Minimum value of expression having $2$ variables 
Minimum value of $$\bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$ for real $x\geq 0,y\in[-2,2]$

Try: Using Partial derivative 
$$f(x,y) = \bigg(x-4-\sqrt{4-y^2}\bigg)^2+\bigg(4\sqrt{x}-y\bigg)^2$$
$$\frac{df(x,y)}{dx}=2\bigg(x-4-\sqrt{4-y^2}\bigg)+\frac{4}{\sqrt{x}}\bigg(4\sqrt{x}-y\bigg)$$
$$\frac{df(x,y)}{dy}=-2y\frac{\bigg(x-4-\sqrt{4-y^2}\bigg)}{\sqrt{4-y^2}}-2\bigg(4\sqrt{x}-y\bigg)$$
Put $\displaystyle \frac{df}{dx} =0$ and $\displaystyle \frac{df}{dy}=0$
But these $2$ equation is tedious work
Could some help me how to solve it , Thanks in advance
 A: For $0\leq x\leq 4$ we have 
$$f(x,y)-f(x,2)=2\left(4\sqrt{x}(2-y)+(4-x)\sqrt{4-y^2}\right)\geq 0 $$
$$f(x,y)\geq  f(x,2)\geq  \underset{0\leq x\leq 4}{\min }f(x,2)= \underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$$
For $x\geq 4$, use the Cauchy–Schwarz inequality, to show $f(x,y) \geq 36$ with $f(4,2)=36$ as follows
$$\left(x-4,4\sqrt{x}\right).\left(\sqrt{4-y^2},y\right)\leq  2(x+4)$$
$$f(x,y)-36=-16+8 x+x^2-2\left((x-4) \sqrt{4-y^2}+4 \sqrt{x} y\right)$$
$$\geq  -16+8 x+x^2-4(x+4)=(x-4)(x+8)\geq 0$$
EDIT:
$\underset{0\leq x\leq 4}{\min }\left(20-16 \sqrt{x}+8 x+x^2\right)$ occurs at the zero of the cubic polynomial $x^3+8x^2+16x-16$ with $x \approx 0.7186$ and the minimum value $\approx 12.7019$
A: I actually get another answer from @Lozenges
If we look at the given expression carefully, we will notice that it is kind of like a distance formula. 
Set $ \vec{x}=(x-4,4\sqrt{x})$ and $\vec{y}=(\sqrt{4-y^2},y)$ , the given expression can be written as $|| \vec{x}-\vec{y}||^2$. The point $\vec{x}$ has a locus of a parabola $y^2 =16x+64, x\ge0, y\ge0$ and the point $\vec{y}$ has a locus of a circle$x^2+y^2=4, x\ge0$. Draw their graph and you will see it is very trivial that the closest distance between the two graphs is 6, hence the minimum value of the given expression is 36. 
The idea should be correct, but I might have made a mistake somewhere? Comments are more than welcome.
