If `sqrt(1-x^(2))+sqrt(1-y^(2))=a(x-y)`, then `(dy)/(dx)` equals

Options

A. `sqrt((1-x^(2))(1-y^(2)))`
B. `sqrt((1-y^(2))/(1-x^(2)))`
C. `sqrt((1-x^(2))/(1-y^(2)))`
D. none of these

Answer

To solve the equation \( \sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y) \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y) \]

### Step 2: Differentiate both sides with respect to \( x \)
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx} \left( \sqrt{1-x^2} \right) + \frac{d}{dx} \left( \sqrt{1-y^2} \right) = \frac{d}{dx} \left( a(x-y) \right) \]

Using the chain rule, we differentiate:
\[ \frac{d}{dx} \left( \sqrt{1-x^2} \right) = \frac{-x}{\sqrt{1-x^2}} \]
\[ \frac{d}{dx} \left( \sqrt{1-y^2} \right) = \frac{-y \frac{dy}{dx}}{\sqrt{1-y^2}} \]
\[ \frac{d}{dx} \left( a(x-y) \right) = a(1 - \frac{dy}{dx}) \]

Putting it all together, we have:
\[ \frac{-x}{\sqrt{1-x^2}} - \frac{y \frac{dy}{dx}}{\sqrt{1-y^2}} = a(1 - \frac{dy}{dx}) \]

### Step 3: Rearranging the equation
Now, we rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{-y \frac{dy}{dx}}{\sqrt{1-y^2}} + a \frac{dy}{dx} = a - \frac{-x}{\sqrt{1-x^2}} \]
\[ \left( a - \frac{y}{\sqrt{1-y^2}} \right) \frac{dy}{dx} = a + \frac{x}{\sqrt{1-x^2}} \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{a + \frac{x}{\sqrt{1-x^2}}}{a - \frac{y}{\sqrt{1-y^2}}} \]

### Final Expression
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{a + \frac{x}{\sqrt{1-x^2}}}{a - \frac{y}{\sqrt{1-y^2}}} \]

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