If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?

The options are:

  
*
  
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
  
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
  
*$$\frac{1-x^2}{1-y^2}$$
  
*$$\frac{1-y^2}{1-x^2}$$
  

I tried differentiating the entire expression:
$$\begin{align}
\frac{-2x}{2\sqrt{1-x^2}}+\frac{-2y}{2\sqrt{1-y^2}}\frac{dy}{dx} &= a(1-\frac{dy}{dx}) \\
a + \frac{x}{\sqrt{1-x^2}} &= \frac{dy}{dx}(a-\frac{y}{\sqrt{1-y^2}}) \\
\frac{dy}{dx} &= (a+\frac{x}{\sqrt{1-x^2}})(a-\frac{y}{\sqrt{1-y^2}})^{-1}
\end{align}$$
I am not sure how to proceed now. Is there a better approach? Is it advisable to use my approach?

It looks like there's supposed to be a way to eliminate $a$ from the formula.
Let's try to tackle the term $a+\frac{x}{\sqrt{1-x^2}}$ first, using the fact that
$$a(x-y) = \sqrt{1-x^2} + \sqrt{1-y^2}.$$
In order to avoid dealing with a lot of fractions, I would
multiply $a + \frac{x}{\sqrt{1-x^2}}$ by the factor $x-y$ from the equation
and by the denominator $\sqrt{1-x^2}$.
Then
\begin{align}
\left(a + \frac{x}{\sqrt{1-x^2}}\right)(x-y)\sqrt{1-x^2}
&= a(x-y)\sqrt{1-x^2} + x(x - y) \\
&= (\sqrt{1-x^2} + \sqrt{1-y^2}) \sqrt{1-x^2} + x^2 - xy \\
&= 1 - x^2 + \sqrt{1-y^2}\sqrt{1-x^2} + x^2 - xy \\
&= 1 - xy + \sqrt{1-x^2}\sqrt{1-y^2} \\
\end{align}
Next, work out the value of
$$
\left(a - \frac{y}{\sqrt{1-y^2}}\right)(x-y)\sqrt{1-y^2}
$$
in a similar manner to get an expression in $x$ and $y$ only.
Compare the two results and see if that allows you to simplify your expression to one of the desired answers.
