How do I simplify this expression? How do I simplify:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
If simplification is possible, it should be possible with elementary algebra, but I'm completely lost as to how to go about it.

What I've done so far:
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+x^2-y^2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}=\frac{-α^2 x^2 - 2 α x^2 + α^2 y^2 + 2 α y^2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$\implies d^2=\frac{(-α^2 x^2 - 2 α x^2 + α^2 y^2 + 2 α y^2 - 2)^2}{\alpha^2x^2+\alpha^2y^2}$$
 A: I believe your third equality is incorrect.  Numerator distribution won't lead to any terms with x and y to the first power.  
Then when you squared d below, you also have additional squares added to a in denominator where there were none as well as a double square for the entire numerator and denominator - essentially you have d to the 4th power with a being squared on bottom where it wasn't in the initial equation.
$$d=\frac{-(x+\alpha x)^2+(y+\alpha y)^2+2+x^2-y^2-2}{\sqrt{\alpha^2 x^2+\alpha^2 y^2}}$$
$$=\frac{-x^2(1+\alpha)^2+y^2(1+\alpha)^2+x^2-y^2}{\sqrt{\alpha x^2+\alpha y^2}}=\frac{(y^2-x^2)(1+ 2\alpha + \alpha^2)+ x^2 - y^2}{\sqrt{\alpha^2} \sqrt{x^2+ y^2}}$$
$$ =\frac{(2\alpha+\alpha^2)(y^2-x^2)}{\alpha \sqrt{x^2+ y^2}}$$
Can continue from there perhaps, factor an alpha, square both sides if that helps.
A: $$\frac{-\alpha^2x^2+\alpha^2y^2-2\alpha x^2+2\alpha y^2}{\sqrt{\alpha^2x^2+\alpha^2y^2}}=\frac{\alpha^2(y^2-x^2)+2\alpha(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}$$
$$\frac{\alpha^2(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}+\frac{2\alpha(y^2-x^2)}{\alpha\sqrt{x^2+y^2}}=\alpha\frac{(y^2-x^2)}{\sqrt{x^2+y^2}}+2\frac{y^2-x^2}{\sqrt{x^2+y^2}}$$
Thanks: @Timmay & @David Diaz
