13.3.27 Given two concentric circles in a plane with radii $R$ and $r$ ($R > r$). Let $P$ be a fixed point on the smaller circle, $B$ be a moving point on the larger circle, and the line $BP$ intersects the larger circle at another point $C$. The line $l$ through point $P$ and perpendicular to $BP$ intersects the smaller circle at another point $A$ (if $l$ is tangent to the smaller circle at point $P$, then point $A$ coincides with point $P$).
(1) Find the set of all possible values of the expression $\left| B C \right|^{2} + |C A|^{2} + |A B|^{2$;
(2) Find the locus of the midpoint of segment $A B$.
(1) As shown in the figure, establish a Cartesian coordinate system with the centers of the two circles as the origin, so that point $P$ is on the negative half-axis of the $x$-axis. Then the coordinates of point $P$ are $(-r, 0)$, and the equations of the two circles are: $x^{2}+y^{2}=R^{2}$, $x^{2}+y^{2}=r^{2}$. Let the coordinates of points $A$, $B$, and $C$ be $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $C\left(x_{3}, y_{3}\right)$, respectively. Then
$$
\begin{aligned}
& |B C|^{2}+|C A|^{2}+|A B|^{2} \\
= & \left(x_{2}-x_{3}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}+\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2} \\
= & 4 R^{2}+2 r^{2}-2\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}+y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1}\right),
\end{aligned}
$$

Since $P A \perp P B$, we have $O=\overrightarrow{P A} \cdot \overrightarrow{P B}=\left(x_{1}+r\right)\left(x_{2}+r\right)+y_{1} y_{2}$, which means $x_{1} x_{2}+y_{1} y_{2}=$ $-r^{2}-r\left(x_{1}+x_{2}\right)$, (1) Therefore,
$$
\begin{aligned}
& x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}+y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1} \\
= & -r^{2}-r\left(x_{1}+x_{2}\right)+x_{3}\left(x_{1}+x_{2}\right)+y_{3}\left(y_{1}+y_{2}\right) \\
= & -\left(x_{1}+x_{2}\right)\left(r-x_{3}\right)+y_{3}\left(y_{1}+y_{2}\right)-r^{2},
\end{aligned}
$$

Let the other intersection point of $B C$ with the smaller circle be $A^{\prime}$. From $P B \perp P A$, we get the coordinates of $A^{\prime}$ as $\left(-x_{1},-y_{1}\right)$. By the property of concentric circles, the line segment $B C$ and $A^{\prime} P$ have the same midpoint. Therefore,
$$
\left\{\begin{array} { l } 
{ x _ { 2 } + x _ { 3 } = - r - x _ { 1 } , } \\
{ y _ { 2 } + y _ { 3 } = - y _ { 1 } , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
x_{1}+x_{2}=-r-x_{3}, \\
y_{1}+y_{2}=-y_{3} .
\end{array}\right.\right.
$$

Substituting into (2) gives
$$
\begin{aligned}
& x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}+y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1} \\
= & -\left(-r-x_{3}\right)\left(r-x_{3}\right)+y_{3}\left(-y_{3}\right)-r^{2}=r^{2}-x_{3}^{2}-y_{3}^{2}-r^{2}=-R^{2} .
\end{aligned}
$$

Therefore, $|B C|^{2}+|C A|^{2}+|A B|^{2}=4 R^{2}+2 r^{2}+2 R^{2}=6 R^{2}+2 r^{2}$.
(2) Let the coordinates of the midpoint of $A B$ be $(x, y)$. From (3), we have
$$
\left\{\begin{array} { l } 
{ 2 x = x _ { 1 } + x _ { 2 } = - r - x _ { 3 } , } \\
{ 2 y = y _ { 1 } + y _ { 2 } = - y _ { 3 } , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
x_{3}=-r-2 x, \\
y_{3}=-2 y .
\end{array}\right.\right.
$$

Substituting into the equation of the larger circle gives $(-r-2 x)^{2}+(-2 y)^{2}=R^{2}$, which simplifies to $\left(x+\frac{r}{2}\right)^{2}+y^{2}=\frac{R^{2}}{4}$. Therefore, the trajectory of the midpoint of $A B$ is a circle with center $\left(-\frac{r}{2}, 0\right)$ and radius $\frac{R}{2}$.