The degree of the differential equation satisfying $\sqrt{1-x^2}+\sqrt{1-y^2}=a\left(x-y\right)$ is

The correct option is A

$1$

Explanation for the correct option:

To find the degree of the equation:

Given equation is $\sqrt{1-x^2}+\sqrt{1-y^2}=a\left(x-y\right)$

Differentiating with respect to $x$ we have

$$\begin{gathered} \Rightarrow \frac{1}{2}{\left(1-x^2\right)}^{\frac{1}{2}-1}\left(-2x\right)+\frac{1}{2}{\left(1-y^2\right)}^{\frac{1}{2}-1}\left(-2y\frac{dy}{dx}\right)=a\left(1-\frac{dy}{dx}\right)\left[\because \frac{d}{dy}{x}^n=n{x}^{n-1}\right] \\ \Rightarrow \frac{-x}{\sqrt{1-x^2}}-\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx}=a\left(1-\frac{dy}{dx}\right) \end{gathered}$$

Degree:

The power of the highest order derivative is $1$.

Therefore, the degree of the above differential equation is $1$

Hence, option (A) is correct .