Example 8 Given 4 spheres in space, their radii are $2, 2, 3, 3$, and each sphere is externally tangent to the other 3 spheres. There is another smaller sphere that is externally tangent to these 4 spheres. Find the radius of the smaller sphere.
Solve as shown in Figure 16-13,
Let the centers of the four spheres be $A, B, C, D$, with radii $3, 3, 2, 2$ respectively.
According to the problem, $A B=6, C D=4, A C=B C=A D=B D=5$.
Let the midpoints of $A B$ and $C D$ be $F$ and $E$ respectively, and the center of the small sphere be $O$ with radius $r$. By symmetry, $O$ lies on the line segment $E F$, and $E F \perp C D, E F \perp A B$.
Thus, $E F=\sqrt{A E^{2}-\overline{A F^{2}}}=\sqrt{A C^{2}-C E^{2}-\overline{A F^{2}}}$
$$
\begin{aligned}
& =\sqrt{5^{2}-2^{2}-3^{2}}=\sqrt{12} \\
O E & =\sqrt{O C^{2}-C E^{2}}=\sqrt{(r+2)^{2}-2^{2}}=\sqrt{r^{2}+4 r} \\
O F & =\sqrt{O B^{2}-B F^{2}}=\sqrt{(r+3)^{2}-3^{2}}=\sqrt{r^{2}+6 r}
\end{aligned}
$$

Since $E F=O F+O E$, we have
$$
\sqrt{r^{2}+4 r}+\sqrt{r^{2}+6 r}=\sqrt{12} \Leftrightarrow \sqrt{r+4}+\sqrt{r+6}=\sqrt{\frac{12}{r}}
$$

Because $(\sqrt{r+6})^{2}-(\sqrt{r+4})^{2}=(r+6)-(r+4)=2$
i.e., $(\sqrt{r+6}+\sqrt{r+4})(\sqrt{r+6}-\sqrt{r+4})=2$
Combining with (1), we get $\sqrt{r+6}-\sqrt{r+4}=2 \sqrt{\frac{r}{12}}=\sqrt{\frac{r}{3}}$. Combining (1) and (2), we get $2 \sqrt{r+6}=\frac{1}{\sqrt{3 r}}(r+6)$, thus $2 \sqrt{3 r}=\sqrt{r+6}$, solving for $r$ gives $r=\frac{6}{11}$ as the solution.