JEE Main 2025 — Circle Question with Solution

By solving $\mathrm{x}=\mathrm{y}$ with circle We get $\begin{aligned} & \mathrm{C}(\sqrt{2}, \sqrt{2}) \\ & \mathrm{D}(-\sqrt{2},-\sqrt{2}) \end{aligned}$ By solving $\mathrm{x}+\mathrm{y}=1$ with circle $x^2+y^2=4$ we set $\begin{aligned} & \mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right) \\ & \& \mathrm{~B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right) \end{aligned}$ $\therefore$ Area of Quadrilateral ACBD $=2\times \text{ Area of }△\mathrm{BCD}$ $\begin{array}{rl}& =2\times \frac{1}{2}\left|\begin{array}{ccc}\sqrt{2}& \sqrt{2}& 1\\ \frac{1-\sqrt{7}}{2}& \frac{1+\sqrt{7}}{2}& 1\\ -\sqrt{2}& -\sqrt{2}& 1\end{array}\right|\\ & =2\sqrt{14}\end{array}$