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 $\overrightarrow{OP}$ .
   \begin{pspicture}(-3,-3)(6,3) \rput[a](0,3){.} \rput[l](-3,0){.} \rput[b](0,-3){.} \rput[r](6,0){.} \pscircle(0,0){3} \psline[linecolor=blue]{*->}(0,0)(6,0) \psdots(0,0)(2,0) \rput[a](0,-0.3){$O$} \rput[a](2,-0.3){$P$} \end{pspicture}
2. Determine $Q \in \overrightarrow{OP}$ such that $Q \neq O$ and $\overline{OP} \cong \overline{PQ}$ .
   \begin{pspicture}(-3,-3)(6,3) \rput[a](0,3){.} \rput[l](-3,0){.} \rput[b](0,-3){.} \rput[r](6,0){.} \pscircle(0,0){3} \psline{*->}(0,0)(6,0) \psline[linecolor=blue](2,0)(4,0) \psdots(0,0)(2,0)(4,0) \rput[a](0,-0.3){$O$} \rput[a](2,-0.3){$P$} \rput[a](4,-0.3){$Q$} \end{pspicture}
3. Construct the perpendicular bisector of $\overline{OQ}$ in order to find one point $T$ where it intersects $c$ .
   \begin{pspicture}(-3,-3)(6,3) \rput[a](0,3){.} \rput[l](-3,0){.} \rput[b](0,-3){.} \rput[r](6,0){.} \pscircle(0,0){3} \psline{*->}(0,0)(6,0) \psarc[linecolor=blue](0,0){2.5}{-40}{40} \psarc[linecolor=blue](4,0){2.5}{140}{220} \psline[linecolor=blue](2,-2)(2,2.236) \psdots(0,0)(2,0)(4,0)(2,2.236) \rput[a](0,-0.3){$O$} \rput[a](2,-0.3){$P$} \rput[a](4,-0.3){$Q$} \rput[b](2.1,2.4){$T$} \end{pspicture}
4. Draw the ray $\overrightarrow{OT}$ .
   \begin{pspicture}(-3,-3)(6,5) \rput[a](4.4723,5){.} \rput[l](-3,0){.} \rput[b](0,-3){.} \rput[r](6,0){.} \pscircle(0,0){3} \psline{*->}(0,0)(6,0) \psarc(0,0){2.5}{-40}{40} \psarc(4,0){2.5}{140}{220} \psline(2,-2)(2,2.236) \psline[linecolor=blue]{*->}(0,0)(4.4723,5) \psdots(0,0)(2,0)(4,0)(2,2.236) \rput[a](0,-0.3){$O$} \rput[a](2,-0.3){$P$} \rput[a](4,-0.3){$Q$} \rput[b](2,2.4){$T$} \end{pspicture}
5. Determine $U \in \overrightarrow{OP}$ such that $U \neq O$ and $\overline{OT} \cong \overline{TU}$ .
   \begin{pspicture}(-3,-3)(6,5) \rput[a](4.4723,5){.} \rput[l](-3,0){.} \rput[b](0,-3){.} \rput[r](6,0){.} \pscircle(0,0){3} \psline{*->}(0,0)(6,0) \psarc(0,0){2.5}{-40}{40} \psarc(4,0){2.5}{140}{220} \psline(2,-2)(2,2.236) \psline{*->}(0,0)(4.4723,5) \psline[linecolor=blue](2,2.236)(4,4.472) \psdots(0,0)(2,0)(4,0)(2,2.236)(4,4.472) \rput[a](0,-0.3){$O$} \rput[a](2,-0.3){$P$} \rput[a](4,-0.3){$Q$} \rput[b](2,2.4){$T$} \rput[a](4,4.2){$U$} \end{pspicture}
6. Construct the perpendicular bisector of $\overline{OU}$ in order to find the point where it intersects $\overrightarrow{OP}$ . This is $P'$ .
   \begin{pspicture}(-3,-3)(6,5) \rput[a](4.4723,5){.} \rput[l](-3,0){.} \rput[b](0,-3){.} \rput[r](6,0){.} \pscircle(0,0){3} \psline{*->}(0,0)(6,0) \psarc(0,0){2.5}{-40}{40} \psarc(4,0){2.5}{140}{220} \psline(2,-2)(2,2.236) \psline{*->}(0,0)(4.4723,5) \psline(2,2.236)(4,4.472) \psarc[linecolor=blue](0,0){3.5}{10}{90} \psarc[linecolor=blue](4,4.472){3.5}{190}{270} \psline[linecolor=blue]{*->}(2,2.236)(6,-1.3416) \psdots(0,0)(2,0)(4,0)(2,2.236)(4,4.472)(4.5,0) \rput[a](0,-0.3){$O$} \rput[a](2,-0.3){$P$} \rput[a](4,-0.3){$Q$} \rput[b](2,2.4){$T$} \rput[a](4,4.2){$U$} \rput[a](4.5,-0.3){$P'$} \end{pspicture}

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.
   \begin{pspicture}(-2,-2)(4,2) \rput[b](0,-2){.} \rput[l](-2,0){.} \rput[a](0,2){.} \pscircle(0,0){2} \psline[linecolor=blue](0,0)(4,0) \psdots(0,0)(4,0) \rput[a](0,-0.3){$O$} \rput[a](4,-0.3){$P$} \end{pspicture}
2. Construct the perpendicular bisector of $\overline{OP}$ in order to determine the midpoint $M$ of $\overline{OP}$ .
   \begin{pspicture}(-2,-2)(4,2) \rput[b](0,-2){.} \rput[l](-2,0){.} \rput[a](0,2){.} \pscircle(0,0){2} \psline(0,0)(4,0) \psarc[linecolor=blue](0,0){2.5}{-40}{40} \psarc[linecolor=blue](4,0){2.5}{140}{220} \psline[linecolor=blue]{<->}(2,-2)(2,2) \psdots(0,0)(4,0)(2,0) \rput[a](0,-0.3){$O$} \rput[a](4,-0.3){$P$} \rput[a](2.3,-0.3){$M$} \end{pspicture}
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.
   \begin{pspicture}(-2,-2)(4,2) \rput[b](0,-2){.} \rput[l](-2,0){.} \rput[a](0,2){.} \pscircle(0,0){2} \psline(0,0)(4,0) \psarc(0,0){2.5}{-40}{40} \psarc(4,0){2.5}{140}{220} \psline{<->}(2,-2)(2,2) \psarc[linecolor=blue](2,0){2}{110}{130} \psdots(0,0)(4,0)(2,0)(1,1.732) \rput[a](0,-0.3){$O$} \rput[a](4,-0.3){$P$} \rput[a](2.3,-0.3){$M$} \rput[l](1.12,1.83){$T$} \end{pspicture}
4. Drop the perpendicular from $T$ to $\overline{OP}$ . The point of intersection is $P'$ .
   \begin{pspicture}(-2,-2)(4,2) \rput[b](0,-2){.} \rput[l](-2,0){.} \rput[a](0,2){.} \pscircle(0,0){2} \psline(0,0)(4,0) \psarc(0,0){2.5}{-40}{40} \psarc(4,0){2.5}{140}{220} \psline{<->}(2,-2)(2,2) \psarc(2,0){2}{110}{130} \psarc[linecolor=blue](1,1.732){1.9}{240}{300} \psarc[linecolor=blue](0.35,0){0.8}{-45}{45} \psarc[linecolor=blue](1.65,0){0.8}{135}{225} \psline[linecolor=blue]{<->}(1,2)(1,-1.5) \psdots(0,0)(1,0)(1,1.732)(2,0)(4,0) \rput[a](0,-0.3){$O$} \rput[a](4,-0.3){$P$} \rput[a](2.3,-0.3){$M$} \rput[l](1.12,1.83){$T$} \rput[a](0.6,0.2){$P'$} \end{pspicture}

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.