Herbert Pang, age 18, Sha Tin College, Hong Kong sent a very good solution to this problem. Well done Herbert. The point $P$ is a fixed point on the smaller circle. The point $P_o$ is the position of $P$ when $P$ is at the point of contact between the two circles. Consider the general position where the point of contact is the point $C$ but here we do not assume that $P_1$ is the position of the point $P$.

By showing that the lengths of the arcs $P_{o}C$ and $P_{1}C$ are equal, we shall prove that $P_1$ must be the position of the point $P$ when the point of contact is at $C$. Hence we shall show that $P$ must always lie on the diameter of the large circle through ${\mathrm{OP}}_o$.

Let $M$ be the centre of the small circle, then $\mathrm{MO}={\mathrm{MP}}_1=\mathrm{MC}=r$ and the triangle ${\mathrm{OMP}}_1$ is isosceles. Hence

${\displaystyle \angle {\mathrm{MOP}}_1=\angle {\mathrm{MP}}_{1}O=\theta }$

${\displaystyle \angle P_1\mathrm{MC}=\pi -(\pi -2\theta )=2\theta .}$

Hence, using the formula " arc length = radius x angle at the centre of the circle":

${\displaystyle P_{0}C=(2r)(\theta )=2r\theta }$

and

${\displaystyle P_{1}C=(r)(2\theta )=2r\theta .}$

Hence $P$ must be at the point $P_1$ because the circle rolls without slipping, which shows that the locus of P is the diameter of the larger circle.