Circle $C_{1}$, midpoint $O_{1}(0,0)$ and radius $r$. Circle $C_{2}$, midpoint $O_{2}(r\sqrt{2-\sqrt{3}},0)$ and radius $r\sqrt{2-\sqrt{3}}$. Points $A(\frac{r\sqrt{2+\sqrt{3}}}{2},\frac{r\sqrt{2-\sqrt{3}}}{2})$ and $A(\frac{r\sqrt{2+\sqrt{3}}}{2},-\frac{r\sqrt{2-\sqrt{3}}}{2})$. Area circle $C_{1}$ equals $\pi r^{2}$. Area $ABO_{1}$ equals $\frac{r^{2}}{4}$.

Attachments:

$(a)$ Since $r(C_1):O_1O_2>1$ $\Rightarrow$ $C_2$ lies inside $C_1$. Now observe that the ratio between radius and distance is equivalent to the ratio between radii. Also $r(C_1):r(C_2)<2$ (it's easy to show) $\Rightarrow$ $r(C_1)<2r(C_2)$ which implies that $C_1$ and $C_2$ intersect at two distinct points. $\Box$ $(b)$ Let $\alpha$ be an exterior angle $\angle AO_1B$. Then the ratio between the areas must be $\frac{2\pi(2+\sqrt 3)}{\alpha}$ (by the sector's formula). By condition: $\angle AO_2O_1=\angle BO_2O_1$ because $AO_2=O_1O_2=BO_2$ and $AO_1=BO_1$. Apply the cosine law in $\triangle AO_2O_1$: $cos(\angle AO_2O_1)=\frac{AO_2^2+O_1O_2^2-AO_1^2}{2AO_2\cdot O_1O_2}=\frac{2r^2-k^2r^2}{2r^2}=\frac{2-k^2}{2}=\frac{2-(2+\sqrt 3)}{2}=\frac{-\sqrt 3}{2}$. There are denote $r$ as radius of $C_2$ and $k$ as ratio by condition. Hence $\angle AO_2O_1=\frac{5\pi}{6} \Rightarrow \alpha = \frac{5\pi}{3}$. Therefore the ratio between areas equal $\frac{6(2+\sqrt 3)}{5}$.