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

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections, contains solutions for all exercise 11.1 questions. The reason for practising NCERT Solution is always to score more in the final exam. Download NCERT Class 11 maths solutions to access the solutions for all the exercises of all the chapters.

Conic section is a chapter that deals wiferent sections of cones. There are a wide variety of topics covered in its first exercise. The Exercise 11.1 of NCERT Solutions for Class 11 Maths Chapter 11- 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

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Solutions for Class 11 Maths Chapter 11 – Exercise 11.1

In each of the following Exercise 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

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) + (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 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 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)

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)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)

Since, 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 = 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 = 130/4

4x2 -28x + 49 +4y2 + 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 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 = 2

(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)2 + y2 = 25

x2 -12x + 36 + y2 = 25

x2 + y2 – 4x + 21 = 0


When h = 6, the equation of the circle becomes

(x – 6)2 + y2 = 25

x2 -12x + 36 + y2 = 25

x2 + y2 -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

h2 + k2 = 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/4)2 = (a/2)2 + (b/2)2

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

4x2 – 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 point (4,5), the radius (r) of the circle is the distance between the 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.

x2 + y2 = 25

(x – 0)2 + (y – 0)2 = 52 [which is of the form (x – h)2 + (y – k)2 = r2]

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

So the distance between 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 point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle, 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

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