The circle $\displaystyle S_1$ with center $\displaystyle C_1 (a_1,b_1)$ and radius $\displaystyle r_1$ touches externally the circle $\displaystyle s_2$ center $\displaystyle C_2 (a_2,b_2)$ radius $\displaystyle r_2$. The tangent at their common point passes through the origin. Show that,

$\displaystyle ({a_1}^2-{a_2}^2)+({b_1}^2-{b_2}^2)=({r_1}^2-{r_2}^2)$

If also, the other two tangents from the origin to $\displaystyle S_1$ and $\displaystyle S_2$ are perpendicular, prove that $\displaystyle |a_2b_1-a_1b_2|=|a_1a_2+b_1b_2|$.

Hence show that if $\displaystyle C_1$ remains fixed but $\displaystyle S_1$ and $\displaystyle S_2$ vary, then $\displaystyle C_2$ lies on the curve

$\displaystyle ({a_1}^2-{b_1}^2)(x^2-y^2)+4a_1b_1xy=0$.

I'm quite sure if I know how to do the first part, I should be able to do the rest, but I don't even know how to start. Thanks for any help.