3 circles internal tangent
I you perform a circular inversion w.r.t the black circle, the red circle becomes the red tangent line in the picture below, while the blue circle gets reflected to another circle, tangent to the black circle, the black line and the red line. The diameter of this new circle must be $1$. Therefore $\overline{AB}=1+1=2$ and $\overline{AC}=1/\overline{AB}=1/2$.
\begin{align}
\triangle ADB:\quad
|DB|^2&=
(\tfrac{R}2+r)^2
-(\tfrac{R}2-r)^2
=2rR
,\\
\triangle BDO:\quad
|DB|^2&=
(R-r)^2-r^2
=R(R-2r)
.
\end{align}
Hence,
\begin{align} R(R-2r)&=2rR ,\\ R&=4r . \end{align}