The circle $S_1$with centre $C_1\left(a_1,b_1\right)$ and radius $r_1$ touches externally the circle $S_2$ with centre $C_2\left(a_2,b_2\right)$ and radius $r_2$. If the tangent at their common point passes through the origin, then 

The two circles are

$S_1={\left(x-a_1\right)}^2+{\left(y-b_1\right)}^2=r_1^2$                  ..(i)

$S_2={\left(x-a_2\right)}^2+{\left(y-b_2\right)}^2=r_2^2$                 ..(ii)

The equation of the common tangent of these two circles is

given by $S_1-S_2=0$ i.e.

$2x\left(a_1-a_2\right)+2y\left(b_1-b_2\right)+\left(a_2^2+b_2^2\right)-\left(a_1^2+b_1^2\right)$

                                                                                $+r_1^2-r_2^2=0$

If this passes through the origin, then

$\begin{array}{l}\left(a_2^{ 2}+b_2^2\right)-\left(a_1^{ 2}+b_1^2\right)+r_1^2-r_2^{ 2}=0\\ \Rightarrow \left(a_2^{ 2}-a_1^2\right)+\left(b_2^{ 2}-b_1^{ 2}\right)=r_2^{ 2}-r_1^{ 2}\end{array}$