Prove that the radical axis of the circles x²+y²+2gx+2fy+c=0 and x²+y²+2g'x+2f'y+c'=0 is the diameter of the later circle (or the former bisects the circumference of the later) if $2{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{g}-{\mathrm{g}}^{\mathrm{\prime }}\right)+2{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{f}-{\mathrm{f}}^{\mathrm{\prime }}\right)=\mathrm{c}-{\mathrm{c}}^{\mathrm{\prime }}$

Given circles are
$\begin{array}{l}\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 ...\left(1\right)\\ {\mathrm{S}}^{\mathrm{\prime }}={\mathrm{x}}^2+{\mathrm{y}}^2+2{\mathrm{g}}^{\mathrm{\prime }}\mathrm{x}+2{\mathrm{f}}^{\mathrm{\prime }}\mathrm{y}+{\mathrm{c}}^{\mathrm{\prime } }=0 ...\left(2\right)\end{array}$
$\text{ The radical axis of (1) \& (2) is }\mathrm{S}-{\mathrm{S}}^{\mathrm{\prime }}=0$
$\Rightarrow 2\left(\mathrm{g}-{\mathrm{g}}^{\mathrm{\prime }}\right)\mathrm{x}+2\left(\mathrm{f}-{\mathrm{f}}^{\mathrm{\prime }}\right)\mathrm{y}+\left(\mathrm{c}-{\mathrm{c}}^{\mathrm{\prime }}\right)=0 ...\left(3\right)$
$\text{ Centre of }\left(2\right),\mathrm{C}=\left(-{\mathrm{g}}^{\mathrm{\prime }},-{\mathrm{f}}^{\mathrm{\prime }}\right)$
Radical axis is the diameter of (2) if C lies on (3)
$\text{ i.e., if }2\left(\mathrm{g}-{\mathrm{g}}^{\mathrm{\prime }}\right)\left(-{\mathrm{g}}^{\mathrm{\prime }}\right)+2\left(\mathrm{f}-{\mathrm{f}}^{\mathrm{\prime }}\right)\left(-{\mathrm{f}}^{\mathrm{\prime }}\right)+\mathrm{c}-{\mathrm{c}}^{\mathrm{\prime }}=0$
$\text{ i.e., if }2{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{g}-{\mathrm{g}}^{\mathrm{\prime }}\right)+2{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{f}-{\mathrm{f}}^{\mathrm{\prime }}\right)=\mathrm{c}-{\mathrm{c}}^{\mathrm{\prime }}$
which is true.