This is a little exercise found in Robin Hartshorne's Euclid: Geometry and Beyond:
I believe I have found a solution, which only exists when $AB$ has length at least that of the chord bisected by $P$, and length at most the diameter of the circle. So suppose that is the case. I do the following construction. $\odot cXrYZ$ is an abbreviation for the circle centered at $X$ with radius $YZ$. (I assume the compass is not collapsible.) Also, it's not necessary to know my solution to answer my question, so feel free to skip.
- Pick $C$ at random on circle $O$. Describe the circle with center $C$ and radius $AB$. Get intersection $D$.
- Describe the circle with center $D$ and radius $DC$. This will intersect with the previous circle at $E$ and $F$, say.
- Connect $EF$.
- Connect $CD$, get intersection $G$.
- Draw circle with center $O$ and radius $OG$. (This gives the circle to which all chords of length $AB$ will be tangent to.)
- Connect $PO$. Get $H$, the intersection of $PO$ with $\odot cRrOG$. We know this exists, since $P$ must fall outside $\odot cOrOG$.
- $\odot cHrPH$. Get $J$.
- $\odot cPrPJ$.
- $\odot cJrJP$, which will intersect with $\odot cPrPJ$ at $K$ and $L$.
- $\odot cOrPO$.
- Extend $LK$ through to $M$, the intersection with $\odot cOrPO$.
- Connect $MO$, get $N$, the intersection on $\odot cOrOH$.
- Extend $PN$ through to intersect the original circle to get $Q$ and $R$. This line is congruent to $AB$ and passes through $P$.
I apologize that this construction may be hard to follow without the picture. This takes me $13$ steps, but I see that Hartshorne suggests it's possible in $5$, hence the (par=$5$). Can anyone produce a more efficient solution in only $5$ steps? I've been thinking about it, but don't see how to make this any simpler from a compass and straightedge construction.