Consider a circle with center $O$ and radius $r$, and a tangent $t$ at point $A$ of the circle. Let $M$ be a point on the tangent $t$. The intersection point of segment $OM$ with the circle is denoted by $N$. In point $M$, we measure the distance $MP = MN$ on the perpendicular to $t$ raised at $M$, for example, on the side of $t$ where the circle is not located. What is the geometric locus of point $P$ as $M$ moves along the tangent $t$?
Place our circle in a Cartesian coordinate system such that its center is at the origin, and point $A$ is on the positive part of the $x$-axis (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_9bfab9ad1d531dd99897g-1.jpg?height=615&width=577&top_left_y=267&top_left_x=752)

Denoting the coordinates of point $P$ by $x, y$:

$$
x=O A+A P^{\prime}=O N+M P=O N+N M=O M
$$

and thus, applying the Pythagorean theorem to the right triangle $O A M$,

$$
x^{2}=y^{2}+r^{2}
$$

or equivalently,

$$
\frac{x^{2}}{r^{2}}-\frac{y^{2}}{r^{2}}=1
$$

This is the equation of an equilateral hyperbola, whose center is $O$, and whose real semi-axis is the segment $O A=r$. Since $x>r>0$, only the branch of the hyperbola with vertex at the given $A$ is relevant.

Conversely: for any point $P(x, y)$ on this branch of the hyperbola,

$$
x^{2}=y^{2}+r^{2},
$$

and on the other hand,

$$
O M^{2}=y^{2}+r^{2}
$$

From the equality of these two relationships,

$$
x=O M
$$

and thus,

$$
x-r=O M-r
$$

or

$$
A P^{\prime}=M P=M N
$$

Therefore, point $P$ satisfies the conditions of the problem, meaning that the indicated branch of the hyperbola is the desired geometric locus.

Mihály Makkai (Bp., V., Eötvös J. g. III. o. t.)