jmc

We can determine the distance from $O$ to $P$ by dropping a perpendicular from $P$ to $T$ on the $x$-axis. We have $OT=8$ and $PT=6$, so by the Pythagorean Theorem, $$O{P}^2=O{T}^2+P{T}^2=8^2+6^2=64+36=100$$Since $OP>0$, then $OP=\sqrt{100}=10$. Therefore, the radius of the larger circle is $10$. Thus, $OR=10$.

Since $QR=3$, then $OQ=OR-QR=10-3=7$. Therefore, the radius of the smaller circle is $7$.

Since $S$ is on the positive $y$-axis and is 7 units from the origin, then the coordinates of $S$ are $(0,7)$, which means that $k=\boxed{7}$.