<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] $\frac{x^2+y^2}{x+y}=4$ is equivalent to $(x-2)^2+(y-2)^2 = 8$, hence $(x,y)$ lies on a circle centered at $(2,2)$ with radius $2\sqrt{2}$. The tangents at the points $(0,4)$ and $(4,0)$ are parallel to the $y=x$ line, so the right answer is $(C)$.
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