The area of the smallest square that will contain a circle of radius 4 is
$\text{(A)}\ 8 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$
Draw a square circumscribed around the circle.  (Alternately, the circle is inscribed in the square.)  If the circle has radius $4$, it has diameter $8$.  Two of the diameters of the circle will run parallel to the sides of the square.  Thus, the smallest square that contains it has side length $8$, and area $8\times8=64$.
$\boxed{\textbf{(D)}}$