A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?
$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{2}$
Let the side of the largest square be $x$. It follows that the diameter of the inscribed circle is also $x$. Therefore, the diagonal of the square inscribed inscribed in the circle is $x$. The side length of the smaller square is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$. Similarly, the diameter of the smaller inscribed circle is $\dfrac{x\sqrt{2}}{2}$. Hence, its radius is $\dfrac{x\sqrt{2}}{4}$. The area of this circle is $\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}$, and the area of the largest square is $x^2$. The ratio of the areas is $\dfrac{\dfrac{x^2\pi}{8}}{x^2}=\frac{\cancel{x^2}\pi}{8}\cdot\frac{1}{\cancel{x^2}}=\boxed{\textbf{(B) }\frac{\pi}{8}}$.