smc

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 $$r=\frac{n}{\sqrt{2}-1}\ast \frac{\sqrt{2}+1}{\sqrt{2}+1}=n(\sqrt{2}+1)$$ Therefore, the answer is $\boxed{n(\sqrt{2}+1)}$