Two Parent Circles and Their Baby Circle

Using the well-known formula $\frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{r_1}}+\frac{1}{\sqrt{r_2}}$ , where $r_3$ is the radius of the smallest circle, we find $r_3=\frac{r_1r_2}{\left(\sqrt{r_1}+\sqrt{r_2}\right)^2}=\frac{216}{49}\approx \boxed{4.4081}$ .

Image from GeoGebra .

Consider the above diagram which only considers the important points, namely the centres.

Let $r$ be the radius of the circle.

Let $C$ and $D$ be the intersections of the line $y=r$ with the radii from $A$ and $B$ to the horizontal line.

By Pythagoras' Theorem, $CD=\sqrt{222^2-210^2}=72$ $CO=\sqrt{(6+r)^2-(6-r)^2}=2\sqrt{6}\sqrt{r}$ $OD=\sqrt{(216+r)^2-(216-r)^2}=12\sqrt{6}\sqrt{r}$

We have $CO+OD=CD$ $2\sqrt{6}\sqrt{r}+12\sqrt{6}\sqrt{r}=72$ $\sqrt{r}=\dfrac {72}{14\sqrt{6}}$ $r=\dfrac {216}{49} \approx \boxed {4.4081}$

Since all other solutions are taken, I will use Descarte's Circle Theorem which states for four mutually tangent circles of radii $r_1, r_2, r_3, r_4$ , the identity holds $\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4}\right)^2 = 2\left(\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r_4^2}\right)$

We will call the desired radii $x$ . Thus, we have four mutually tangent circles with radii $x, 6, 216, \infty$ . Substituting these values in we have $\left(\frac{1}{x} + \frac{1}{6} + \frac{1}{216}\right)^2 = 2\left(\frac{1}{x^2} + \frac{1}{36} + \frac{1}{216^2}\right)$

which is easy to solve.