JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 9

Curves. The circle has centre $(-2\sqrt3,0)$ and radius 4. Solving with $y^2=8x+4$ gives $x^2+(8+4\sqrt3)x=0$, so they meet at $x=0$, $y=\pm2$.

Smaller region (inside the circle, right of the parabola):
$$2\int_0^2\left[\left(\sqrt{16-y^2}-2\sqrt3\right)-\frac{y^2-4}8\right]dy.$$

Evaluate. $\int_0^2\sqrt{16-y^2}\,dy=2\sqrt3+\frac{4\pi}3$, so the area is $2\left[\frac{4\pi}3-2\sqrt3+\frac23\right]=\frac13\left(4-12\sqrt3+8\pi\right)$.

(C)