4050.12 – Area External to Two Circles

Let $\mathcal{A}$ be the shaded area. Then

$\mathcal{A} = 2(\Delta ADB - \text{I} - \text{II}).$

For the area of $\Delta ADB$, we need the length $AD$:

$AD = \sqrt{(3r)^2 - (2r)^2} = r\sqrt{5},$

$\Delta ADB = \frac{r\sqrt{5} \cdot (2r)}{2} = r^2\sqrt{5}.$

For the area of sectors $\text{I}$ and $\text{II}$ we need the angle $\Theta$:

$\Theta = \arcsin{(\frac{2r}{3r})} = \arcsin{(\frac{2}{3})},$

$I = \frac{\Theta}{360} \pi r^2, II = \frac{90-\Theta}{360} \pi (2r)^2.$

Putting this all together:

$\mathcal{A} = r^2 \left( 2 \sqrt{5} + \pi \left(\frac{\arcsin{(\frac{2}{3})}}{360} + \frac{90 - \arcsin{(\frac{2}{3})}}{90}\right) \right).$