Centres and constant terms in the circles $x^{2}+y^{2}-5=0, x^{2}+y^{2}-8 x-6 y+10=0$ and
$x^{2}+y^{2}-4 x+2 y-2=0$ are $C_{1}^{\prime}(0,0), c_{1}=-5$
$C_{2}^{\prime}(4,3), c_{2}=10$ and $C_{3}^{\prime}(2,-1), c_{3}=-2$
Also, centre and constant of circle $x^{2}+y^{2}+2 g x+2 f y+c=0$
is $\quad C_{4}^{\prime}(-g,-f)$ and $c_{4}=c$
since, the first circle intersect all the three at the
extremities of diameter, therefore they are orthogonal to each other.
$\therefore \quad 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c$
$\therefore \quad 2[-g(0)+(-f)(0)]=c-5$
$\Rightarrow \quad c=5$
$$
2[-g(4)+(-f)(3)]=c+10
$$
$\Rightarrow \quad-2(4 g+3 f)=5+10$
$\Rightarrow \quad 4 g+3 f=-\frac{15}{2}$
and $2[-g(2)+(-f)(-1)]=c-2$
$\Rightarrow \quad 2[-2 g+f]=5-2$
$\Rightarrow \quad-2 g+f=\frac{3}{2}$
On solving Eqs. (i) and (ii), we get $f=-\frac{9}{10}$ and $g=-\frac{12}{10}$
$\therefore \quad f g=\frac{-9}{10} \times-\frac{12}{10}=\frac{27}{25}$
and $\quad 4 f=4 \times \frac{-9}{10}$
$\Rightarrow \quad 4 f=\frac{-36}{10}$
$\Rightarrow \quad 4 f=3 g$