smc

The pathway from $A(0,0)$ to $D(12,16)$ will consist of three segments: 1) $\overline{AB}$, where $AB$ is tangent to the circle at point $B$. 2) $\overline{CD}$, where $CD$ is tangent to the circle at point $C$. 3) $\widehat{BC}$, where $BC$ is an arc around the circle. The actual path will go $A\to B\to C\to D$, so the actual segments will be in order $1,3,2$. Let $O$ be the center of the circle at $(6,8)$. $OA=10$ and $OB=5$ since $B$ is on the circle. Since $△OAB$ is a right triangle with right angle $B$, we find that $AB=\sqrt{10^2-5^2}=5\sqrt{3}$. This means that $△OAB$ is a $30-60-90$ triangle with sides $5:5\sqrt{3}:10$. Thus, $\angle DOC=60^{\circ }$, and $\angle AOB=60^{\circ }$. Since $\angle AOB$, $\angle DOC$ and $\angle BOC$ lie on a straight line, $\angle BOC$ must be $60^{\circ }$ as well. Thus, the arc that we travel is a $60^{\circ }$ arc, and we travel $\frac{C}{6}=\frac{2\pi r}{6}=\frac{2\pi \cdot 5}{6}=\frac{5\pi }{3}$ around the circle. Thus, $AB=5\sqrt{3}$, $\widehat{BC}=\frac{5\pi }{3}$, and $CD=5\sqrt{3}$. The total distance is $10\sqrt{3}+\frac{5\pi }{3}$, which is option $\boxed{10\sqrt{3}+\frac{5\pi }{3}}$.