Let P = \ z ∈ ℂ: |z + 2 - 3i| ≤ 1\ and Q = \ z ∈ ℂ: z(1 + i) + ¯ z...

Given: $\left|z + 2 - 3i\right| \le 1$

Putting, $z = x + iy$.

$⇒ {\left(x + 2\right)}^{2} + {\left(y - 3\right)}^{2} \le 1 . . . \left(i\right)$

For the circle represented in equation \left(i\right)$, centre is \left(- 2, 3\right) and radius,$r = 1$.</p><p>It is given that,$z\left(1 + i\right) + z \left(1 - i\right) \le - 8$</p><p>$⇒ \left(x + iy\right)\left(1 + i\right) + \left(x - iy\right)\left(1 - i\right) \le - 8$</p><p>$⇒ x + ix - y + iy + x - ix - iy - y \le - 8$</p><p>$⇒ 2x - 2y \le - 8$</p><p>$⇒ x - y + 4 = 0$</p><p>So, the line passing through \left(3, - 2\right) and perpendicular to$x - y + 4 = 0$is given by,$x + y - 1 = 0$.</p><p><img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/6a48608a-82e8-4389-92ee-b95c00c27545.png" style="width: 350px; height: 272px;"/></p><p>So, the distance of line$x - y + 4 = 0$from the centre \left(- 2, 3\right)$ is given by,

$d = \left|\dfrac{- 2 - 3 + 4}{\sqrt{2}}\right|$

$⇒ d = \dfrac{1}{\sqrt{2}}$

Now, $x + y - 1 = 0$ can be rewritten as,

$\dfrac{x + 2}{\dfrac{- 1}{\sqrt{2}}} = \dfrac{y - 3}{\dfrac{1}{\sqrt{2}}} = CA\text{or}CB$

For $CA = \dfrac{- 1}{\sqrt{2}}$ and $CB = 1$, $A ≡ \left(\dfrac{- 3}{2}, \dfrac{5}{2}\right)$ and $B ≡ \left(\dfrac{- 1}{\sqrt{2}} - 2, \dfrac{1}{\sqrt{2}} + 3\right)$.

$⇒ {\left|{z}_{1}\right|}^{2} = {\left(2 + \dfrac{1}{\sqrt{2}}\right)}^{2} + {\left(3 + \dfrac{1}{\sqrt{2}}\right)}^{2}$

$⇒ {\left|{z}_{1}\right|}^{2} = 4 + \dfrac{1}{2} + 2\sqrt{2} + 9 + \dfrac{1}{2} + 3\sqrt{2}$

$⇒ {\left|{z}_{1}\right|}^{2} = 14 + 5\sqrt{2}$

$⇒ {\left|{z}_{2}\right|}^{2} = 2\left(\dfrac{9}{4} + \dfrac{25}{4}\right)$

$⇒ {\left|{z}_{2}\right|}^{2} = 17$

$⇒ {\left|{z}_{1}\right|}^{2} + 2{\left|{z}_{2}\right|}^{2} = 31 + 5\sqrt{2}$

$⇒ \alpha = 31, \beta = 5$

$⇒ \alpha + \beta = 36$