Points $ A$ and $ B$ lie on a circle centered at $ O$, and $ \angle AOB\equal{}60^\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 that of the larger circle?

$ \textbf{(A)}\ \frac{1}{16} \qquad
\textbf{(B)}\ \frac{1}{9} \qquad
\textbf{(C)}\ \frac{1}{8} \qquad
\textbf{(D)}\ \frac{1}{6} \qquad
\textbf{(E)}\ \frac{1}{4}$
1. Let the radius of the larger circle be \( R \). The center of the larger circle is \( O \), and the points \( A \) and \( B \) lie on the circumference of the larger circle such that \( \angle AOB = 60^\circ \).

2. The second circle is internally tangent to the larger circle and tangent to both \( \overline{OA} \) and \( \overline{OB} \). Let the radius of the smaller circle be \( r \), and its center be \( P \).

3. Since the smaller circle is tangent to \( \overline{OA} \) and \( \overline{OB} \), the center \( P \) of the smaller circle lies on the angle bisector of \( \angle AOB \). Therefore, \( \angle AOP = \angle BOP = 30^\circ \).

4. The distance from \( O \) to \( P \) is \( R - r \) because the smaller circle is internally tangent to the larger circle.

5. In the right triangle \( \triangle OAP \), where \( \angle AOP = 30^\circ \), we have:
   \[
   \cos(30^\circ) = \frac{OA}{OP} = \frac{R}{R - r}
   \]
   Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), we get:
   \[
   \frac{\sqrt{3}}{2} = \frac{R}{R - r}
   \]

6. Solving for \( R - r \):
   \[
   \frac{\sqrt{3}}{2} (R - r) = R
   \]
   \[
   \sqrt{3} (R - r) = 2R
   \]
   \[
   \sqrt{3} R - \sqrt{3} r = 2R
   \]
   \[
   \sqrt{3} R - 2R = \sqrt{3} r
   \]
   \[
   R (\sqrt{3} - 2) = \sqrt{3} r
   \]
   \[
   r = \frac{R (\sqrt{3} - 2)}{\sqrt{3}}
   \]
   \[
   r = R \left( \frac{\sqrt{3} - 2}{\sqrt{3}} \right)
   \]
   \[
   r = R \left( \frac{\sqrt{3}}{\sqrt{3}} - \frac{2}{\sqrt{3}} \right)
   \]
   \[
   r = R \left( 1 - \frac{2}{\sqrt{3}} \right)
   \]
   \[
   r = R \left( 1 - \frac{2\sqrt{3}}{3} \right)
   \]
   \[
   r = R \left( \frac{3 - 2\sqrt{3}}{3} \right)
   \]

7. The ratio of the areas of the smaller circle to the larger circle is:
   \[
   \left( \frac{r}{R} \right)^2 = \left( \frac{\frac{3 - 2\sqrt{3}}{3}}{1} \right)^2 = \left( \frac{3 - 2\sqrt{3}}{3} \right)^2
   \]

8. Simplifying the expression:
   \[
   \left( \frac{3 - 2\sqrt{3}}{3} \right)^2 = \left( \frac{3 - 2\sqrt{3}}{3} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9}
   \]

The final answer is \(\boxed{\frac{1}{9}}\)