Points $A,B$ lie on a circle centered at $O$, and $\angle AOB=90^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to the larger?

1. Let $\theta=\angle AOB$, $R$ the radius of the large circle and $r$ that of the small one, with center $E$. As in the problem, $E$ lies on the bisector of $\angle AOB$ and $r=OE\sin\frac\theta2$; internal tangency gives $OE+r=R$. So $\dfrac rR=\dfrac{\sin(\theta/2)}{1+\sin(\theta/2)}$.
2. With $\theta=90^\circ$: $\sin45^\circ=\frac{\sqrt2}2$, so $\dfrac rR=\dfrac{\sqrt2/2}{1+\sqrt2/2}=\sqrt2-1$.
3. Area ratio $=(\sqrt2-1)^2=\boxed{3-2\sqrt2}$.