$26 \cdot 33$ On the $x y$ coordinate plane, what is the length of the shortest path between the points $(0,0)$ and $(12,16)$, given that the path must go around the circle $(x-6)^{2}+(y-8)^{2}=25$?
(A) $10 \sqrt{3}$.
(B) $10 \sqrt{5}$.
(C) $10 \sqrt{3}+\frac{5 \pi}{3}$.
(D) $40 \frac{\sqrt{3}}{3}$.
(E) $10+5 \pi$.
(47th American High School Mathematics Examination, 1996)
[Solution] The shortest path sought is obviously as shown in the right figure, which includes two line segments and an arc, all of which are tangent to the given circle.
It is easy to see from $AB=5, OB=10$,
and $AB \perp OA$, that $\angle OBA=60^{\circ}$,
Similarly, $\angle CBD=60^{\circ}$.
And $OA=5 \sqrt{3}$.
Similarly, $CD=5 \sqrt{3}$, noting that $B$ is the midpoint of $OC$.

Then $\angle ABD=60^{\circ}$ and $\overparen{AD}=\frac{5}{3} \pi$.
$\therefore$ The length of the shortest path is $10 \sqrt{3}+\frac{5 \pi}{3}$. Therefore, the answer is $(C)$.