JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Area Under Curves

Both circles have radius $2$; the first is centred at $(0,0)$ and the second at $(0,2)$, so the centres are $2$ apart. They meet where $y^2=(y-2{)}^2$, giving $y=1$ and $x=\pm \sqrt{3}$. The region common to the two circles is a symmetric lens, with area $A=4{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx$. Using $\int \sqrt{4-x^2}dx=\frac{x\sqrt{4-x^2}}{2}+2{\sin }^{-1}\frac{x}{2}$, $A=4\left[\frac{\sqrt{3}}{2}+2\cdot \frac{\pi }{3}-\sqrt{3}\right]=\frac{8\pi }{3}-2\sqrt{3}=\frac{2}{3}(4\pi -3\sqrt{3})$.