We can determine the distance from $O$ to $P$ by dropping a perpendicular from $P$ to $T$ on the $x$-axis. We have $O T=8$ and $P T=6$, so by the Pythagorean Theorem, $O P^{2}=O T^{2}+P T^{2}=8^{2}+6^{2}=64+36=100$. Since $O P>0$, then $O P=\sqrt{100}=10$. Therefore, the radius of the larger circle is 10. Thus, $O R=10$. Since $Q R=3$, then $O Q=O R-Q R=10-3=7$. 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 $k=7$.