Given equation of circles are
\(\begin{aligned}
(x+a)^2+(y+b)^2 & =a^2 \\
\Rightarrow \quad x^2+y^2+2 a x+2 b y+b^2 & =0 \quad \ldots (i) \\
(x+c)^2+(y+d)^2 & =d^2 \\
\Rightarrow \quad x^2+y^2+2 c x+2 d y+c^2 & =0 \quad \ldots (ii)
\end{aligned}\)
From Eq. (i) \(g_1=a ; f_1=b, c_1=b^2\)
From Eq. (ii) \(g_2=c ; f_2=d ; c_2=c^2\)
If circles (i) and (ii) are orthogonal,
\(\begin{array}{rlrl}
\Rightarrow & & 2\left[g_1 g_2+f_1 f_2\right] & =c_1+c_2 \\
\Rightarrow & 2 & 2(a c+b d) & =b^2+c^2 \\
& \Rightarrow & 2 a c+2 b d & =b^2+c^2 \\
& \Rightarrow & 2 a c-c^2 & =b^2-2 b d \\
& \Rightarrow & c(2 a-c) & =b(b-2 d) \\
& \therefore & b(b-2 d) & =c(2 a-c)
\end{array}\)
\(\therefore\) Hence, answer is (b).