JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 17)
Let $$S = \{ z \in C:|z - 2| \le 1,\,z(1 + i) + \overline z (1 - i) \le 2\} $$. Let $$|z - 4i|$$ attains minimum and maximum values, respectively, at z1 $$\in$$ S and z2 $$\in$$ S. If $$5(|{z_1}{|^2} + |{z_2}{|^2}) = \alpha + \beta \sqrt 5 $$, where $$\alpha$$ and $$\beta$$ are integers, then the value of $$\alpha$$ + $$\beta$$ is equal to ___________.
Answer
26
Explanation
_29th_June_Morning_Shift_en_17_1.png)
$S$ represents the shaded region shown in the diagram.
Clearly $z_{1}$ will be the point of intersection of $P A$ and given circle.
$P A: 2 x+y=4$ and given circle has equation $(x-2)^{2}+y^{2}=1$
On solving we get
$z_{1}=\left(2-\frac{1}{\sqrt{5}}\right)+\frac{2}{\sqrt{5}} i \Rightarrow\left|z_{1}\right|^{2}=5-\frac{4}{\sqrt{5}}$
$z_{2}$ will be either $B$ or $C$.
$\because P B=\sqrt{17}$ and $P C=\sqrt{13}$ hence $z_{2}=1$
So $5\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)=30-4 \sqrt{5}$
Clearly $\alpha=30$ and $\beta=-4 \Rightarrow \alpha+\beta=26$
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