JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Area Under Curves

The circle $(x-2\sqrt{3}{)}^2+y^2=12$ has center $(2\sqrt{3},0)$ and radius $2\sqrt{3}$, so it passes through the origin. The parabola $y^2=2\sqrt{3}x$ also passes through the origin, and substituting into the circle equation:


So the curves meet at $x=0$ and $x=2\sqrt{3}$ — the parabola's arc lies inside the circle between these two intersections, and the full circle has area $\pi (2\sqrt{3}{)}^2=12\pi$, split into two halves by symmetry about the $x$-axis with the chord through the intersection points.

Required area = area of circle to the right of the chord at $x=2\sqrt{3}$ (a semicircle) minus the area between the parabola and this chord line, doubled for symmetry above and below the axis:

where $\frac{\pi r^2}{2}=\frac{\pi (12)}{2}=6\pi$ is the semicircular area beyond the chord $x=2\sqrt{3}$, and the second term is the area under the parabola from $x=0$ to $x=2\sqrt{3}$ (doubled for both branches $\pm y$).



Since $(2\sqrt{3}{)}^{3/2}=(2\sqrt{3})\sqrt{2\sqrt{3}}$, this becomes $\frac{2}{3}(2\sqrt{3}){\left(\sqrt{2\sqrt{3}}\right)}^2=\frac{2}{3}(2\sqrt{3})(2\sqrt{3})=\frac{2}{3}\cdot 12=8$.

So