<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] The condition $\frac{x^2+y^2}{x+y}=4$ is equivalent to $(x+y)^2+(x-y)^2=8(x+y)$. Let $x+y=s$ and $x-y=d$. Note that this induces a bijection from $\Bbb R^2$ to itself, meaning that for every pair $(s,d)$ there exist corresponding $x,y$. We have $0\leq d^2=8s-s^2$. The nonnegative values that $8s-s^2$ can take are $[0,16]$, so $d$ takes values in $[-4,4]$.
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