Question -
Answer -
Let the equation of the requiredcircle be (x – h)2 + (y – k)2 = r2.
Sincethe circle passes through (0, 0),
(0– h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
Theequation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.
Itis given that the circle makes intercepts a and b onthe coordinate axes. This 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)
Fromequation (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 =
.
Fromequation (2), we obtain
h2 + b2 –2bk + k2 = h2 + k2
⇒ b2 – 2bk = 0
⇒ b(b –2k)= 0
⇒ b =0 or(b –2k)= 0
However, b ≠0; hence, (b –2k)= 0 ⇒ k = 
.
Thus,the equation of the required circle is
