imc

Let the radius of the circle be $r$, and let its center be $O$. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle $O$, then $\angle OBA=\angle OCA=90^{\circ }$, so $\angle BOC=120^{\circ }$. Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to $r$, then (pick your favorite method) $\overline{BC}=r\sqrt{3}$. The area of the equilateral triangle is $\frac{(r\sqrt{3}{)}^2\sqrt{3}}{4}=\frac{3r^2\sqrt{3}}{4}$, and the area of the sector we are subtracting from it is $\frac{1}{3}\pi r^2-\frac{1}{2}\cdot r\sqrt{3}\cdot \frac{r}{2}=\frac{\pi r^2}{3}-\frac{r^2\sqrt{3}}{4}$. The area outside of the circle is $\frac{3r^2\sqrt{3}}{4}-\left(\frac{\pi r^2}{3}-\frac{r^2\sqrt{3}}{4}\right)=r^2\sqrt{3}-\frac{\pi r^2}{3}$. Therefore, the answer is $$\frac{r^2\sqrt{3}-\frac{\pi r^2}{3}}{\frac{3r^2\sqrt{3}}{4}}=\boxed{\frac{4}{3}-\frac{4\sqrt{3}\pi }{27}}$$