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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?

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5 Answers 5

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$\frac{x^2+y^2}{x+y}=4$ 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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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 $$-4 \leq x-y\leq 4\Rightarrow x-y\in \left[-4,4\right]$$

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  • $\begingroup$ ,nice solution,but there is a typo.$x=2+2\sqrt2\cos\phi,y=2+2\sqrt2\sin\phi$ and $x-y=2\sqrt2(\cos\phi-\sin\phi)$.Anyways,thanks for the solution. $\endgroup$
    – diya
    Sep 20, 2015 at 12:20
  • $\begingroup$ sorry Diya, I have edited it. $\endgroup$
    – juantheron
    Sep 20, 2015 at 12:25
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Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$

So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = 8$$

Now we can write $$x-y = (x-2)-(y-2) = \left[(x-2)+(2-y)\right]$$

Now Using $\bf{Cauchy\; Schwartz\; Inequality}$

$$\displaystyle \left[(x-2)^2+(2-y)^2\right]\cdot [1^2+1^2]\geq \left[x-2+2-y\right]^2$$

So $$8\times 2 \geq (x-y)^2$$

So $$(x-y)^2\leq 4^2$$

So $$-4 \leq (x-y)\leq 4\Rightarrow x-y\in \left[-4,4\right]$$

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  • $\begingroup$ You also have to argue that every value in $[-4,4]$ is possible. $\endgroup$ Sep 20, 2015 at 13:25
  • $\begingroup$ Yes barto you are Right. $\endgroup$
    – juantheron
    Sep 20, 2015 at 13:49
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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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${x^2+y^2\over x+y}=4 \implies x^2+y^2=4x+4y \implies x^2+y^2-4x-4y=0 \implies (x-2)^2+(y-2)^2=(2\sqrt{2})^2$ which is a circle with center $(2,2)$ and radius $2\sqrt{2}$

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