Prove that the four vertices of a quadrilateral,the combined equation of whose sides is $|x^2-y^2|-2(|x+y|+|x-y|)+4=0$,are concyclic.

Question

Prove that the four vertices of a quadrilateral,the combined equation of whose sides is $|x^2-y^2|-2(|x+y|+|x-y|)+4=0$,are concyclic.

I need to find the equations of the sides in order to get the vertices.So tried to fetch the individual equations of the edges.
I took four cases.

Case$(1):$When $x+y>0,x-y>0$
$x^2-y^2-4x+4=0\Rightarrow (x-2)^2-y^2=0\Rightarrow (x-2-y)(x-2+y)=0$

either $x+y=2$ or $x-y=2$

Case$(2):$When $x+y<0,x-y<0$
$x^2-y^2+4x+4=0\Rightarrow (x+2)^2-y^2=0\Rightarrow (x+2-y)(x+2+y)=0$

either $x+y=-2$ or $x-y=-2$

Case$(3):$When $x+y>0,x-y<0$
$y^2-x^2-4y+4=0\Rightarrow (y-2)^2-x^2=0\Rightarrow (y-2-x)(y-2+x)=0$

either $x+y=2$ or $x-y=-2$

Case$(4):$When $x+y<0,x-y>0$
$y^2-x^2+4y+4=0\Rightarrow (y+2)^2-x^2=0\Rightarrow (y+2-x)(y+2+x)=0$

either $x+y=-2$ or $x-y=2$

But i cant find the vertices of the quadrilateral and proceed further.

But in the above cases,in each case i am getting 2 equations,not 4 equations.Is my method correct,if not what is the correct method.Please help me.

Answer 1

The question here is how do you define "the combined equation of the sides" of a quadrilateral ? The interpretation I used in this answer is: the quadrilateral can be determined by four lines and the set of points $(x,y)$ in these lines verifies : $$|x^2−y^2|−2(|x+y|+|x−y|)+4 $$ but we have the following equivalences : $$\begin{align}|x^2−y^2|−2(|x+y|+|x−y|)+4 =0 &\iff \left(\left|x-y\right|-2\right)\left(\left|x+y\right|-2\right)=0 \\ & \iff \left(x-y-2\right)\left(x-y+2\right)\left(x+y-2\right)\left(x+y+2\right)=0\end{align}$$ The set of points verifying the last equation is the union of four lines which can be illustrated below :

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad $

Finally the quadrilateral defined by the four lines is concyclic.