If $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$, then $OP$ is—

Since $PA = PB$ and $OA = OB$, both $P$ and $O$ are equidistant from the touch points $A$ and $B$, so each lies on the perpendicular bisector of the chord $AB$. Hence the line $OP$ is the perpendicular bisector of the chord of contact.