Given a circle, a point $P$ is inside it, around which we rotate a right angle. The sides of the right angle intersect the circle at points $A$ and $B$. Determine the geometric locus of the midpoint of the chord $AB$!
Let the center of the circle be $O$. The origin of our Cartesian coordinate system is the point $P$, the line $PO$ is the $X$-axis; $PO=a$. In this system, the equation of the circle is

$$
(x-a)^{2}+y^{2}-R^{2}=0 \ldots
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_57d061e52ab129fcddf1g-1.jpg?height=477&width=480&top_left_y=344&top_left_x=817)

The coordinates of points $A, B$ are $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. These satisfy the following:

$$
\begin{aligned}
& x_{1}^{2}+y_{1}^{2}-2 a x_{1}+a^{2}-R^{2}=0 \ldots \\
& x_{2}^{2}+y_{2}^{2}-2 a x_{2}+a^{2}-R^{2}=0 \ldots
\end{aligned}
$$

Since $P A \perp P B$,

$$
x_{1} x_{2}+y_{1} y_{2}=0 \ldots
$$

The coordinates of the midpoint $M$ of the chord $AB$ are:

$$
x=\frac{x_{1}+x_{2}}{2}, \quad y=\frac{y_{1}+y_{2}}{2} \ldots
$$

The geometric locus of point $M$ can be obtained by eliminating $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ from equations 2), 3), 4), and 5). For this purpose, we add the equations 2), 3), and 4), multiplying the terms of equation 4) by 2. This results in:

$$
\left(x_{1}+x_{2}\right)^{2}+\left(y_{1}+y_{2}\right)^{2}-2 a\left(x_{1}+x_{2}\right)+2\left(a^{2}-R^{2}\right)=0 \ldots
$$

Considering the relationships in 5):

$$
4 x^{2}+4 y^{2}-4 a x+2\left(a^{2}-R^{2}\right)=0, \text { or } x^{2}+y^{2}-a x+\frac{1}{2}\left(a^{2}-R^{2}\right)=0
$$

or

$$
\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{2 R^{2}-a^{2}}{4} \ldots
$$

We have obtained the equation of a circle whose center is the midpoint of the distance $PO$ and whose radius is $\frac{1}{2} \sqrt{2 R^{2}-a^{2}}$. This circle is the geometric locus of point $M$.

Jenő Krisztonosich (Szent-László g. VII. o. Bp. X.)

Note. In one of the solutions, the question arises whether the circle 7) lies inside the given circle $O$. This is determined by the condition that

$$
\frac{1}{2} \sqrt{2 R^{2}-a^{2}}<R \text { and } O \omega=\frac{a}{2}<R-\frac{\sqrt{2 R^{2}-a^{2}}}{2} \cdots
$$

where $\omega$ is the center of the circle that serves as the geometric locus of point $M$; $O \omega$, the distance between the centers of the two circles, is less than the difference of the two radii. (The inequalities under 8) can be easily verified.)

II. Solution. In the right triangle $OMA$, $\overline{O M}^{2}+\overline{M A}^{2}=R^{2}$. In the right triangle $APB$, $\overline{M P}=\overline{M A}$. Thus,

$$
\overline{M O}^{2}+\overline{M P}^{2}=R^{2}
$$

which means that for point $M$, the sum of the squares of the distances from two fixed points is constant, so the geometric locus of $M$ is a circle whose center is the midpoint of the distance $PO$.