Let $S_1: x^2+y^2-4 x+2 f y+1=0$ ...(i)
$S_2: x^2+y^2+2 g x-4 y-1=0$ ...(ii)
$\begin{aligned} & \mathrm{g}_1=2, \mathrm{f}_1=\mathrm{f}, \mathrm{c}_1=1 \\ & \mathrm{~g}_2=\mathrm{g}, \mathrm{f}_2=2, \mathrm{c}_2=-1\end{aligned}$
$\because$ Circle (i) \& (ii) are orthogonal :
$2\left(g_1 g_2+f_1 f_2\right)=c_1+c_2$ and $r_1^2+r_2^2=d^2$ where $d$ is distance between the centers
$\begin{aligned} & \text { Now, } 2\left(\mathrm{f}_1 \mathrm{f}_2+\mathrm{g}_1 \mathrm{~g}_2\right)=\mathrm{c}_1+\mathrm{c}_2 \\ & \Rightarrow 2(2 \mathrm{f}+2 \mathrm{~g})=1-1 \Rightarrow \mathrm{f}=-\mathrm{g} \\ & \text { Also, } \mathrm{r}_1{ }^2+\mathrm{r}_2{ }^2=\mathrm{d}^2\end{aligned}$
$\begin{aligned} & \Rightarrow r_1^2+r_2^2=(g-2)^2+(g+2)^2 \\ & \Rightarrow r_1^2+r_2^2=2 g^2+8 \Rightarrow r_1^2+r_2^2-8=2 g^2\end{aligned}$