7. As shown in Figure 2, the radii of circles $A$ and $B$ are both 1, and they are externally tangent to each other. The radii of circles $P, Q, R, S$ are all $r$, and circle $P$ is externally tangent to circles $A, B, Q, S$, circle $Q$ is externally tangent to circles $P, B, R$, circle $R$ is externally tangent to circles $A, B, Q, S$, and circle $S$ is externally tangent to circles $P, A, R$. Then $r=$ $\qquad$
7. $\frac{3+\sqrt{17}}{2}$

As shown in Figure 7,
let $M_{1} 、 M_{2}$, and
$M_{3}$ represent the
centers of circles $A 、 P 、 S$,
respectively. Suppose circle $A$
is tangent to circle $B$ at
point $T$. By the Pythagorean
theorem, we have
$$
\begin{array}{l}
\left|M_{1} M_{2}\right|^{2}-\left|M_{1} T\right|^{2} \\
=\left|M_{2} T\right|^{2} \\
=\left|M_{3} M_{2}\right|^{2}-\left|M_{3} T\right|^{2} .
\end{array}
$$

Then $(r+1)^{2}-1^{2}=(r+r)^{2}-(r+2)^{2}$
$$
\Rightarrow r^{2}-3 r-2=0 \Rightarrow r=\frac{3+\sqrt{17}}{2} \text {. }
$$