Question -
Answer -
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passesthrough 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 circlelies on line x – 3y – 11 = 0,
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 
.
On substituting the values of h and k inequation (1), we obtain
Thus, the equation of therequired circle is
