Let the circles with centers $O_{1}$ and $O_{2}$ intersect at points $A$ and $C$. We draw an arbitrary secant $B_{1} B_{2}$ through point $A$. What is the geometric locus of the intersection point $M$ of the lines $B_{1} O_{1}$ and $B_{2} O_{2}$?
$$
B_{1} M B_{2} \varangle=180^{\circ}-\left(M B_{1} B_{2} \varangle+M B_{2} B_{1} \varangle\right)
$$

Since

$$
M B_{1} B_{2} \varangle+M B_{2} B_{1} \varangle=B_{1} A O_{1} \varangle+B_{2} A O_{2} \varangle=180^{\circ}-O_{1} A O_{2} \varangle,
$$

therefore

$$
B_{1} M B_{2} \varangle=O_{1} A O_{2} \varangle=\text{const.}
$$

Since the angle $B_{1} M B_{2} \varangle$ subtended by the chord $O_{1} O_{2}$ is constant, the point $M$ traces a circle whose radius is

$$
r=\frac{O_{1} O_{2}}{2 \sin O_{1} A O_{2} \varangle}
$$

(Kirchknopf Ervin, Budapest.)