Given two circles externaly tangent to each other and the common tangent line. Draw a third tangent circle. We are given $\Gamma_A$ centered at $A$ and $\Gamma_B$ centered at $B$ tangent to each other externally at $C$. Line $DE$ is one common tangent to both not through $C$.

Is there a nice way to draw the red circle externally tangent to both and to line $DE$ without using the complete apollonius solution for the $CCL$? Are there any nice symetries in this problem?
I only saw the homotheties which would mean a few tangency points would be colinear and of course Monge-D'Alembert theorem implies one more colinearity between the tangency points and the exterior homothetic center
 A: Use inversion with center at $O$ (touching point of given circle) and arbitrary radius (I've choose $r$ = distance of $O$ to given line for ease). So we are looking at inversion with respect to red circle.
Mark all given objects with dots.
Now both circles goes to parallel and line goes to a circle touching both lines. Now the image of a circle we are searching (blue) is a circle touching these twi parallels and circle between them and on a picture is green circle (and it is easy to draw it). Now map this green and you get blue circle, the one we are searching.

A: If the radius of the circle $A$ is $a$ and of $B$ is $b$, $DE = 2\sqrt{ab}$.
Say, the center of red circle is $O$, the radius $= r$, and parallel line to $DE$ through $O$ meets $AE$ at $A'$ and $BD$ at $B'$.
$DE = OA' + OB'$
or, $2\sqrt{ab} = 2\sqrt{ar} + 2\sqrt{br}$
or, $r = \frac{ab}{a+b+2\sqrt{ab}}$
Now that we know the value of $r$, draw A'B' parallel line to $DE$ at a distance $r$.
Use a compass to draw an arc from center $A$ with radius $a+r$. Wherever it cuts the line $A'B'$ is point $O$, the center of the red circle. Now you can draw the circle of radius $r$ from the center.
A: I did it. It uses a generalization of a result involving a chain of tangent circles that look like this:
Given the following tangent circles chain to draw the next circle of the chain
To a degenerate case. Anyway here it is:
Let the two given circles $\Gamma _A$ and $\Gamma _B$ touch the line $\ell$ at points $A \in \Gamma _A$ and $B \in \Gamma _B$. Take $C$ to be the antipode of $A$ on $\Gamma _A$ and $D$ the antipode of $B$ on $\Gamma _B$.
Draw the tangent line from $D$ to $\Gamma _A$ (there are two tangent lines, right? Take the one such that the contact point $T$ is in the small region between $\Gamma _A$ and $\Gamma _B$).
Take $G = CT \cap \ell$. To end it just draw the perpendicular bissector of $GT$ and let it meet with a line perpendicular to $\ell$ through $G$ to get the center $O$ of the third circle and draw it centered at $O$ passing through $G$.
