NCERT Solutions for Class 11th Mathematics
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