# What will be the value of $r$?

Question: Two Circles of radius 36 and 9 touch each other externally, a third circle of radius r touches the two given circles externally and also their common tangent, then the value of r is ?

My attempt: So, I tried taking a Circle $A$ at orgin (radius = 36 units) and then a circle B at center(45,0). I tried the classical approach of finding the length of common tangent by dropping foot of perpendicular and stuff, but that doesn't really works out.

In my second attempt, I tried finding the equation of common tangent: $$y-y_o=m(x-x_o)+r \sqrt{1+m^2}$$ I successfully derived the value of $m^2$ but then faltered at the step as what should be the calculation to get the Center of my new little circle with radius $r$?

The answer given here is 4 units but I have no idea how. Please suggest your solutions.

Cheers!

Descartes' Circle Theorem (in the special case that the 4th circle is a line) provides an answer. You can calculate the curvature of each of the circles by $k_i = \frac{1}{r_i}$ and then use $k_{4}=k_{1}+k_{2}\pm 2{\sqrt {k_{1}k_{2}}}$ to calculate the curvature of the 3rd circle. Notice that $k_4$ is the curvature of the third circle because we actually have a fourth circle (whose curvature is associated with $k_3 = 0$) that is just the tangent line.