JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 27)
Let $\mathrm{P}=\{\mathrm{z} \in \mathbb{C}:|z+2-3 i| \leq 1\}$ and $\mathrm{Q}=\{\mathrm{z} \in \mathbb{C}: z(1+i)+\bar{z}(1-i) \leq-8\}$. Let in $\mathrm{P} \cap \mathrm{Q}$, $|z-3+2 i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $\left|z_1\right|^2+2\left|z_2\right|^2=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals _____________.
Answer
36
Explanation
_1st_February_Morning_Shift_en_27_1.png)
Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $\mathrm{P}\left(\mathrm{L}_1\right)$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$
Circle : $(x+2)^2+(y-3)^2=1$
$$ \begin{aligned} & \mathrm{L}_1: \mathrm{x}+\mathrm{y}-1=0 \\\\ & \mathrm{~L}_2: \mathrm{x}-\mathrm{y}+4=0 \end{aligned} $$
On solving circle $\& \mathrm{~L}_1$ we get
$$ \mathrm{z}_1:\left(-2-\frac{1}{\sqrt{2}}, 3+\frac{1}{\sqrt{2}}\right) $$
On solving $\mathrm{L}_1$ and $\mathrm{z}_2$ is intersection of line $\mathrm{L}_1 \& \mathrm{~L}_2$ we get $z_2:\left(\frac{-3}{2}, \frac{5}{2}\right)$
$\begin{aligned} & \left|z_1\right|^2+2\left|z_2\right|^2=14+5 \sqrt{2}+17 \\\\ & =31+5 \sqrt{2}\end{aligned}$
$\begin{array}{ll}\text { So } & \alpha=31 \\\\ & \beta=5 \\\\ & \alpha+\beta=36\end{array}$
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