compass and straightedge construction of inverse point

Let $c$ be a circle in the Euclidean plane with center $O$ and let $P\ne O$. One can construct the inverse point $P^{\prime }$ of $P$ using compass and straightedge.

If $P\in c$, then $P=P^{\prime }$. Thus, it will be assumed that $P\notin c$.

The construction of $P^{\prime }$ depends on whether $P$ is in the interior of $c$ or not. The case that $P$ is in the interior of $c$ will be dealt with first.

1. Draw the ray $\overrightarrow{OP}$.

2. Determine $Q\in \overrightarrow{OP}$ such that $Q\ne O$ and $\overline{OP}\cong \overline{PQ}$.

3. Construct the perpendicular bisector of $\overline{OQ}$ in order to find one point $T$ where it intersects $c$.

4. Draw the ray $\overrightarrow{OT}$.

5. Determine $U\in \overrightarrow{OP}$ such that $U\ne O$ and $\overline{OT}\cong \overline{TU}$.

6. Construct the perpendicular bisector of $\overline{OU}$ in order to find the point where it intersects $\overrightarrow{OP}$. This is $P^{\prime }$.

Now the case in which $P$ is not in the interior of $c$ will be dealt with.

1. Connect $O$ and $P$ with a line segment.

2. Construct the perpendicular bisector of $\overline{OP}$ in order to determine the midpoint $M$ of $\overline{OP}$.

3. Draw an arc of the circle with center $M$ and radius $\overline{OM}$ in order to find one point $T$ where it intersects $C$. By Thales’ theorem, the angle $\mathrm{\angle }OTP$ is a right angle; however, it does not need to be drawn.

4. Drop the perpendicular from $T$ to $\overline{OP}$. The point of intersection is $P^{\prime }$.

A justification for these constructions is supplied in the entry inversion of plane.

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.