The area of a circle is doubled when its radius $r$ is increased by $n$. Then $r$ equals:
$\textbf{(A)}\ n(\sqrt{2} + 1)\qquad  \textbf{(B)}\ n(\sqrt{2} - 1)\qquad  \textbf{(C)}\ n\qquad  \textbf{(D)}\ n(2 - \sqrt{2})\qquad  \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}$
Since the new circle has twice the area of the original circle, its radius is $\sqrt{2}$ times the old radius. 
Thus, 
\[r + n = r\sqrt{2}\]
\[n = r\sqrt{2} - r\]
\[n = r(\sqrt{2} - 1)\]
\[r = \frac{n}{\sqrt{2} - 1}\]
Rationalizing the denominator yields
\[r = \frac{n}{\sqrt{2} - 1} *  \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = n(\sqrt{2} + 1)\]
Therefore, the answer is $\fbox{(A)}$