Problem 828

Let a circle $l$ and a circle $S$ with center $O$ be given. It is required to construct the image of circle $l$ under inversion with respect to circle $S$.

It is known that a circle passing through the center of inversion transforms into a line not passing through the center of inversion, and a circle not passing through the center of inversion transforms into a circle.

Suppose circle $l$ is located inside circle $S$ (Fig.1), passes through point $O$, and intersects the line of centers of the circles at point $C$, different from $O$. Draw a line through point $C$ perpendicular to the line of centers. Draw a tangent to circle $S$ through the point of intersection $A$ of this line with circle $S$. Let $P$ be the point of intersection of this tangent with the line of centers of circles $l$ and $S$. Then the line passing through point $P$ perpendicular to the line of centers is the desired image of circle $l$ under the considered inversion. If circle $l$ passes through point $O$ and intersects circle $S$ at different points $A$ and $B$ (Fig.2), then under inversion with respect to circle $S$, these points remain in place, so the desired image of circle $l$ is the line $AB$.

If circle $l$ passes through point $O$ and is tangent to circle $S$ at point $A$, then under inversion with respect to circle $S$, circle $l$ transforms into the common tangent to circles $l$ and $S$ passing through point $A$. Now suppose circle $l$ does not pass through point $O$ (Fig.3). If the line of centers of circles $l$ and $S$ intersects circle $l$ at different points $A$ and $B$, then $AB$ is the diameter of circle $l$. Then the desired image of circle $l$ under the considered inversion is a circle with diameter $A' B'$, where $A'$ and $B'$ are the images of points $A$ and $B$ under this inversion.