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

Given that
$\sqrt{1-x^2}+\sqrt{1-y^2}=b(x-y)\dots \left(1\right)$
Put $x=\sin \alpha ,y=\sin \beta$ in equation $\left(1\right)$
$\alpha ={\sin }^{-1}x,\beta ={\sin }^{-1}y$
$\cos \alpha +\cos \beta =b(\sin \alpha -\sin \beta )$
Use identities
$2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)=b\cdot 2\cos \frac{\alpha +\beta }{2}\cdot \sin \frac{\alpha -\beta }{2}$
$\Rightarrow \alpha -\beta =2{\cot }^{-1}b$
$\Rightarrow {\sin }^{-1}x-{\cos }^{-1}y=2{\cot }^{-1}b$
Differentiating w.r.t $x$, we get
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}\frac{dy}{dx}=0$
Degree of above differential equation is $1$.