compass and straightedge construction of inverse point

Let $c$ be a circle in the Euclidean plane with center $O$ and let $P \neq O$ . One can construct the inverse point $P'$ of $P$ using compass and straightedge.

If $P \in c$ , then $P=P'$ . Thus, it will be assumed that $P \notin c$ .

The construction of $P'$ 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 $Q \neq O$ 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 $U \neq O$ 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 $P'$ .
   
   
   

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 $\angle OTP$ 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 $P'$ .
   
   
   

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 link provided.