NCERT Solutions
Class 11 Maths
Conic Sections
Ex.11.1 Q.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.
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2
Since 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,
So, h – 3k = 11 ..................... (3)
From equations (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)
On solving equations (3) and (4), we obtain h = 
and k = −
On substituting the values of h and k in equation (1), we obtain
(2 – 
)2 + (3 + )2 = r2
=> (-
)2 + (
)2 = r2
=> 
+ 
= r2
=> r2 = 
Thus, the equation of the required circle is
(x – )2 + (y + )2 =
=> {(2x – 7) ÷ 2}2 + {(2y + 5) ÷ 2}2 =
=> 4x2 – 28x + 49 + 4y2 + 20y + 25 = 130
=> 4x2 – 28x + 49 + 4y2 + 20y + 25 – 130 = 0
=> 4x2 – 28x + 4y2 + 20y - 56 = 0
=> 4(x2 + y2 – 7x + 5y – 14) = 0
=> x2 + y2 – 7x + 5y – 14 = 0