Two circles $O_1$ (with radius $r_1$) and $O_2$ (with radius $r_2$), $r_1>r_2$, touch each other externally, and the line $m$ is a common tangent. The line $l$ is parallel to $m$ and touches $O_1$ and the circle $O_3$( with radius $r_3$) touches $l$ and the two given circles externally. If $r_2$ is $3$ and $r_3$ is $5$, find $r_1$.
What I tried: I found the common tangent length between $O_1$, $O_3$ and $O_2$, $O_1$. However, I couldn't proceed further.
Thanks for reading! I appreciate your help.
As shown in the picture with Pythagorean triangles, we can form the equation
$$\sqrt{12r} + \sqrt{32r-4r^2} = \sqrt{20r}$$
Simplfy,
$$\sqrt{32-4r} = \sqrt{20} - \sqrt{12}$$
$$32 - 4r = 32 - 2\sqrt{240}$$
$$r = \sqrt{60}$$