Given the unit sphere, and a point $P$ inside it, along with a number $0<r \leq 1$. Determine the geometric locus of the centers of $r$-radius disks that contain the point $P$ and whose circumference lies on the surface of the sphere.
Consider a circle $K$ with radius $r$ and center $Q$, whose circumference lies on the surface of the unit sphere with center $O$, and let $T$ be a point on the circumference of this circle.

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Figure 1

Since the line $OQ$ is perpendicular to the plane of the circle, the triangle $OQT$ is a right triangle, and by the Pythagorean theorem, the distance from point $O$ to the plane of circle $K$ is:

$$
OQ = \sqrt{OT^2 - TQ^2} = \sqrt{1 - r^2}
$$

Thus, the point $Q$ lies on the sphere $S$ with center $O$ and radius $\sqrt{1 - r^2}$ (which degenerates to a single point $O$ when $r = 1$), and the plane of circle $K$ is tangent to this sphere at point $Q$ (Figure 1).

Conversely: if $Q'$ is any point on the surface of sphere $S$, then the plane tangent to sphere $S$ at point $Q'$ intersects the unit sphere in a circle of radius $r$.

Therefore, the geometric locus of the centers of circles of radius $r$ whose circumferences lie on the surface of the unit sphere is the surface of sphere $S$.

If $r = 1$ (the degenerate sphere $S$ consists only of the point $O$), then the sought unit-radius circles are precisely the plane sections of the unit sphere by planes containing the line $OP$, and the centers of these circles are the point $O$.

For the rest of the discussion, we can assume that $0 < r < 1$. Let $P$ be a point outside the sphere $S$, i.e., $OP > \sqrt{1 - r^2}$. A point $U$ on the surface of sphere $S$ is precisely the point of tangency of a plane passing through point $P$ if and only if the triangle $PUO$ is a right triangle (Figure 2), i.e., the Pythagorean theorem holds for it:

$$
PU = \sqrt{OP^2 - OU^2} = \sqrt{OP^2 + r^2 - 1}
$$

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Figure 2

Now, the geometric locus of the centers of the appropriate circles is the intersection of the surface of sphere $S$ and the surface of the sphere with center $P$ and radius $\sqrt{OP^2 + r^2 - 1}$, which is a circle whose plane is perpendicular to the central axis of the two spheres, i.e., the line $OP$. Let the center of this circle be $O'$ and its radius be $r'$ (Figure 3).

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Figure 3

Considering the above, we can see that the foot of the altitude from vertex $U$ in the right triangle $OUP$ is $O'$, and the altitude $UO'$ is precisely the radius of the resulting circle; hence, $OO' = \frac{OU^2}{OP} = \frac{1 - r^2}{OP}$ and

$$
r' = UO' = \frac{OU \cdot UP}{OP} = \frac{\sqrt{1 - r^2} \cdot \sqrt{OP^2 + r^2 - 1}}{OP}
$$

In this case, the sought geometric locus is a circle with center $O'$ and radius $r'$, whose plane is perpendicular to the line $OP$.