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?

1. Let the side length be $x$ and let $E$ be the midpoint of $AB$. By symmetry $E$ lies on the $y$-axis with $OE\perp AB$, $OE=1+x$ and $EB=\frac x2$.
2. In the right triangle $OEB$, $OB=3$, so $(x+1)^2+\left(\frac x2\right)^2=9$, i.e. $5x^2+8x-32=0$.
3. $x=\dfrac{-8+\sqrt{64+640}}{10}=\dfrac{-8+8\sqrt{11}}{10}=\boxed{\dfrac{4\sqrt{11}-4}5}\approx1.85$.