Question

[Key Concept: Two circles touching externally with a common tangent: the small circle's radius obeys $\frac{1}{\sqrt{\ }r} = \frac{1}{\sqrt{r_{1}}} + \frac{1}{\sqrt{r_{2}}}$.]

$$\begin{matrix} & 2\sqrt{rr_{1}} + 2\sqrt{rr_{2}} = 2\sqrt{r_{1}r_{2}} \\ \Rightarrow & \sqrt{r}(6 + 3) = 6 \times 3 \\ \Rightarrow & r = 4 \end{matrix}$$