About minimization of length Problem :
Let there are two points : $\mathrm{P}(-2, 0), \mathrm{Q}(2, 0)$.
And a point T moves on xy-plane with these two restrictions :
(i) : T moves from $\mathrm{P}$ to $\mathrm{Q}$.
(ii) : $\mathrm{T}(x, y)$ always safisfies : $x^2 + y^2 \ge 1$.
Find the minimum length of trajectory where $\mathrm{T}$ moved.

From condition (ii), T can be exterior of $x^2 + y^2 =1$ or on of it.
And we can assume $y>0$ for everytime.
Intuitively, It seems to have minimum when it moves like (tangent line of the circle) - (circle) - (tangent line of the circle).
To show that case is minimum, I tried to :
A : show there is no path such that has lower length.
B : show the length exists and its finite.
B is easy, but how can I prove A?
At first, I assumed trajectory of $\mathrm{T}$ is symmetric, but the trajectory is free about symmetric.
Is there any nice approach?
Thank you.
 A: Shortest path between two points lying on circle $x^2+y^2=1$ is shortest arc between these points. That's why we need only one arc on circle.
When segment $AB$ has no point inside circle $x^2+y^2=1$, then segment $AB$ is shortest path between $A$ and $B$.
So our shortest path consists of three parts: segment $PR$ to circle, arc $RS$ on circle and segment $SQ$ to final point.
There is no sense in arc $RS$ which intersects line $PQ$, because mirroring point $S$ about line $PQ$  we can decrease arc length without changing segment length. So $S$ and $R$  are in the same semi-plane with respect to line $PQ$. Without loss of generality, we can take that this semi-plane is $y\geq 0$.
Let $R=(\cos t;\sin t)$, $S=(\cos u;\sin u)$, $0\leq u \leq \frac{\pi}{3} < \frac{2\pi}{3}\leq t \leq \pi$. $u \leq \frac{\pi}{3}$ is required for $SQ$ not intersecting circle, similar reason for $\frac{2\pi}{3}\leq t$. Then $L=PR+(RS)+SQ=$ $\sqrt{(2+\cos t)^2+\sin^2 t}+(t-u)+\sqrt{(2-\cos u)^2+\sin^2 u}=$ $(t+\sqrt{5+4\cos t})+(\sqrt{5-4 \cos u}-u)$. We can divide $L$ on two parts and optimize each part separately and find that $u=\frac{\pi}{3}$ and $t=\frac{2\pi}{3}$ make part the shortest.
$$L_{min}=2\sqrt{3}+\frac{\pi}{3}$$
