Problem 1269

## Solution

Let $\mathrm{C}, \mathrm{C}^{\prime}$ have center $\mathrm{O}, \mathrm{O}^{\prime}$ respectively and radius $\mathrm{r}, \mathrm{r}^{\prime}$ respectively. Let a circle center $\mathrm{P}$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $A B$ is perpendicular to $O P$ and is a diameter of C. Hence $\mathrm{PA}^{2}=\mathrm{OP}^{2}+\mathrm{r}^{2}$. Conversely, the circle center $\mathrm{P}$, radius $\sqrt{ }\left(\mathrm{OP}^{2}+\mathrm{r}^{2}\right)$ bisects $\mathrm{C}$. So $\mathrm{P}$ will bisect $\mathrm{C}$ and $\mathrm{C}^{\prime}$ iff $\mathrm{OP}^{2}+\mathrm{r}^{2}=\mathrm{OP}^{\prime 2}+\mathrm{r}^{\prime 2}$.

It is well-known that the locus of points $\mathrm{P}^{\prime}$ with equal tangents to $\mathrm{C}$ and $\mathrm{C}^{\prime}$ is the radical axis. Call the radical axis $\mathrm{R}$. For a point $\mathrm{P}^{\prime}$ on the radical axis we have $\mathrm{P}^{\prime} \mathrm{O}^{2}-\mathrm{r}^{2}=\mathrm{P}^{\prime} \mathrm{O}^{\prime 2}-\mathrm{r}^{\prime 2}$. If we reflect $\mathrm{P}^{\prime}$ in the perpendicular bisector of $\mathrm{OO}^{\prime}$ to get $\mathrm{P}$, then $\mathrm{PO}=\mathrm{P}^{\prime} \mathrm{O}^{\prime}$ and $\mathrm{PO}^{\prime}=\mathrm{P}^{\prime} \mathrm{O}$, so $\mathrm{PO}^{\prime 2}$ $-r^{2}=\mathrm{PO}^{2-r^{2}}$ and hence $\mathrm{PO}^{2}+\mathrm{r}^{2}$. Call the reflection of the $\mathrm{R}$ in the perpendicular bisector of OO' the line $\mathrm{R}^{\prime}$. We have established that points on $\mathrm{R}^{\prime}$ form part of the locus. Conversely, if $\mathrm{P}^{\prime}$ is such that there is a circle center $\mathrm{P}^{\prime}$ bisecting both circles, then $\mathrm{OP}^{\prime 2}+\mathrm{r}^{2}=\mathrm{O}^{\prime} \mathrm{P}^{\prime 2}+\mathrm{r}^{\prime 2}$, so if $\mathrm{P}$ is the reflection of $\mathrm{P}^{\prime}$ then $\mathrm{OP}^{2}-\mathrm{r}^{2}=\mathrm{O}^{2}-\mathrm{r}^{\mathrm{r}^{2}}$ and hence $\mathrm{P}$ lies on the radical axis R. Hence $\mathrm{P}^{\prime}$ must lie on $\mathrm{R}^{\prime}$.

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We have $\mathrm{PT}^{2}=\mathrm{PO}^{2}-\mathrm{r}^{2}=\mathrm{PX}^{2}+\mathrm{OX}^{2}-\mathrm{r}^{2}$, and similarly $\mathrm{PT}^{\prime 2}=\mathrm{PX}^{2}+\mathrm{O}^{\prime} \mathrm{X}^{2}-\mathrm{r}^{\prime 2}$. So $\mathrm{PT}=\mathrm{PT}^{\prime}$ iff $O X^{2}-r^{2}=O^{\prime} X^{2}-r^{\prime 2}$. There is evidently a unique point $X$ for which that is true, so the locus of such $\mathrm{P}$ is the line through $\mathrm{X}$ perpendicular to $\mathrm{OO}^{\prime}$

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If the circles intersect, then the point $\mathrm{X}$ evidently lies on the line joining the two common points, because $O X^{2}-r^{2}=-X Y^{2}=O^{\prime} X^{2}-r^{\prime 2}$. In any case the midpoint of each common tangent evidently lies on the line, so that provides a way of constructing it.