The line $MN$ is the radical axis I created.
Because of its properties, we have $EM=MF, HN=NG, IQ=QL$, and it is perpendicular to $AC$.
Everything is as you see on the diagram below.
Here $(ABC)$ is the area of $\triangle ABC$.
Prove that $\frac{(EHI)}{(FGL)}=\frac{AE}{CF}$.
Any proofs are appreciated. You could use coordinates or complex numbers if you wish. It could teach me some new approaches.
I'm using very different labels in my figure.
Let circles $\bigcirc A$ and $\bigcirc B$ have radii $a$ and $b$. With elements labeled as in the figure, we can write $$\frac{|\triangle PTR|}{|\triangle QUS|} = \frac{\frac12 |\overline{PR}|\;|\overline{HK}|}{\frac12 |\overline{QS}|\;|\overline{JL}|} = \frac{a\;|\overline{HK}|}{b\;|\overline{JL}|}\tag{$\star$}$$
Now, the radical axis (through $M$) bisects external tangents $\overline{PQ}$ and $\overline{RS}$, and therefore also segment $\overline{KL}$. Likewise, the axis bisects internal tangent $\overline{TU}$, and also $\overline{HJ}$. Thus, $M$ serves as the common midpoint of both $\overline{KL}$ and $\overline{HJ}$, and we have $$|\overline{HK}| = |\overline{MK}| - |\overline{MH}| = |\overline{ML}|-|\overline{MJ}| = |\overline{JL}|$$ so that $(\star)$ reduces to $a/b$, as desired. $\square$