JEE Main 2026 Jan 23 Shift 2, Mathematics Q12: Area as Definite Integral

Both circles have radius $2$; centres at $(0,0)$ and $(0,2)$, distance between centres $=2=r$.

The intersection points satisfy both circles: subtracting gives $y=1$, then $x=\pm \sqrt{3}$. So they meet at $(\pm \sqrt{3},1)$.

By symmetry the lens-shaped overlap area is

$A=2{\int }_0^{\sqrt{3}}\left[\sqrt{4-x^2}-\left(2-\sqrt{4-x^2}\right)\right]dx=4{\int }_0^{\sqrt{3}}\left(\sqrt{4-x^2}-1\right)dx$.

Using $\int \sqrt{4-x^2}dx=\frac{x\sqrt{4-x^2}}{2}+2{\sin }^{-1}\frac{x}{2}$:

At $x=\sqrt{3}$: $\frac{\sqrt{3}\cdot 1}{2}+2\cdot \frac{\pi }{3}=\frac{\sqrt{3}}{2}+\frac{2\pi }{3}$.

$A=4\left[\frac{\sqrt{3}}{2}+\frac{2\pi }{3}-\sqrt{3}\right]=4\left[\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right]=\frac{8\pi }{3}-2\sqrt{3}=\frac{2}{3}(4\pi -3\sqrt{3})$.