Example 6 Consider two concentric circles in the same plane with radii $R$ and $r$ $(R>r)$. Let $P$ be a fixed point on the circumference of the smaller circle, $B$ be a moving point on the circumference of the larger circle, and the line $BP$ intersects the larger circle at another point $C$. The line $l$ passing 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 $A=P$).
(1) Find the set of values taken by the expression $BC^2 + CA^2 + AB^2$;
(2) Find the locus of the midpoint of segment $AB$.
(29th IMO)
Analysis:
(1) As shown in Figure 7, establish a Cartesian coordinate system. Let $\angle B P x = \alpha, A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3})$. Suppose $P B$ intersects the smaller circle at another point $A'$, with $A'(-x_{1}, -y_{1})$, then
$$
\begin{array}{l}
x_{1} = -r \cos 2 \alpha, y_{1} = -r \sin 2 \alpha, \\
x_{2} + x_{3} = 2 x_{D} = -2 r \sin ^{2} \alpha, \\
y_{2} + y_{3} = 2 y_{D} = 2 r \cos \alpha \cdot \sin \alpha.
\end{array}
$$

Substitute the equation of $B C$, $y = \tan \alpha \cdot (x + r)$, into the equation of the larger circle, $x^{2} + y^{2} = R^{2}$, to get
$$
\begin{array}{l}
\left(1 + \tan ^{2} \alpha\right) x^{2} + 2 r \tan ^{2} \alpha \cdot x + \tan ^{2} \alpha \cdot r^{2} - \\
\text{ then } x_{2} x_{3} = \frac{\tan ^{2} \alpha \cdot r^{2} - R^{2}}{1 + \tan ^{2} \alpha} \\
= r^{2} \sin ^{2} \alpha - R^{2} \cos ^{2} \alpha, \\
y_{2} y_{3} = \tan ^{2} \alpha \cdot (x_{2} + r)(x_{3} + r) \\
= (r^{2} - R^{2}) \sin ^{2} \alpha.
\end{array}
$$

Thus, $B C^{2} + C A^{2} + A B^{2}$
$$
\begin{aligned}
= & (x_{2} - x_{3})^{2} + (y_{2} - y_{3})^{2} + (x_{3} - x_{1})^{2} + \\
& (y_{3} - y_{1})^{2} + (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} \\
= & 4 R^{2} + 2 r^{2} - 2 (x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1} + \\
& y_{1} y_{2} + y_{2} y_{3} + y_{3} y_{1}) \\
= & 4 R^{2} + 2 r^{2} - 4 r^{2} \cos 2 \alpha \cdot \sin ^{2} \alpha - \\
& 2 (r^{2} \sin ^{2} \alpha - R^{2} \cos ^{2} \alpha) - 2 (r^{2} - R^{2}) \sin ^{2} \alpha + \\
& 4 r^{2} \sin 2 \alpha \cdot \sin \alpha \cdot \cos \alpha \\
= & 6 R^{2} + 2 r^{2}.
\end{aligned}
$$
(2) Let the midpoint of $A B$ be $Q(x, y)$. As shown in Figure 7, since $A P B B'$ is a rectangle, we have $x_{B'} = 2 x - r, y_{B'} = 2 y$.
Since $B'$ lies on the circle $x^{2} + y^{2} = R^{2}$, we have
$$
\left(x - \frac{r}{2}\right)^{2} + y^{2} = \frac{R^{2}}{4}.
$$

Therefore, this is a circle with center at $\left(\frac{r}{2}, 0\right)$ and radius $\frac{R}{2}$.