Correct option (c) a2 – 4ab + b2 = 0
Explanation:
Equation of the two circles be
(x – r)2 + (y – r)2 = r2 ⇒ x2 + y2 – 2xr – 2yr + r2 =0
These two circles passes through (a,b)
(a – r)2 + (b – r)2 = r2
a2 + r2 – 2ar + b2 + r2 – 2br – r2 = 0
r2– 2r(a + b) +(a2 + b2 ) = 0
It is a quadratic equation in r
r1 + r2 = 2(a + b)and r1 r2 = a2 + b2
Condition for orthogonality is
2g1 g2 +2f1f2 = C1 + C2
2r1r2 + 2r1 r2 = r1 2 + r22
4r1 r2 = r12 + r22
6r1r2 = r12 + r22 + 2r1r2
6r1r2 = (r1 + r2)2
6(a2 + b2) = 4 (a + b)2
6a2 + 6b2 = 4a2 + 4b2 + 8ab
2a2 + 2b2 – 8ab = 0
a2 + b2 – 4ab = 0