5. In the plane, what is the length of the shortest path from $(-2,0)$ to $(2,0)$ that avoids the interior of the unit circle (i.e., circle of radius 1 ) centered at the origin?
Answer: $\quad 2 \sqrt{3}+\frac{\pi}{3}$
Solution: The path goes in a line segment tangent to the circle, then an arc of the circle, then another line segment tangent to the circle. Since one of these tangent lines and a radius of the circle give two legs of a right triangle with hypotenuse the line from $(0,0)$ to $(-2,0)$ or $(2,0)$, the length of each tangent line is $\sqrt{2^{2}-1^{2}}=\sqrt{3}$. Also, because these are $30^{\circ}-60^{\circ}-90^{\circ}$ right triangles, the angle of the arc is $60^{\circ}$ and has length $\pi / 3$.