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NCERT Solutions for Class 11th Mathematics

 

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Chapter 11. Conic Sections

Exercise 11.1

 

 

 

Exercise 11.1 (Conic Sections)


Q1. Find the equation of the circle with centre (0, 2) and radius 2.

Solution: 

The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
⟹ x2 + y2 + 4 – 4 y = 4
⟹ x2 + y2 – 4y = 0

Q2. Find the equation of the circle with centre (–2, 3) and radius 4.

Solution: 

The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, 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

Q3. 

Q10. 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 the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (4, 1) and (6, 5),
(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 equations (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 + 25 – 10k + k2
⇒ 16 – 8h + 1 – 2k = 36 – 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
⇒ r2 = 10
⇒ 𝑟=√10
Thus, the equation of the required circle is
(x – 3)2 + (y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 = 10
x2 + y2 – 6x – 8y + 15 = 0

The required equation for given circle is x2 + y2 – 6x – 8y + 15 = 0

 

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