# NCERT Solutions for Class 11 Maths Chapter 11 - Conic Sections Exercise 11.1

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections contain solutions for all exercise 11.1 questions. Chapter 11 Conic Sections of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. Practising the NCERT Solution boosts studentsâ€™ confidence to score more in the board exam. Download NCERT Class 11 Maths Solutions to access the solutions for all the exercises of all the chapters.

A conic section is a chapter that deals with different sections of cones. There are a wide variety of topics covered in its first exercise. Exercise 11.1 of NCERT Solutions for Class 11 Maths Chapter 11 â€“ Conic Sections is based on the following topics:

1. Introduction
2. Sections of a Cone
1. Circle, ellipse, parabola and hyperbola
2. Degenerated conic sections
3. Circle

## NCERT Solutions for Class 11 Maths Chapter 11 â€“ Conic Sections Exercise 11.1

### Solutions for Class 11 Maths Chapter 11 â€“ Exercise 11.1

In each of the following Exercises 1 to 5, find the equation of the circle with
1. Centre (0, 2) and radius 2.

Solution:

Given:

Centre (0, 2) and radius 2.

Let us consider the equation of a circle with centre (h, k), and

Radius r is given as (x â€“ h)2 + (y â€“ k)2 = r2

So, centre (h, k) = (0, 2) and radius (r) = 2

The equation of the circle is

(x â€“ 0)2 + (y â€“ 2)2 = 22

x2 + y2 + 4 â€“ 4y = 4

x2 + y2Â â€“ 4y = 0

âˆ´ The equation of the circle is x2 + y2Â â€“ 4y = 0

2. Centre (â€“2, 3) and radius 4.

Solution:

Given:

Centre (-2, 3) and radius 4

Let us consider the equation of a circle with centre (h, k), and

Radius r is given as (x â€“ h)2 + (y â€“ k)2 = r2

So, centre (h, k) = (-2, 3) and radius (r) = 4

The equation of the circle is

(x + 2)2Â + (y â€“ 3)2Â = (4)2

x2Â + 4x + 4 + y2Â â€“ 6y + 9 = 16

x2Â + y2Â + 4x â€“ 6y â€“ 3 = 0

âˆ´ The equation of the circle is x2Â + y2Â + 4x â€“ 6y â€“ 3 = 0

3. Centre (1/2, 1/4) and radius (1/12).

Solution:

Given:

Centre (1/2, 1/4) and radius 1/12

Let us consider the equation of a circle with centre (h, k), and

Radius r is given as (x â€“ h)2 + (y â€“ k)2 = r2

So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12

The equation of the circle is

(x â€“ 1/2)2 + (y â€“ 1/4)2 = (1/12)2

x2 â€“ x + Â¼ + y2 â€“ y/2 + 1/16 = 1/144

x2 â€“ x + Â¼ + y2 â€“ y/2 + 1/16 = 1/144

144x2Â â€“ 144x + 36 + 144y2Â â€“ 72y + 9 â€“ 1 = 0

144x2Â â€“ 144x + 144y2Â â€“ 72y + 44 = 0

36x2Â + 36x + 36y2 â€“ 18y + 11 = 0

36x2Â + 36y2Â â€“ 36x â€“ 18y + 11= 0

âˆ´ The equation of the circle is 36x2Â + 36y2Â â€“ 36x â€“ 18y + 11= 0

4. Centre (1, 1) and radius âˆš2

Solution:

Given:

Centre (1, 1) and radius âˆš2

Let us consider the equation of a circle with centre (h, k), and

Radius r is given as (x â€“ h)2 + (y â€“ k)2 = r2

So, centre (h, k) = (1, 1) and radius (r) = âˆš2

The equation of the circle is

(x-1)2Â + (y-1)2 = (âˆš2)2

x2Â â€“ 2x + 1 + y2Â -2y + 1 = 2

x2Â + y2Â â€“ 2x -2y = 0

âˆ´ The equation of the circle is x2Â + y2Â â€“ 2x -2y = 0

5. Centre (â€“a, â€“b) and radius âˆš(a2 â€“ b2)

Solution:

Given:

Centre (-a, -b) and radius âˆš(a2 â€“ b2)

Let us consider the equation of a circle with centre (h, k), and

Radius r is given as (x â€“ h)2 + (y â€“ k)2 = r2

So, centre (h, k) = (-a, -b) and radius (r) = âˆš(a2 â€“ b2)

The equation of the circle is

(x + a)2Â + (y + b)2Â = (âˆš(a2 â€“ b2)2)

x2Â + 2ax + a2Â + y2Â + 2by + b2Â = a2Â â€“ b2

x2Â + y2Â +2ax + 2by + 2b2Â = 0

âˆ´ The equation of the circle is x2Â + y2Â +2ax + 2by + 2b2Â = 0

In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)2Â + (y â€“ 3)2Â = 36

Solution:

Given:

The equation of the given circle is (x + 5)2Â + (y â€“ 3)2 = 36

(x â€“ (-5))2 + (y â€“ 3)2Â = 62 [which is of the form (x â€“ h)2Â + (y â€“ k )2Â = r2]

Where, h = -5, k = 3 and r = 6

âˆ´ The centre of the given circle is (-5, 3), and its radius is 6.

7. x2Â + y2Â â€“ 4x â€“ 8y â€“ 45 = 0

Solution:

Given:

The equation of the given circle is x2Â + y2 â€“ 4x â€“ 8y â€“ 45 = 0

x2Â + y2Â â€“ 4x â€“ 8y â€“ 45 = 0

(x2Â â€“ 4x) + (y2Â -8y) = 45

(x2 â€“ 2(x) (2) + 22) + (y2 â€“ 2(y) (4) + 42) â€“ 4 â€“ 16 = 45

(x â€“ 2)2Â + (y â€“ 4)2Â = 65

(x â€“ 2)2Â + (y â€“ 4)2Â = (âˆš65)2 [which is in the form (x-h)2Â +(y-k)2Â = r2]

Where h = 2, K = 4 and r = âˆš65

âˆ´ The centre of the given circle is (2, 4), and its radius is âˆš65.

8. x2Â + y2Â â€“ 8x + 10y â€“ 12 = 0

Solution:

Given:

The equation of the given circle is x2Â + y2 -8x + 10y -12 = 0

x2Â + y2Â â€“ 8x + 10y â€“ 12 = 0

(x2Â â€“ 8x) + (y2 + 10y) = 12

(x2 â€“ 2(x) (4) + 42) + (y2 â€“ 2(y) (5) + 52) â€“ 16 â€“ 25 = 12

(x â€“ 4)2Â + (y + 5)2Â = 53

(x â€“ 4)2Â + (y â€“ (-5))2Â = (âˆš53)2 [which is in the form (x-h)2Â +(y-k)2Â = r2]

Where h = 4, K= -5 and r = âˆš53

âˆ´ The centre of the given circle is (4, -5), and its radius is âˆš53.

9. 2x2 + 2y2 â€“ x = 0

Solution:

The equation of the given circle is 2x2Â + 2y2 â€“x = 0

2x2Â + 2y2Â â€“x = 0

(2x2Â + x) + 2y2Â = 0

(x2 â€“ 2 (x) (1/4) + (1/4)2) + y2 â€“ (1/4)2 = 0

(x â€“ 1/4)2 + (y â€“ 0)2 = (1/4)2 [which is in the form (x-h)2Â +(y-k)2Â = r2]

Where, h = Â¼, K = 0, and r = Â¼

âˆ´ The centre of the given circle is (1/4, 0), and its radius is 1/4.

10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

Solution:

Let us consider the equation of the required circle be (x â€“ h)2+ (y â€“ k)2Â = r2

We know that the circle passes through points (4,1) and (6,5).

So,

(4 â€“ h)2 + (1 â€“ k)2Â = r2Â â€¦â€¦â€¦â€¦â€¦..(1)

(6â€“ h)2+ (5 â€“ k)2Â = r2Â â€¦â€¦â€¦â€¦â€¦â€¦(2)

The centre (h, k) of the circle lies on line 4x + y = 16.

4h + k =16â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

From the equation (1) and (2), we obtain

(4 â€“ h)2+ (1 â€“ k)2Â =(6 â€“ h)2Â + (5 â€“ k)2

16 â€“ 8h + h2Â +1 -2k +k2Â = 36 -12h +h2+15 â€“ 10k + k2

16 â€“ 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11â€¦â€¦â€¦â€¦â€¦. (4)

On solving equations (3) and (4), we obtain h=3 and k= 4.

On substituting the values of h and k in equation (1), we obtain

(4 â€“ 3)2+ (1 â€“ 4)2Â = r2

(1)2Â + (-3)2Â = r2

1+9 = r2

r =Â âˆš10

So now, (x â€“ 3)2 + (y â€“ 4)2Â = (âˆš10)2

x2Â â€“ 6x + 9 + y2Â â€“ 8y + 16 =10

x2Â + y2Â â€“ 6x â€“ 8y + 15 = 0

âˆ´ The equation of the required circle is x2Â + y2Â â€“ 6x â€“ 8y + 15 = 0

11. Find the equation of the circle passing through the points (2, 3) and (â€“1, 1) and whose centre is on the line x â€“ 3y â€“ 11 = 0.

Solution:

Let us consider the equation of the required circle be (x â€“ h)2 + (y â€“ k)2Â = r2

We know that the circle passes through points (2,3) and (-1,1).

(2 â€“ h)2+ (3 â€“ k)2Â =r2Â â€¦â€¦â€¦â€¦â€¦..(1)

(-1 â€“ h)2+ (1â€“ k)2Â =r2Â â€¦â€¦â€¦â€¦â€¦â€¦(2)

The centre (h, k) of the circle lies on line x â€“ 3y â€“ 11= 0.

h â€“ 3k =11â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

From the equation (1) and (2), we obtain

(2 â€“ h)2+ (3 â€“ k)2Â =(-1 â€“ h)2Â + (1 â€“ k)2

4 â€“ 4h + h2Â +9 -6k +k2Â = 1 + 2h +h2+1 â€“ 2k + k2

4 â€“ 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11â€¦â€¦â€¦â€¦â€¦. (4)

Now, let us multiply equation (3) by 6 and subtract it from equation (4) to get

6h+ 4k â€“ 6(h-3k) = 11 â€“ 66

6h + 4k â€“ 6h + 18k = 11 â€“ 66

22 k = â€“ 55

K = -5/2

Substitute this value of K in equation (4) to get

6h + 4(-5/2) = 11

6h â€“ 10 = 11

6h = 21

h = 21/6

h = 7/2

We obtain h =Â 7/2and k = -5/2

On substituting the values of h and k in equation (1), we get

(2 â€“ 7/2)2 + (3 + 5/2)2 = r2

[(4-7)/2]2 + [(6+5)/2]2 = r2

(-3/2)2 + (11/2)2 = r2

9/4 + 121/4 = r2

130/4 = r2

The equation of the required circle is

(x â€“ 7/2)2 + (y + 5/2)2 = 130/4

[(2x-7)/2]2 + [(2y+5)/2]2 = 130/4

4x2Â -28x + 49 +4y2Â + 20y + 25 =130

4x2Â +4y2Â -28x + 20y â€“ 56 = 0

4(x2Â +y2Â -7x + 5y â€“ 14) = 0

x2 + y2 â€“ 7x + 5y â€“ 14 = 0

âˆ´ The equation of the required circle is x2 + y2 â€“ 7x + 5y â€“ 14 = 0

12. Find the equation of the circle with radius 5 whose centre lies on the x-axis and passes through the point (2, 3).

Solution:

Let us consider the equation of the required circle be (x â€“ h)2+ (y â€“ k)2Â = r2

We know that the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

So now, the equation of the circle is (x â€“ h)2Â + y2Â = 25.

It is given that the circle passes through the point (2, 3), so the point will satisfy the equation of the circle.

(2 â€“ h)2+ 32Â = 25

(2 â€“ h)2Â = 25-9

(2 â€“ h)2Â = 16

2 â€“ h = Â± âˆš16 = Â± 4

If 2-h = 4, then h = -2

If 2-h = -4, then h = 6

Then, when h = -2, the equation of the circle becomes

(x + 2)2Â + y2Â = 25

x2Â + 12x + 36 + y2Â = 25

x2Â + y2Â + 4x â€“ 21 = 0

When h = 6, the equation of the circle becomes

(x â€“ 6)2Â + y2Â = 25

x2Â -12x + 36 + y2Â = 25

x2Â + y2Â -12x + 11 = 0

âˆ´ The equation of the required circle is x2Â + y2Â + 4x â€“ 21 = 0 and x2Â + y2Â -12x + 11 = 0

13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Solution:

Let us consider the equation of the required circle be (x â€“ h)2+ (y â€“ k)2Â =r2

We know that the circle passes through (0, 0).

So, (0 â€“ h)2+ (0 â€“ k)2Â = r2

h2Â + k2Â =Â r2

Now, the equation of the circle is (x â€“ h)2 + (y â€“ k)2Â = h2Â + k2.

It is given that the circle intercepts a and b on the coordinate axes.

i.e., the circle passes through points (a, 0) and (0, b).

So, (a â€“ h)2+ (0 â€“ k)2Â =h2Â +k2â€¦â€¦â€¦â€¦â€¦..(1)

(0 â€“ h)2+ (bâ€“ k)2Â =h2Â +k2â€¦â€¦â€¦â€¦â€¦â€¦(2)

From equation (1), we obtain

a2Â â€“ 2ah + h2Â +k2Â = h2Â +k2

a2Â â€“ 2ah = 0

a(a â€“ 2h) =0

a = 0 or (a -2h) = 0

However, a â‰ Â 0; hence, (a -2h) = 0

h = a/2

From equation (2), we obtain

h2Â â€“ 2bk + k2Â + b2= h2Â +k2

b2Â â€“ 2bk = 0

b(bâ€“ 2k) = 0

b= 0 or (b-2k) =0

However, aÂ â‰  0; hence, (b -2k) = 0

kÂ =b/2

So, the equation is

(x â€“ a/2)2 + (y â€“ b/2)2 = (a/2)2 + (b/2)2

[(2x-a)/2]2 + [(2y-b)/2]2 = (a2 + b2)/4

4x2Â â€“ 4ax + a2Â +4y2Â â€“ 4by + b2Â = a2Â + b2

4x2Â + 4y2Â -4ax â€“ 4by = 0

4(x2Â +y2Â -7x + 5y â€“ 14) = 0

x2 + y2 â€“ axÂ â€“ by = 0

âˆ´ The equation of the required circle is x2 + y2 â€“ ax â€“ by = 0

14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Solution:

Given:

The centre of the circle is given as (h, k) = (2,2).

We know that the circle passes through the point (4,5), and the radius (r) of the circle is the distance between points (2,2) and (4,5).

r = âˆš[(2-4)2 + (2-5)2]

= âˆš[(-2)2 + (-3)2]

= âˆš[4+9]

= âˆš13

The equation of the circle is given as

(xâ€“ h)2+ (y â€“ k)2Â = r2

(x â€“h)2Â + (y â€“ k)2Â = (âˆš13)2

(x â€“2)2Â + (y â€“ 2)2Â = (âˆš13)2

x2Â â€“ 4x + 4 + y2 â€“ 4y + 4 = 13

x2Â + y2Â â€“ 4x â€“ 4y = 5

âˆ´ The equation of the required circle is x2Â + y2Â â€“ 4x â€“ 4y = 5

15. Does the point (â€“2.5, 3.5) lie inside, outside or on the circle x2Â + y2Â = 25?

Solution:

Given:

The equation of the given circle isÂ x2Â +y2 = 25

x2Â + y2Â = 25

(x â€“ 0)2Â + (y â€“ 0)2Â = 52 [which is in the form (x â€“ h)2 + (y â€“ k)2Â = r2]

Where, h = 0, k = 0 and r = 5

So, the distance between the point (-2.5, 3.5) and the centre (0,0) is

= âˆš[(-2.5 â€“ 0)2 + (-3.5 â€“ 0)2]

= âˆš(6.25 + 12.25)

= âˆš18.5

= 4.3 [which is < 5]

Since the distance between the point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle, the point (-2.5, -3.5) lies inside the circle.

### Access Other Exercise Solutions of Class 11 Maths Chapter 11 â€“ Conic Sections

Exercise 11.2 Solutions: 12 Questions

Exercise 11.3 Solutions: 20 Questions

Exercise 11.4 Solutions: 15 Questions

Miscellaneous Exercise on Chapter 11 Solutions: 8 Questions

Also explore â€“ NCERT Class 11 Solutions