*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 students’ 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 – Conic Sections is based on the following topics:
- Introduction
- Sections of a Cone
- Circle, ellipse, parabola and hyperbola
- Degenerated conic sections
- Circle
NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Exercise 11.1
Solutions for Class 11 Maths Chapter 11 – 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 – h)2 + (y – k)2 = r2
So, centre (h, k) = (0, 2) and radius (r) = 2
The equation of the circle is
(x – 0)2 + (y – 2)2 = 22
x2 + y2 + 4 – 4y = 4
x2 + y2 – 4y = 0
∴ The equation of the circle is x2 + 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)2 + (y – k)2 = 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 centre (h, k), and
Radius r is given as (x – h)2 + (y – k)2 = 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)2 + (y – k)2 = r2
So, centre (h, k) = (1, 1) and radius (r) = √2
The equation of the circle is
(x-1)2 + (y-1)2 = (√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)2 + (y – k)2 = 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
x2Â + y2Â +2ax + 2by + 2b2Â = 0
∴ The equation of the circle is x2 + y2 +2ax + 2by + 2b2 = 0
In each of the following Exercises 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 in the 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 x2Â + y2 -8x + 10y -12 = 0
x2 + y2 – 8x + 10y – 12 = 0
(x2 – 8x) + (y2 + 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 in the 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 circle is 2x2 + 2y2 –x = 0
2x2 + 2y2 –x = 0
(2x2Â + x) + 2y2Â = 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 in the form (x-h)2 +(y-k)2 = r2]
Where, h = ¼, K = 0, and r = ¼
∴ 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 – h)2+ (y – k)2 = r2
We know that the circle passes through points (4,1) and (6,5).
So,
(4 – h)2 + (1 – k)2 = r2 ……………..(1)
(6– h)2+ (5 – k)2 = r2 ………………(2)
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)2Â + (-3)2Â = r2
1+9 = r2
r = √10
So now, (x – 3)2 + (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
11. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
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 (2,3) and (-1,1).
(2 – h)2+ (3 – k)2 =r2 ……………..(1)
(-1 – h)2+ (1– k)2 =r2 ………………(2)
The centre (h, k) of the circle lies on line x – 3y – 11= 0.
h – 3k =11………………… (3)
From the equation (1) and (2), we obtain
(2 – h)2+ (3 – k)2 =(-1 – h)2 + (1 – k)2
4 – 4h + h2 +9 -6k +k2 = 1 + 2h +h2+1 – 2k + k2
4 – 4h +9 -6k = 1 + 2h + 1 -2k
6h + 4k =11……………. (4)
Now, let us multiply equation (3) by 6 and subtract it from equation (4) to get
6h+ 4k – 6(h-3k) = 11 – 66
6h + 4k – 6h + 18k = 11 – 66
22 k = – 55
K = -5/2
Substitute this value of K in equation (4) to get
6h + 4(-5/2) = 11
6h – 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 – 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 – 7/2)2 + (y + 5/2)2 = 130/4
[(2x-7)/2]2 + [(2y+5)/2]2 = 130/44x2Â -28x + 49 +4y2Â + 20y + 25 =130
4x2 +4y2 -28x + 20y – 56 = 0
4(x2 +y2 -7x + 5y – 14) = 0
x2 + y2 – 7x + 5y – 14 = 0
∴ The equation of the required circle is x2 + y2 – 7x + 5y – 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 – h)2+ (y – k)2 = 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 – h)2 + y2 = 25.
It is given that the circle passes through the point (2, 3), so the point will satisfy the equation of the circle.
(2 – h)2+ 32 = 25
(2 – h)2 = 25-9
(2 – h)2 = 16
2 – h = ± √16 = ± 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
x2 + y2 + 4x – 21 = 0
When h = 6, the equation of the circle becomes
(x – 6)2 + y2 = 25
x2Â -12x + 36 + y2Â = 25
x2Â + y2Â -12x + 11 = 0
∴ The equation of the required circle is x2 + y2 + 4x – 21 = 0 and x2 + y2 -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 – h)2+ (y – k)2 =r2
We know that the circle passes through (0, 0).
So, (0 – h)2+ (0 – k)2 = r2
h2Â + k2Â =Â r2
Now, the equation of the circle is (x – h)2 + (y – k)2 = h2 + 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 – h)2+ (0 – k)2 =h2 +k2……………..(1)
(0 – h)2+ (b– k)2 =h2 +k2………………(2)
From equation (1), we obtain
a2 – 2ah + h2 +k2 = h2 +k2
a2 – 2ah = 0
a(a – 2h) =0
a = 0 or (a -2h) = 0
However, a ≠ 0; hence, (a -2h) = 0
h = a/2
From equation (2), we obtain
h2 – 2bk + k2 + b2= h2 +k2
b2 – 2bk = 0
b(b– 2k) = 0
b= 0 or (b-2k) =0
However, a ≠0; hence, (b -2k) = 0
k =b/2
So, the equation is
(x – a/2)2 + (y – b/2)2 = (a/2)2 + (b/2)2
[(2x-a)/2]2 + [(2y-b)/2]2 = (a2 + b2)/44x2 – 4ax + a2 +4y2 – 4by + b2 = a2 + b2
4x2 + 4y2 -4ax – 4by = 0
4(x2 +y2 -7x + 5y – 14) = 0
x2 + y2 – ax – by = 0
∴ The equation of the required circle is x2 + y2 – ax – 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 = √[(2-4)2 + (2-5)2]
= √[(-2)2 + (-3)2]
= √[4+9]
= √13
The equation of the circle is given as
(x– h)2+ (y – k)2 = r2
(x –h)2 + (y – k)2 = (√13)2
(x –2)2 + (y – 2)2 = (√13)2
x2 – 4x + 4 + y2 – 4y + 4 = 13
x2 + y2 – 4x – 4y = 5
∴ The equation of the required circle is x2 + y2 – 4x – 4y = 5
15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution:
Given:
The equation of the given circle is x2 +y2 = 25
x2Â + y2Â = 25
(x – 0)2 + (y – 0)2 = 52 [which is in the form (x – h)2 + (y – k)2 = 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
= √[(-2.5 – 0)2 + (-3.5 – 0)2]
= √(6.25 + 12.25)
= √18.5
= 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 – 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 – NCERT Class 11 Solutions
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