Short answer:
The circumference centered at $T$ with radius $TP$ is orthogonal to the circumference centered at $O$ with radius $OP$.
Long answer:
Let me first recall this very elemental geometry Theorem: Let $C$ be a circumference and let $P$, $Q$ and $R$ be points in $C$ with $P$ and $Q$ diametrically opposite. Then the angle $\angle{PRQ}=90^º.$
With this being said, in your problem, the angle $\angle{OPT}=90^º$ (by the way you defined $T$).
By the Theorem, $O$, $T$ and $P$ are in a circumference ($C_1$) where $OT$ is a diameter. By the same reasoning, the $\angle{OQT}=90^º$ and therefore $O$, $T$ and $Q$ are in a circumference ($C_2$) where $OT$ is a diameter. Finally, since $C_1$ and $C_2$ contain $O$ and $T$ and have the same diameter, then $C_1$ and $C_2$ are the same. Therefore $O$, $P$, $T$, and $Q$ lay on the same circumference.
Note: This is very related to the concept of ''power of a point resect to a circumference''. Check https://en.wikipedia.org/wiki/Power_of_a_point for some interesting information about it.