jmc

Let $O$ and $O^{\prime }$ denote the centers of the smaller and larger circles, respectively. Let $D$ and $D^{\prime }$ be the points on $\overline{AC}$ that are also on the smaller and larger circles, respectively. Since $△ADO$ and $△A{D}^{\prime }{O}^{\prime }$ are similar right triangles, we have $$\frac{AO}{1}=\frac{A{O}^{\prime }}{2}=\frac{AO+3}{2},\text{so}AO=3.$$As a consequence, $$AD=\sqrt{A{O}^2-O{D}^2}=\sqrt{9-1}=2\sqrt{2}.$$

Let $F$ be the midpoint of $\overline{BC}$. Since $△ADO$ and $△AFC$ are similar right triangles, we have $$\frac{FC}{1}=\frac{AF}{AD}=\frac{AO+O{O}^{\prime }+O^{\prime }F}{AD}=\frac{3+3+2}{2\sqrt{2}}=2\sqrt{2}.$$So the area of $△ABC$ is $$\frac{1}{2}\cdot BC\cdot AF=\frac{1}{2}\cdot 4\sqrt{2}\cdot 8=\boxed{16\sqrt{2}}.$$