# Conic Sections Chapter 11 Class 11 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 11 Maths Chapter 11‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Maths Chapter 11 Exercise Solution’ using the download button.

### Chapter 11: Conic Sections

#### 11.1 Introduction

In the preceding Chapter 10, we have studied various forms of the equations of a line. In this chapter, we shall study some other curves, viz., circles, ellipses, parabolas, and hyperbolas.

The names parabola and hyperbola are given by Apollonius.

These curves are in fact, known as conic sections or more commonly conics because they can be obtained as intersections of a plane with a double-napped right circular cone.

These curves have a very wide range of applications in fields such as planetary motion, design of telescopes and antennas, reflectors in flashlights and automobile headlights, etc.

Now, in the subsequent sections, we will see how the intersection of a plane with a double-napped right circular cone results in different types of curves.

#### 11.3 Circle

Definition 1 A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.

The fixed point is called the center of the circle and the distance from the center to a point on the circle is called the radius of the circle.

#### 11.4 Parabola

Definition 2 A parabola is the set of all points in a plane that is equidistant from a fixed line and a fixed point (not on the line) in the plane.

### NCERT Solutions Class 11 Maths Chapter 11 Conic Sections

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 – h)+ (y – k)= r2

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

The equation of the circle is

(x – 0)+ (y – 2)= 22

x+ y+ 4 – 4y = 4

x+ y2 – 4y = 0

∴ The equation of the circle is x+ y2 – 4y = 0

2. Centre (–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 – h)+ (y – k)= r2

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

The equation of the circle is

(x + 2)2 + (y – 3)2 = (4)2

x2 + 4x + 4 + y2 – 6y + 9 = 16

x2 + y2 + 4x – 6y – 3 = 0

∴ The equation of the circle is x2 + y2 + 4x – 6y – 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 center (h, k) and

Radius r is given as (x – h)+ (y – k)= r2

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

The equation of the circle is

(x – 1/2)2 + (y – 1/4)2 = (1/12)2

x2 – x + ¼ + y2 – y/2 + 1/16 = 1/144

x2 – x + ¼ + y2 – y/2 + 1/16 = 1/144

144x2 – 144x + 36 + 144y2 – 72y + 9 – 1 = 0

144x2 – 144x + 144y2 – 72y + 44 = 0

36x2 + 36x + 36y2 – 18y + 11 = 0

36x2 + 36y2 – 36x – 18y + 11= 0

∴ The equation of the circle is 36x2 + 36y2 – 36x – 18y + 11= 0

4. Centre (1, 1) and radius √2

Solution:

Given:

Centre (1, 1) and radius √2

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

Radius r is given as (x – h)+ (y – k)= r2

So, centre (h, k) = (1, 1) and radius (r) = √2

The equation of the circle is

(x-1)2 + (y-1)= (√2)2

x2 – 2x + 1 + y2 -2y + 1 = 2

x2 + y2 – 2x -2y = 0

∴ The equation of the circle is x2 + y2 – 2x -2y = 0

5. Centre (–a, –b) and radius √(a2 – b2)

Solution:

Given:

Centre (-a, -b) and radius √(a2 – b2)

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

Radius r is given as (x – h)+ (y – k)= r2

So, centre (h, k) = (-a, -b) and radius (r) = √(a2 – b2)

The equation of the circle is

(x + a)2 + (y + b)2 = (√(a2 – b2)2)

x2 + 2ax + a2 + y2 + 2by + b2 = a2 – b2

x2 + y2 +2ax + 2by + 2b2 = 0

∴ The equation of the circle is x2 + y2 +2ax + 2by + 2b2 = 0

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

6. (x + 5)2 + (y – 3)2 = 36

Solution:

Given:

The equation of the given circle is (x + 5)2 + (y – 3)2 = 36

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

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

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

7. x2 + y2 – 4x – 8y – 45 = 0

Solution:

Given:

The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.

x2 + y2 – 4x – 8y – 45 = 0

(x2 – 4x) + (y2 -8y) = 45

(x2 – 2(x) (2) + 22) + (y2 – 2(y) (4) + 42) – 4 – 16 = 45

(x – 2)2 + (y – 4)2 = 65

(x – 2)2 + (y – 4)2 = (√65)2 [which is form (x-h)2 +(y-k)2 = r2]

Where h = 2, K = 4 and r = √65

∴ The centre of the given circle is (2, 4) and its radius is √65.

8. x2 + y2 – 8x + 10y – 12 = 0

Solution:

Given:

The equation of the given circle is x2 + y2 -8x + 10y -12 = 0.

x2 + y2 – 8x + 10y – 12 = 0

(x2 – 8x) + (y+ 10y) = 12

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

(x – 4)2 + (y + 5)2 = 53

(x – 4)2 + (y – (-5))2 = (√53)2 [which is form (x-h)2 +(y-k)2 = r2]

Where h = 4, K= -5 and r = √53

∴ The centre of the given circle is (4, -5) and its radius is √53.

9. 2x2 + 2y2 – x = 0

Solution:

The equation of the given of the circle is 2x2 + 2y2 –x = 0.

2x2 + 2y2 –x = 0

(2x2 + x) + 2y2 = 0

(x2 – 2 (x) (1/4) + (1/4)2) + y2 – (1/4)2 = 0

(x – 1/4)2 + (y – 0)2 = (1/4)2 [which is form (x-h)2 +(y-k)2 = r2]

Where, h = ¼, K = 0, and r = ¼

∴ The center 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 center is on the line 4x + y = 16.

Solution:

Let us consider the equation of the required circle be (x – h)2+ (y – k)2 = r2

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

So,

(4 – h)+ (1 – k)2 = r2 ……………..(1)

(6– h)2+ (5 – k)2 = r2 ………………(2)

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

4h + k =16………………… (3)

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

(4 – h)2+ (1 – k)2 =(6 – h)2 + (5 – k)2

16 – 8h + h2 +1 -2k +k2 = 36 -12h +h2+15 – 10k + k2

16 – 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11……………. (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 – 3)2+ (1 – 4)2 = r2

(1)2 + (-3)2 = r2

1+9 = r2

r = √10

so now, (x – 3)+ (y – 4)2 = (√10)2

x2 – 6x + 9 + y2 – 8y + 16 =10

x2 + y2 – 6x – 8y + 15 = 0

∴ The equation of the required circle is x2 + y2 – 6x – 8y + 15 = 0