1. In the plane, there are two circles with the same center, with radii $R$ and $r$ ($R > r$). $P$ is a fixed point on the smaller circle, and $B$ runs along the larger circle. The other intersection point of the line $BP$ with the larger circle is $C$. The other intersection point of the perpendicular line "l" to $BP$ at $P$ with the smaller circle is $A$ (if "l" is the tangent to the circle at $P$, let $A = P$).

I. Determine the set of values taken by $BC^2 + CA^2 + AB^2$.

II. Determine the geometric locus of the midpoint of segment $AB$.
I. Let $O$ be the center of the two circles. Suppose that $P, O, B$ are not collinear. Let $PB$ intersect the circle of radius $r$ again at $Q$. Since the roles of $B$ and $C$ are interchangeable, we can assume that $Q$ lies between $P$ and $B$. From $P$, $QA$ is seen at a $90^{\circ}$ angle, so $QA$ is a diameter. $CP = QB = a$ (since, for example, $\triangle CPO \simeq \triangle BQO$, two sides and the larger opposite angle are equal).

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Applying the Pythagorean theorem:

$$
\begin{gathered}
AB^{2} + BC^{2} + CA^{2} = PA^{2} + (PQ + a)^{2} + (PQ + 2a)^{2} + PA^{2} + a^{2} = \\
= 2(PA^{2} + PQ^{2}) + 6a(a + PQ)
\end{gathered}
$$

However, from $\triangle PQA$, $PA^{2} + PQ^{2} = 4r^{2}$, and the power of $P$ with respect to the circle of radius $R$ is

$$
a(a + PQ) = CP \cdot PB = R^{2} - r^{2}
$$

Thus, $AB^{2} + BC^{2} + CA^{2} = 6R^{2} + 2r^{2}$, even if $C, P, O, B$ are collinear. The value in question is therefore constant.

II. Let the reflection of $B$ over the perpendicular bisector of $PA$ be $B'$. $B'$ lies on the circle of radius $R$, since $f$ is a symmetry axis for both circles. $PAB'B$ is a rectangle, $PB = AB'$, and $\angle APB = \angle PAB' = 90^{\circ}$, so the midpoint of $AB$ coincides with the midpoint of $PB'$. As $B$ moves around the circle, so does $B'$, and thus the midpoint of $AB$ lies on the image of the circle of radius $R$ scaled down by a factor of $\frac{1}{2}$ from $P$. In the extreme case, this circle has a diameter whose endpoints are the desired points.