A point $P$ is inside a circle of radius $r$. The legs of a right angle with vertex $P$ intersect the circle at $A$ and $B$. Extend the triangle $B P A$ to form a rectangle $P A Q B$. What path does the point $Q$ follow as the right angle rotates around $P$?
Solution. Let the center of the circle be $O$, and $OP=d$. Let the feet of the perpendiculars from $O$ to the lines $AP$ and $BP$ be $F$ and $G$, respectively. By applying the Pythagorean theorem multiple times:

$$
OQ^2 = AF^2 + BG^2 = (r^2 - OF^2) + (r^2 - OG^2) = 2r^2 - OP^2
$$

Thus, the point $Q$ lies on the circumference of the circle centered at $O$ with radius $\sqrt{2r^2 - d^2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f24fd5f034ae1d9bdfb3g-1.jpg?height=501&width=553&top_left_y=427&top_left_x=772)

Furthermore, it is also true that $Q$ traces the entire circumference of the circle as the right angle rotates around $P$. For this, let $Q'$ be any point on the circle, i.e., $OQ' = \sqrt{2r^2 - d^2}$. Let $A'PB'Q$ be the rectangle where $OA' = OB' =: r'$ (and $A'$ and $B'$ are the intersection points of the Thales circle of $PQ$ and the line perpendicular to $OK$ through its center $K$). Applying our previous result to the circle with radius $r'$ instead of $r$, we get that $OQ' = \sqrt{2r'^2 - d^2}$, so $r = r'$, which means $Q'$ originates from the right angle $A'PB'$.