If the least and the largest real values of

$x+i y+\alpha \sqrt{(x-1)^{2}+y^{2}}+2 i=0$

$\therefore y+2=0$ and $x+\alpha \sqrt{(x-1)^{2}+y^{2}}=0$

$y=-2 \& x^{2}=\alpha^{2}\left(x^{2}-2 x+1+4\right)$

$\alpha^{2}=\frac{x^{2}}{x^{2}-2 x+5} \Rightarrow x^{2}\left(\alpha^{2}-1\right)-2 x \alpha^{2}+5 \alpha^{2}=0$

$x \in R \Rightarrow D \geq 0$

$\alpha \alpha^{4}-4\left(\alpha^{2}-1\right) 5 \alpha^{2} \geq 0$

$\alpha^{2}\left[4 \alpha^{2}-2 \alpha^{2}+20\right] \geq 0$

$\alpha^{2}\left[-16 \alpha^{2}+20\right] \geq 0$

$\alpha^{2}\left[\alpha^{2}-\frac{5}{4}\right] \leq 0$

$0 \leq \alpha^{2} \leq \frac{5}{4}$

$\therefore \alpha^{2} \in\left[0, \frac{5}{4}\right]$

$\therefore \alpha \in\left[-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{2}\right]$

then $4\left[(q)^{2}+(p)^{2}\right]=4\left[\frac{5}{4}+\frac{5}{4}\right]=10$