If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$? 
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$

I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?

Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$
So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = (2\sqrt{2})^2$$
Now Put $$x-2 = 2\sqrt{2}\cos \phi\Rightarrow x = 2+2\sqrt{2}\cos \phi$$
and $$y-2 = 2\sqrt{2}\sin \phi\Rightarrow y = 2+2\sqrt{2}\sin \phi$$
So $$\displaystyle x-y = 2\sqrt{2}\left(\cos\phi-\sin \phi\right) = 4\cdot \left[\cos \phi \cdot \frac{1}{\sqrt{2}}-\sin \phi\cdot \frac{1}{\sqrt{2}}\right] = 4\cos\left(\phi+\frac{\pi}{4}\right)$$
So we know that $$\displaystyle -4 \leq 4\cos\left(\phi+\frac{\pi}{4}\right)\leq 4$$
So we get $$-4 \leq x-y\leq 4\Rightarrow x-y\in \left[-4,4\right]$$
