Solve quadratic equation with two variables

Question

I can't find a solution for this equation

$$\frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2}= -\frac{1}{2}$$

Any help?

Answer 1

If you try to solve as always you will find:

$$2(xy-2016y-2017x+(2016\cdot 2017))=(-1)(x^2-4032x+2016^2+y^2-4034y+2017^2)$$

That is:

$$x^2+2xy+y^2-8066(x+y)+ 4033^2=0$$

$$(x+y)^2-8066(x+y)+4033^2=0$$

$$((x+y) - 4033)^2=0$$

Hence $x+y=4033$.

You can see that last step by doing $t=x+y$ and solving $t^2-8066t +4033^2=0$.

Answer 2

HINT

In the first place, let us set that $a = x - 2016$ and $b = y - 2017$, from whence we conclude that \begin{align*} \frac{ab}{a^{2}+b^{2}} = -\frac{1}{2} \Longleftrightarrow 2ab = -a^{2} - b^{2} \Longleftrightarrow (a + b)^{2} = 0 \Longleftrightarrow a + b = 0 \end{align*}

Can you proceed from here?

Answer 3

$$ \frac{(x-2016)(y-2017)}{(x-2016)^2 + (y-2017)^2} $$

$ Let: \ t=x-2016, s=y-2017 $, then:

$$ \frac{ts}{t^2+s^2} = -\frac{1}{2} $$

If $t=0$ we have no solution, so multiply left side by $\frac{t^2}{t^2} $ getting:

$$ \frac{\frac{s}{t}}{(\frac{s}{t})^2 + 1} = -\frac{1}{2} $$

Now substituting $ z=\frac{s}{t} $ we finally get:

$$\frac{z}{z^2+1} = -\frac{1}{2}$$

$ -2z = z^2 + 1 $

$ (z+1)^2 = 0 $

so $ z=-1 $, that is $t = -s$, in terms of $x,y$ it means:

$x-2016 = -y + 2017$

$x+y = 4033 $