Finding the minimum value of $(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2$ We assume $\alpha,\beta$ are real numbers, I would like to find the minimum value of 

$$(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2.$$

After expanding $(\alpha+5)^2+9\cos^2 \beta-2(\alpha+5)|3\cos \beta|+\alpha^2+4\sin^2 \beta-4\alpha|2\sin \beta|$, I would have some help with this.
 A: 
Find the minimum value of $(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2$

Hint. By expanding as you did then factoring the quadratic expression in $\alpha$ one gets
$$ \small
\begin{align}
(\alpha+5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2&=2\left(\alpha+\frac{5-3|\cos \beta|-2|\sin \beta|}2\right)^2 +\frac{\left(5-3|\cos \beta|+2|\sin \beta|\right)^2}2
\\\\&\ge\frac{\left(5-3|\cos \beta|+2|\sin \beta|\right)^2}2
\\\\&\ge 2.
\end{align}
$$
A: Using Geometry
Let $(\alpha +5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2 = d^2$ 
and Let $x_{1}=\alpha+5\;\;,x_{2}=3|\cos \alpha|$ and $y_{1}=\alpha$ and $y_{2} = 2|\sin \beta|$
$\Rightarrow (x_{1},y_{1})$ lie on the line $x-y-5=0$ and $(x_{2},y_{2})$ lie in the first quadrant,s
well on the ellipse $\displaystyle \frac{x^2}{9}+\frac{y^2}{4} = 1$
and $d$ is the distance between $(x_{1},y_{2})$ and $(x_{2},y_{2})$ 
Now minimum $$d = \bf{Distance\; b/w\; (3,0)\; and \; line  = \frac{|3-0-5|}{\sqrt{1^2+1^2}}} = \sqrt{2}$$
So $$\min (\alpha +5-3|\cos \beta|)^2+(\alpha-2|\sin \beta|)^2 = \min(d^2) = 2$$
