Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} ={a}(x-y)$ Finding $\frac{ \mathrm{d}y}{ \mathrm{d}x}$ when given that $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$. $a$ is a constant.
I have the final answer, which is $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \sqrt{\frac{1-y^2}{1-x^2}}.$$ 
But I've only been able to get till $$\frac{ \mathrm{d}y}{ \mathrm{d}x} = \frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}.\sqrt{\frac{1-y^2}{1-x^2}}.$$ 
How can the two terms be cancelled off? Thanks.

Solve explicitly for $y$:
$y = \frac{a^2 x-2 a \sqrt{1-x^2}-x}{a^2+1}$.
Then take the derivative:
${d y \over d x} = \frac{a^2+\frac{2 a x}{\sqrt{1-x^2}}-1}{a^2+1}$
or
$-\frac{2 a \sqrt{1-x^2}}{a^2+1}-\frac{2 x}{a^2+1}+x$
