If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$ 
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$

My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\
\frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}
$$
How do I poceed further and find the derivative ?
 A: Both $x,y\in[-1,1]$. So, $x=\sin\theta$ and $y=\sin\phi$ for some $\displaystyle \theta,\phi\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$. Note that $\cos\theta, \cos\phi\ge0$.
We have $$\cos\theta+\cos\phi=a(\sin\theta-\sin\phi)$$
So, $$2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}=2a\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2}$$
$\displaystyle \tan\frac{\theta-\phi}{2}=\frac{1}{a}$ is a constant. So, $\displaystyle \frac{d\phi}{d\theta}=1$.
$$\frac{dy}{dx}=\frac{\cos\phi}{\cos\theta}\frac{d\phi}{d\theta}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$$
A: HINT :
There is no parameter $a$ in the formula to be proved. So, first transform the initial equation into an equation where $a$ will be immediately eliminated by differentiation:
$$\frac{\sqrt{1-x^2}+\sqrt{1-y^2}}{x-y}=a$$
Differentiate and simplify.
A: After my comment you will get $$a\sqrt{1-x^2}+x=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and
$$a\sqrt{1-y^2}-y=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and you will get the desired result!
A: Move $a(x-y)$ to the LHS to make $F(x,y)=0$. Then:
$$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{-\frac{x}{\sqrt{1-x^2}}-a}{-\frac{y}{\sqrt{1-y^2}}+a}=\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}\cdot \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}},$$
because:
$$\begin{align}&\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}= \\
&1 \iff a(\sqrt{1-x^2}-\sqrt{1-y^2})=-(x+y) \iff a(y^2-x^2)= \\
&-(x+y)(\sqrt{1-x^2}+\sqrt{1-y^2}) \iff a(x-y)=\\
&\sqrt{1-x^2}+\sqrt{1-y^2}.\end{align}$$
A: Let us remark that the looked for differential equation can be written under the form  
$$\frac{dx}{\sqrt{1-x^2}}=\frac{dy}{\sqrt{1-y^2}}\tag{1}$$
involving solutions of the form :
$$-\arccos(x)=-\arccos(y)+a\tag{2}$$
otherwise said with cartesian equation :
$$y=\cos(\arccos(x)-a)\tag{3}$$
where $a$ is any real.
Setting $\alpha=\arccos(x)$ in (3), we get the equivalent parametric equations :
$$\begin{cases}x&=&\cos(\alpha)\\y&=&\cos(\alpha-a)\end{cases}\tag{4}$$
in which we recognize that we are working with elliptical arcs, as shown on the graphical representation of the family of curves $C_a$ displayed below (we have to pay attention to the domains of variables $x$ and $y$ : in general we will not have the whole ellipses as solutions but arcs of them). 
From (4), it is easy to establish a connection with relationship:
$$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$$
(in the spirit of the solution given by @CY Aries).

Fig. 1:Curves $C_a$ for $a=-\pi$ to $a=\pi$ with step $\pi/8$ (progressively changing from blue to red). The curves are elliptical arcs with two degenerate cases (straight lines).
