Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.
Let the equation of the required circle be (x – h)2+ (y – k)2 =r2
Since, the circle passes through (0, 0),
(0 – h)2+ (0 – k)2 =r2
h2 + k2 = r2
The equation of the circle now becomes (x – h)2+ (y – k)2 =h2 + k2.
It is given that the circle makes intercepts a and b on the coordinate axes.
That means that the circle passes through points (a, 0) and (0,b). Therefore,
(a – h)2+ (0 – k)2 =h2 +k2.................(1)
(0 – h)2+ (b– k)2 =h2 +k2..................(2)
From equation (1), we obtain
a2 – 2ah + h2 +k2 = h2 +k2
a2 – 2ah = 0
a(a – 2h) =0
a = 0 or (a -2h) = 0
However, a 
0; hence, (a -2h) = 0 
h =
.
From the equation (2), we obtain
h2 + – 2bk + k2 + b2= h2 +k2
b2 – 2bk = 0
b(b– 2k) = 0
b= 0 or (b-2k) =0
However, a 
0; hence, (a -2k) = 0 
h =
.
+
= 
⇒ 4x2 -4ax + a2 +4y2 - 4by + b2 =a2 + b2
⇒ 4x2 + 4y2 -4ax – 4by =0
⇒ 4( x2 +y2 -7x + 5y – 14 ) = 0
⇒ x2+y2- ax - by=0