Let S = z ∈ ℂ: |z - 3| ≤ 1 and z(4 + 3i) + ¯ z (4 - 3i) ≤ 24 . If α...

Given,

$\left|z - 3\right| \le 1$

Which represent a circle of radius $1 and centred at \left(3, 0\right)$

Now simplifying $z\left(4 + 3i\right) + \overset{¯}{z}\left(4 - 3i\right) \le 24$

We get, \left(x + iy\right)\left(4 + 3i\right) + \left(x - iy\right)\left(4 - 3i\right) \le 24$</p><p>$⇒ 4x + 3xi + 4iy - 3y + 4x - 3ix - 4iy - 3y \le 24$</p><p>$⇒ 8x - 6y \le 24$</p><p>$⇒ 4x - 3y \le 12$</p><p>Now plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b2d63a67-5377-4864-bf4e-0cc617285a9a-9722930.png" style="width: 245px; height: 206px;" /></p><p>Minimum distance of$\left(0, 4\right)$from circle$= \sqrt{{3}^{2} + {4}^{2}} - 1 = 4$will lie along line joining \left(0, 4\right) \text{ and } \left(3, 0\right)$
$∴$ equation line
$\dfrac{x}{3} + \dfrac{y}{4} = 1 ⇒ 4x + 3y = 12 \ldots \left(i\right)$
Equation of circle ${\left(x - 3\right)}^{2} + {y}^{2} = 1 \ldots \left(ii\right)$

Solving (i) and (ii) we get,

${\left(\dfrac{12 - 3y}{4} - 3\right)}^{2} + {y}^{2} = 1$

${\left(\dfrac{- 3y}{4}\right)}^{2} + {y}^{2} = 1$

$\dfrac{25{y}^{2}}{16} = 1 ⇒ y = \pm \dfrac{4}{5}$

So, for minimum distance $y = \dfrac{4}{5}$

$∴ x = \dfrac{12}{5}$

$∴ 25\left(\alpha + \beta\right) = 25\left(\dfrac{4}{5} + \dfrac{12}{5}\right)$

$⇒ 16 \times 5 = 80$