In a Cartesian coordinate system, two points are given: $A(-m, 0), B(n, 0)$, and let $O(0,0)$ be the origin. Determine the equation of the geometric locus of points from which the segments $A O$ and $O B$ are seen at equal angles.
The task is to determine the geometric locus of points $C$ for which, in the triangle $ABC$, the internal angle bisector from $C$ passes through the point $O$. According to a known theorem, in this case,

$$
\frac{C A}{C B}=\frac{A O}{O B}=\frac{m}{n}
$$

But

$$
C A=\sqrt{(x+m)^{2}+y^{2}} \quad \text { and } \quad C B=\sqrt{(x-n)^{2}+y^{2}}
$$

and thus

$$
\frac{m^{2}}{n^{2}}=\frac{(x+m)^{2}+y^{2}}{(x-n)^{2}+y^{2}}
$$

or

$$
\left(m^{2}-n^{2}\right) x^{2}-2 m n\left(m+n\right) x+\left(m^{2}-n^{2}\right) y^{2}=0
$$

If $m=n$, then the equation of the geometric locus is $x=0$, which leads to the trivial result that in this case, the points on the $y$-axis satisfy the requirement.

If $m \neq n$, then we can divide our equation by $\left(m^{2}-n^{2}\right)$:

$$
x^{2}-\frac{2 m n}{m-n} x+y^{2}=0
$$

which can also be written as

$$
\left(x-\frac{m n}{m-n}\right)^{2}+y^{2}=\left(\frac{m n}{m-n}\right)^{2}
$$

This is the equation of a circle with center at $\left(\frac{m n}{m-n}, 0\right)$ and radius $\left|\frac{m n}{m-n}\right|$. It is evident that every point on this circle satisfies the requirement of the problem, so this circle is the sought geometric locus.

Remark: This provides a coordinate-geometric derivation of the Apollonian circle.

Argay Gyula (Balassagyarmat, Balassa g. III. o. t.)