NCERT Solutions Class 11th Maths Chapter – 11 – Conic Sections Exercise 11.1

NCERT Solutions Class 11th Maths Chapter – 11 – Conic Sections Exercise 11.1

NCERT Solutions Class 11th Maths Chapter – 11 – Conic Sections Exercise 11.1

?Chapter – 11?

✍Conic Sections✍

?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
144×2 – 144x + 36 + 144y² – 72y + 9 – 1 = 0
144×2 – 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. 2×2 + 2y² – x = 0

Solution: The equation of the given of the circle is 2×2 + 2y² –x = 0.
2×2 + 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 = 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)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/2)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.