integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$ 
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$

what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please 

Look at the discriminant which should be positive definite so the $y$ is real for real $x$. This demands 
$$-3x^2+12x+4 \ge 0 \implies 3x^2-12x-4 \le 0 \implies (x-4.3)(x+.3) \le 0  \implies -.3 \le x \le 4.3 ~~(roughly)$$
So the possible integral values of $x$ are $x=0,1,2,3,4$
out of these we accept the ones which give $y$ as integrals.
Check that $x=0,1,2,4$ give integral values of $y$, such pairs of  $x$ and $y$ are the solutions.
