Given a [circle](https://artofproblemsolving.com/wiki/index.php/Circle) of [radius](https://artofproblemsolving.com/wiki/index.php/Radius) $2$, there are many line segments of length $2$ that are [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent_(geometry)) to the circle at their [midpoints](https://artofproblemsolving.com/wiki/index.php/Midpoint). Find the area of the region consisting of all such line segments.
$\text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi$
Let line segment $AB = 2$, and let it be tangent to circle $O$ at point $P$, with radius $OP = 2$.  Let $AP = PB = 1$, so that $P$ is the midpoint of $AB$.
$\triangle OAP$ is a right triangle with right angle at $P$, because $AB$ is tangent to circle $O$ at point $P$, and $OP$ is a radius.
Since $AP^2 + OP^2 = OA^2$ by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), we can find that $OA = \sqrt{1^2 + 2^2} = \sqrt{5}$. Similarly, $OB = \sqrt{5}$ also.
Line segment $APB$ can rotate around the circle.  The closest distance of this segment to the center will always be $OP = 2$, and the longest distance of this segment will always be $PA = PB = \sqrt{5}$.  Thus, the region in question is the [annulus](https://artofproblemsolving.com/wiki/index.php?title=Annulus&action=edit&redlink=1) of a circle with outer radius $\sqrt{5}$ and inner radius $2$.  This area is $\pi \cdot 5 - \pi \cdot 4 = \pi$, and the answer is $\boxed{D}$.