NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1.


In each of the following Exercises 1 to 5, find the equation of the circle with

Ex 11.1 Class 11 Maths Question 1.

centre (0, 2) and radius 2

Solution:
Here, h = 0, k = 2 and r = 2
The equation of the circle is:
(x – h)2 + (y – k)2 = r2
(x – 0)2 + (y – 2)2 = (2)2
x2 + y2 + 4 – 4y = 4
x2 + y2 – 4y = 0

 

Ex 11.1 Class 11 Maths Question 2.

centre (-2, 3) and radius 4

Solution:
Here, h = -2, k = 3 and r = 4
The equation of the circle is:
(x – h)2 + (y – k)2 = r2
(x + 2)2 + (y – 3)2 = (4)2
x2 + 4 + 4x + y2 + 9 – 6y = 16
x2 + y2 + 4x – 6y – 3 = 0

Ex 11.1 Class 11 Maths Question 3.

centre (1/2, 1/4) and radius 1/12

Solution:
Here, h = 1/2, k = 1/4 and r = 1/12
The equation of the circle is:


Ex 11.1 Class 11 Maths Question 4.

centre (1, 1) and radius 2

Solution:
Here, h = 1, k = 1 and r = 2
The equation of the circle is:
(x – h)2 + (y – k)2 = r2
(x – 1)2 + (y – 1)2 = (2)2
x2 + 1 – 2x + y2 + 1 – 2y = 2
x2 + y2 – 2x – 2y = 0

 

Ex 11.1 Class 11 Maths Question 5.

centre (-a, -b) and radius .


Solution:
Here h=-a, k = -b and r = 

The equation of circle is, (x – h)2 + (y – k)2 = r2
(x + a)2 + (y + b)2 = (
)2

x2 + a2 + 2ax + y2 + b2 + 2by = a2 – b2
x2 + y2 + 2ax + 2by + 2b2 = 0

 

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

Ex 11.1 Class 11 Maths Question 6.

(x + 5)2 + (y – 3)2 = 36

Solution:
The given equation of the circle is:
(x + 5)2 + (y – 3)2 = 36
(x + 5)2 + (y – 3)2 = (6)2
Comparing this equation with (x – h)2 + (y – k)2 = r2, we get
h = -5, k = 3 and r = 6.
Thus, the coordinates of the centre are (-5, 3) and radius is 6.

Ex 11.1 Class 11 Maths Question 7.

x2 + y2 – 4x – 8y – 45 = 0

Solution:
The given equation of the circle is:
x2 + y2 – 4x – 8y – 45 = 0
(x2 – 4x) + (y2 – 8y) = 45
[x2 – 4x + (2)2] + [y2 – 8y + (4)2] = 45 + (2)2 + (4)2
(x – 2)2 + (y – 4)2 = 45 + 4 + 16
(x – 2)2 + (y – 4)2 = 65
(x – 2)2 + (y – 4)2(65)2
Comparing this equation with (x – h)2 + (y – k)2 = r2, we get
h = 2, k = 4 and r = 65.
Thus, the coordinates of the centre are (2, 4) and radius is 65.

 

Ex 11.1 Class 11 Maths Question 8.

x2 + y2 – 8x + 10y – 12 = 0

Solution:
The given equation of the circle is:
x2 + y2 – 8x + 10y – 12 = 0
(x2 – 8x) + (y2 + 10y) = 12
[x2 – 8x + (4)2] + [y2 + 10y + (5)2] = 12 + (4)2 + (5)2
(x – 4)2 + (y + 5)2 = 12 + 16 + 25
(x – 4)2 + (y + 5)2 = 53
(x – 4)2 + (y + 5)2 = (53)2
Comparing this equation with (x – h)2 + (y – k)2 = r2, we get

h = 4, k = -5 and r = 53
Thus, the coordinates of the centre are (4, -5) and radius is 53.

 

Ex 11.1 Class 11 Maths Question 9.

2x2 + 2y2 – x = 0

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


Ex 11.1 Class 11 Maths Question 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:
The equation of the circle is:
(x – h)2 + (y – k)2 = r2                 ….(i)
Since the circle passes through the point (4, 1)
(4 – h)2 + (1 – k)2 = r2
16 + h2 – 8h + 1 + k2 – 2k = r2
h2 + k2 – 8h – 2k + 17 = r2       …. (ii)
Also, the circle passes through the point (6, 5)
(6 – h)2 + (5 – k)2 = r2
36 + h2 – 12h + 25 + k2 – 10k = r2
h2 + k2 – 12h – 10k + 61 = r2         …. (iii)
From (ii) and (iii), we have

h2 + k2 – 8h – 2k + 17 = h2 + k2 – 12h – 10k + 61
4h + 8k = 44 h + 2k = 11             ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
4h + k = 16             …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting the value of h and k in eq (ii), we get
(3)2 + (4)2 – 8 × 3 – 2 × 4 + 17 = r2
r2 = 10
Thus, the required equation of circle is:
(x – 3)2 + (y – 4)2 = 10
x2 + 9 – 6x + y2 + 16 – 8y = 10
x2 + y2 – 6x – 8y + 25 = 0

Ex 11.1 Class 11 Maths Question 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:
The equation of the circle is:
(x – h)2 + (y – k)2 = r2               ….(i)
Since the circle passes through the point (2, 3)
(2 – h)2 + (3 – k)2 = r2
4 + h2 – 4h + 9 + k2 – 6k = r2
h2 + k2 – 4h – 6k + 13 = r2          ….(ii)
Also, the circle passes through the point (-1, 1)
(-1 – h)2 + (1 – k)2 = r2
1 + h2 + 2h + 1 + k2 – 2k = r2
h2 + k2 + 2h – 2k + 2 = r2            ….(iii)
From (ii) and (iii), we have
h2 + k2 – 4h – 6k + 13 = h2 + k2 + 2h – 2k + 2
-6h – 4k = -11 6h + 4k = 11          …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y – 11 = 0.
h – 3k – 11 = 0 h – 3k = 11            …(v)
Solving (iv) and (v), we get
h = 7/2 and k = 5/2
Putting these values of h and k in eq (ii), we get
(7/2)2 + (−5/2)24 × 7/2 6 × −5/2 + 13 = r2
 49/4 + 25/4 14 + 15 + 13 = r2  r2 = 65/2
Thus, the required equation of circle is:
 (x 7/2)2 + (y + 5/2)2 = 65/2
 x2 + 49/4 7x + y2 + 25/4 + 5y = 65/2
4x2 + 49 – 28x + 4y2 + 25 + 20y = 130
4x2 + 4y2 – 28x + 20y – 56 = 0
4(x2 + y2 – 7x + 5y – 14) = 0
x2 + y2 – 7x + 5y – 14 = 0

 

Ex 11.1 Class 11 Maths Question 12.

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

Solution:
Since the centre of the circle lies on x-axis, the coordinates of the centre are (h, 0). Now, the circle passes through the point (2, 3).
Radius of circle


Ex 11.1 Class 11 Maths Question 13.

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

Solution:
Let the circle makes intercept a with x-axis and intercept b with y-axis.
OA = a and OB = b
So, the coordinates of A are (a, 0) and the coordinates of B are (0, b).
Now, the circle passes through three points O(0, 0), A(a, 0) and B(0, b).


Ex 11.1 Class 11 Maths Question 14.

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

Solution:
The equation of the circle is:
(x – h)2 + (y – k)2 = r2                     ….(i)
Since the circle passes through the point (4, 5) and the coordinates of centre are (2, 2).
Radius of the circle


Ex 11.1 Class 11 Maths Question 15.

Does the point (-2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Solution:
The equation of the given circle is: x2 + y2 = 25
(x – 0)2 + (y – 0)2 = (5)2
Comparing this equation with (x – h)2 + (y – k)2 = r2, we get
h = 0, k = 0 and r = 5
Now, distance of the point (-2.5, 3.5) from the centre (0, 0)

Thus, the point (-2.5, 3.5) lies inside the circle.


You can also like these:

NCERT Solutions for Maths Class 9

NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 12

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