Let the circles $PXY'$ and $PX'Y$ meet again at $Q$. A suitable inversion of pole $P$ sends the circles $PXY'$ and $PX'Y$ onto the lines $XY'$ and $X'Y$, respectively, while leaving $\gamma$ invariant. The image of $Q$ under this inversion is the point $R$ where the lines $XY'$ and $X'Y$ meet. Upon inversion, the locus of $R$ — the polar of $P$ relative to $\gamma$ — transforms into the circle on diameter $OP$, where $O$ is the centre of $\gamma$. Consequently, the lines $OQ$ and $PQ$ are perpendicular, so the line through the centres of the circles $PXY'$ and $PX'Y$, which is the perpendicular bisector of the segment $PQ$, passes through the midpoint of the segment $OP$. The conclusion follows.