Problem 1045

207. Let $O$ be the center of the sphere, $r$ its radius, $AP$ and $BQ$ the tangents to the sphere ($P$ and $Q$ being the points of tangency), and $M$ the point of intersection of the lines $AP$ and $BQ$. Denote: $|OA|=a, |OB|=b, |PM| = |QM| = x$. Then $|OM|^2 = r^2 + x^2$, $|AM|^2 = (\sqrt{a^2 - r^2} \pm x)^2$, $|BM|^2 = (\sqrt{b^2 - r^2} \pm x)^2$.

If the signs are the same, then the following relation holds:

$$\begin{aligned} & \sqrt{b^2 - r^2}|AM|^2 - \sqrt{a^2 - r^2}|BM|^2 + \\ & + \left( \sqrt{a^2 - r^2} - \sqrt{b^2 - r^2} \right) |OM|^2 = l_1 \end{aligned}$$

If the signs are different, then

$$\begin{aligned} & \sqrt{b^2 - r^2}|AM|^2 + \sqrt{a^2 - r^2}|BM|^2 - \\ & - \left( \sqrt{a^2 - r^2} + \sqrt{b^2 - r^2} \right) |OM|^2 = l_2 \end{aligned}$$

where $l_1$ and $l_2$ are constants depending on $r$, $a$, and $b$.

Since the sum of the coefficients of $|AM|^2$, $|BM|^2$, and $|OM|^2$ in expressions (1) and (2) is zero, the geometric locus of points $M$ for which one of these relations holds is a plane. In both cases, this plane is perpendicular to the plane $OAB$.