JEE Main 2024 — 1 February, Shift 1 — Question 27

Clearly for the shaded region $\mathrm{z}_{1}$ is the intersection of the circle and the line passing through $\mathrm{P}\left(\mathrm{L}_{1}\right)$ and $z_{2}$ is intersection of line $L_{1} \& L_{2}$

Circle : $(x+2)^{2}+(y-3)^{2}=1$

$\mathrm{L}_{1}: \mathrm{x}+\mathrm{y}-1=0$

$\mathrm{L}_{2}: \mathrm{x}-\mathrm{y}+4=0$

On solving circle $\& L_{1}$

we get $z_{1}:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right)$

On solving $L_{1}$ and $z_{2}$ is intersection of line $L_{1} \& L_{2}$

we get $z_{2}:\left(\frac{-3}{2}, \frac{5}{2}\right)$

$\begin{aligned} & \left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2}=14+5 \sqrt{2}+17 & =31+5 \sqrt{2} \end{aligned}$

So $\alpha=31$

$\beta=5$

$\alpha+\beta=36$