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.

Given $$\sqrt{1-x^2}+\sqrt{1-y^2} = a(x-y)\;,$$ Where $a$ is constant.
Now Put $x=\sin \alpha$ and $y = \sin \beta$
So $$\sqrt{1-\sin^2 \alpha}+\sqrt{1-\sin^2 \beta} = a(\sin \alpha-\sin \beta)$$
So $$\displaystyle \cos \alpha+\cos \beta = a(\sin \alpha-\sin \beta)$$
So $$\displaystyle 2\cos \left(\frac{\alpha+\beta}{2}\right)\cdot \cos \left(\frac{\alpha-\beta}{2}\right) = 2a\cdot \cos \left(\frac{\alpha+\beta}{2}\right)\cdot \sin \left(\frac{\alpha-\beta}{2}\right)$$
So we get $$\displaystyle 2\cos \left(\frac{\alpha+\beta}{2}\right)\cdot \left[\cos \left(\frac{\alpha-\beta}{2}\right)-a\cdot \sin \left(\frac{\alpha-\beta}{2}\right)\right] = 0$$
So either $$\displaystyle \cos \left(\frac{\alpha+\beta}{2}\right) = 0\Rightarrow \alpha+\beta = 2n\pi\pm \pi,n\in \mathbb{Z}$$ or $$\displaystyle \tan \left(\frac{\alpha-\beta}{2}\right) = \frac{1}{a}$$
So $\displaystyle \tan \left(\frac{\alpha-\beta}{2}\right) = \frac{1}{a}\Rightarrow \left(\frac{\alpha-\beta}{2}\right) = \tan^{-1}\left(\frac{1}{a}\right)\Rightarrow \alpha-\beta = 2\tan^{-1}\left(\frac{1}{a}\right)$
so $$\displaystyle \sin^{-1}(x)-\sin^{-1}(y) = 2\tan^{-1}\left(\frac{1}{a}\right)$$
Now Differentitae both side w r to $x\;,$ We get $\displaystyle \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx} = 0$
so we get $\displaystyle \frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = \sqrt{\frac{1-y^2}{1-x^2}}$
