What is the geometric place (on the plane) of the points of the circles passing through two given points, which are farthest from the perpendicular bisector of the given points?
The locus of the centers of circles passing through the given points $A, B$ is precisely the perpendicular bisector $f$, so $f$ is the axis of symmetry for every such circle. Therefore, each circle has two points $P_{1}, P_{2}$ that are farthest from $f$, and these points are symmetric pairs with respect to $f$, making $f$ the axis of the desired locus. The points $P_{1}, P_{2}$ in each circle are cut by the diameter perpendicular to $f$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9cb1e7a2634bd15d4ad2g-1.jpg?height=371&width=512&top_left_y=313&top_left_x=796)

Furthermore, it is evident that the line $AB$ is also an axis of the locus, so it is convenient to choose $f$ and $AB$ as the $Y$- and $X$-axes of our coordinate system, respectively. Let $A(-c, 0)$ and $B(c, 0)$ accordingly.

The equation of a circle with center $K(0, t)$ and radius $r$ is: $x^{2}+(y-t)^{2}-r^{2}=0$. This circle passes through $A$ and $B$ if and only if $c^{2}+t^{2}-r^{2}=0$. Therefore, the equation of the circles under consideration is: $x^{2}+(y-t)^{2}-c^{2}-t^{2}=0$, where $t$ is a parameter. The line cutting the farthest points has the equation $y=t$, so the abscissas of $P_{1}, P_{2}$ are: $x_{1,2}=\mp \sqrt{c^{2}+t^{2}}$, and their common ordinate is: $y_{1,2}=t$. This gives us the parametric equations of the desired locus. By eliminating $t$, we find that every point of the locus lies on the curve with the equation: $x=\mp \sqrt{c^{2}+y^{2}}$, or equivalently $x^{2}-y^{2}=c^{2}$. This is an equilateral hyperbola with the real axis being the segment $AB$.

We obtain all points of the desired locus by considering every point of $f$ as $K$, in other words, by letting $t$ take all (real) values. Therefore, every point of the obtained hyperbola belongs to the locus: the locus is this hyperbola.

Alpár András (Veszprém, Lovassy L. g. III. o. t.)