Consider a circle with radius $r$, center $O$, and a fixed point $P$ inside it. $OP=a$. Take a point $M$ randomly inside the circle, and let $M'$ be the point symmetric to $M$ with respect to $OP$. What is the probability that the circle passing through the points $M, P, M'$ is completely inside the circle $O$ and includes the point $O$? For example, $a=0.5r$.
Given that the diameter of the circle $k$ passing through point $P$ is $AB$, as shown in the diagram, draw a perpendicular to $AB$ through point $P$. If we want to construct a circle $\gamma$ that passes through points $P, M, M'$ and includes point $O$, then the point $M$ can only lie on the side of this perpendicular where point $O$ is. Since $M'$ is the reflection of $M$ with respect to $AB$, the center of the circle $\gamma$, denoted as $m$, lies on the line $AB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_51b60789e68ceae7d59fg-1.jpg?height=461&width=469&top_left_y=347&top_left_x=817)

For $\gamma$ to be completely inside the circle $k$, the condition is that its diameter must be at most $AP = r + a$. The midpoint of $AP$, denoted as $C_1$, is the center of the boundary circle $\gamma_1$, which lies between $A$ and $O$. The distance from $C$ to $O$ is

\[
CO = CP - OP = \frac{r + a}{2} - \frac{a}{2} = \frac{r - a}{2}
\]

For $\gamma$ to include point $O$, the condition is that its diameter must be at least $OP = a$. If the center of the circle with diameter $OP = a$ is $D$ and $M$ lies inside this circle $\gamma_2$, then $O$ lies outside the circle passing through points $M, P, M'$.

Therefore, point $M$ must lie in the region bounded by the circles $\gamma_1$ and $\gamma_2$. The probability $v$ is the ratio of the area of this region to the area of the circle $k$:

\[
v = \frac{\left(\frac{r + a}{2}\right)^2 \pi - \left(\frac{a}{2}\right)^2 \pi}{r^2 \pi} = \frac{r + 2a}{4r} = \frac{1}{4} + \frac{a}{2r}
\]

For example, if $a = \frac{r}{2}$, then $v = \frac{1}{2}$.