36 4 = 40 = 2 10

Centroid divides orthocentre and circumcentre in the ratio 2 : 1.

AB : BC = 2 : 1

AC = 3

2 AB

= 3

2 2 10 = 3 10

radius = 1

2 AC =

1

2 3 10 =

53

2

7.

Here, the diagonals AC and BD of rectangleABCD are diameters of the circle passingthrough the vertices A, B, C and D. Considering diagonal AC with end pointsA(a, b) and C (–a, –b), we get Equation of circle in diameter form as,

(x – a) (x – (–a)) + (y – b) (y – (–b)) = 0 x2 – a2 – y2 – b2 = 0 x2 + y2 = a2 + b2

.y

xO

(–8, –6)

2x – y + 5 = 0

A B C

x = – a

B (– a, b)

x = a

A (a, b)

C (– a, – b) D (a, – b)

y = b

X

y = – b

Y

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