The area of the region that is common to the circle $x^2+y^2=16 a^2$ and the parabola $y^2=6 a x$ is
- $\frac{4 a^2}{3}(4 \pi+\sqrt{3})$
- $\frac{2 a^2}{3}(3 \pi+\sqrt{3})$
- $\frac{4 a^2}{3}(2 \pi+\sqrt{2})$
- $\frac{2 a^2}{3}(2 \pi+\sqrt{3})$
Solution
Asked in: AP EAMCET 2017 (25 Apr Shift 2)