Solution:

Let the common tangent meet $C$ at $S$. Let $X$ be the intersection of $C$ and $OP$ lying between $O$ and $P$. $PT = PO$, hence $\angle POT = \angle PTO$, so $\angle OPT = 180^{\circ} - 2\angle POT$. But $PT$ and $OS$ are parallel, because both are perpendicular to the common tangent. Hence $\angle POS = 2\angle POT$, so $\angle SOT = \angle XOT$. Hence $TX$ is tangent to $C$, in other words $T$ lies on the (fixed) tangent to $C$ at $X$. Conversely, it is easy to see that any such point can be obtained (just take $P$ such that $PO = PT$). Thus the required locus is the pair of tangents to $C$ which are perpendicular to $L$.