Prove that two circles are homothetic I am trying to prove that any two circles are homothetic. In general there are two centers of homothety, one at the intersection of the external tangents and one at the intersection of the internal tangents. The external center does not exist when the circles have the same radius, and the internal center does not exist when the circles overlap. I have proved that the intersection of the external tangents is a center of homothety using similar triangles but I am having trouble proving it for the internal tangents. 

Maybe this helps:

Let $l$ be the line through the centres $O_1,O_2$ of your circles $\Gamma_1,\Gamma_2$, $T_1,B_1$ the intersections of the perpendicular to $l$ through $O_1$ with $\Gamma_1$, $T_2,B_2$ the intersections of the perpendicular to $l$ through $O_2$ with $\Gamma_2$. Then the homothety centres lie on $l$ and they are given by $T_1 T_2\cap B_1 B_2$ and $B_1 T_2 \cap B_2 T_1$. Can you figure out why?
