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 - 16= 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)
On solving equations (3) and (4), we obtain h=
and k=
.
On substituting the values of h and k in equation (1), we get
+
= r2
+
= r2
+
= r2
+ 
= r2
r2
Thus, the equation of the required circle is
+
=
+
= 
⇒ 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