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?
 A: Your approach is absolutely fine and the result that you have obtained is correct. If you handed me a homework like this I would be satisfied, since it is clear that you have understood the procedure of implicit differentiation.
Still, the problem is somewhat pedantic and asks you to do a final superfluous simplification: note that $a$ is absent from all of the available options, therefore express $a$ as $\dfrac {\sqrt{1-x^2} + \sqrt{1-y^2}} {x-y}$ and replace it in your own result. After a number of elementary algebraic simplifications you will obtain $\dfrac {\sqrt{1-y^2}} {\sqrt{1-x^2}}$, which is option 2.
A: 
As stated in the comments this is actually a wee-bit lengthier and
  hence not recommended.

Since i always try to resort to trigonometric substitutions whenever fractional powers are involved I'm gonna attempt this by substituting $x=sinA$ and $y=sinB$
$cosA +cosB=a(sinA-sinB)$
$2cos\frac{(A+B)}{2} \times cos\frac{(A-B)}{2} =a\times 2Cos\frac{(A+B)}{2}\times Sin\frac{(A-B)}{2}$
$cot\frac{(A-B)}{2} =a$
$A-B=2cot^{-1}a$
Magically reducing it to 
$sin^{-1}x-sin^{-1}y =2cot^{-1}a$
Now differentiate…
$\frac{1}{\sqrt{1-x^{2}}} - \frac{1}{\sqrt{1-y^{2}}} \times \frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$

If you are preparing for JEE ADV i would recommend practising differential calculus book by cengage , and for theory i would recommend Differential Calculus by AmitMAgarwal

A: 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.
