<s>[INST] 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?
 [/INST] ${x^2+y^2\over x+y}=4 \implies x^2+y^2=4x+4y \implies x^2+y^2-4x-4y=0 \implies (x-2)^2+(y-2)^2=(2\sqrt{2})^2$ which is a circle with center $(2,2)$ and radius $2\sqrt{2}$
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