# Difference between revisions of "1996 AHSME Problems/Problem 11"

## Problem

Given a circle of radius $2$, there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$\text{(A)}\ \frac{\pi} 4\qquad\text{(B)}\ 4-\pi\qquad\text{(C)}\ \frac{\pi} 2\qquad\text{(D)}\ \pi\qquad\text{(E)}\ 2\pi$

## Solution

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, 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 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 $\boxed{D}$.