A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$. Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$. What is the side length of this square? $\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad$ $\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad$ $\textbf{(C)}\ \frac{2\sqrt{2}}{3}\qquad$ $\textbf{(D)}\ \frac{2\sqrt{19}-4}{5}\qquad$ $\textbf{(E)}\ \frac{9-\sqrt{17}}{5}$
Solution
Assume the side length is $x$. Let $E$ be the midpoint of $AB$. By symmetry, $E$ lies on the $y$-axis, so $OE\perp AB$, with $OE=1+x$ (the radius $1$ plus the side $x$) and $EB=\frac x2$.
In the right triangle $\triangle OEB$, $OB=2$, so $(x+1)^2+(\frac{x}2)^2=4$.
A square region $ABCD$ is externally tangent to the circle $x^2+y^2=1$ at $(0,1)$ on side $CD$. Vertices $A,B$ lie on the circle $x^2+y^2=9$. What is the side length of the square?
