JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 16)
Let S = {z $$\in$$ C : |z $$-$$ 3| $$\le$$ 1 and z(4 + 3i) + $$\overline z $$(4 $$-$$ 3i) $$\le$$ 24}. If $$\alpha$$ + i$$\beta$$ is the point in S which is closest to 4i, then 25($$\alpha$$ + $$\beta$$) is equal to ___________.
Answer
80
Explanation
Here $$|z - 3| < 1$$
$$ \Rightarrow {(x - 3)^2} + {y^2} < 1$$
and $$z = (4 + 3i) + \overline z (4 - 3i) \le 24$$
$$ \Rightarrow 4x - 3y \le 12$$
$$\tan \theta = {4 \over 3}$$
$$\therefore$$ Coordinate of $$P = (3 - \cos \theta ,\sin \theta )$$
$$ = \left( {3 - {3 \over 5},{4 \over 5}} \right)$$
$$\therefore$$ $$\alpha + i\beta = {{12} \over 5} + {4 \over 5}i$$
$$\therefore$$ $$25(\alpha + \beta ) = 80$$
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