The differential equation satisfied by the curves 1−x2+1−y2=m(x−y) where m is a parameter, is

x=sin⁡θ,y=sin⁡ϕ,dx=cos⁡θdθ,dy=cos⁡ϕdϕ⇒m=cos⁡θ+cos⁡ϕsin⁡θ−sin⁡ϕ=cot⁡θ−ϕ2⇒θ−ϕ=2cot−1⁡mDifferentiating w.r. t. , θ,1−dϕdθ=0⇒dθ=dϕdydx=cos⁡ϕdϕcos⁡θdθ=cos⁡ϕcos⁡θ=1−y21−x2⇒1−x2y1=1−y2

The differential equation satisfied by the curves 1−x2+1−y2=m(x−y) where m is a parameter, is

x=sin⁡θ,y=sin⁡ϕ,dx=cos⁡θdθ,dy=cos⁡ϕdϕ⇒m=cos⁡θ+cos⁡ϕsin⁡θ−sin⁡ϕ=cot⁡θ−ϕ2⇒θ−ϕ=2cot−1⁡mDifferentiating w.r. t. , θ,1−dϕdθ=0⇒dθ=dϕdydx=cos⁡ϕdϕcos⁡θdθ=cos⁡ϕcos⁡θ=1−y21−x2⇒1−x2y1=1−y2

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The differential equation satisfied by the curves $\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = m (x - y)$ where $m$ is a parameter, is

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$\sqrt{1 - x^{2}} y_{1} + \sqrt{1 - y^{2}} = 0$

$\sqrt{1 - x^{2}} y_{1} - \sqrt{1 - y^{2}} = 0$

$\sqrt{1 - y^{2}} y_{1} + \sqrt{1 - x^{2}} = 0$

$\sqrt{1 - y^{2}} y_{1} - \sqrt{1 - x^{2}} = 0$

answer is C.

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Detailed Solution

$x = \sin θ, y = \sin ϕ, d x = \cos θ d θ, d y = \cos ϕ d ϕ$

$\begin{array}{l} \Rightarrow m = \frac{\cos θ + \cos ϕ}{\sin θ - \sin ϕ} = \cot (\frac{θ - ϕ}{2}) \\ \Rightarrow θ - ϕ = 2 \cot^{- 1} m \end{array}$

Differentiating w.r. t. , $θ, 1 - \frac{d ϕ}{d θ} = 0 \Rightarrow d θ = d ϕ$

$\begin{array}{l} \frac{d y}{d x} = \frac{\cos ϕ d ϕ}{\cos θ d θ} = \frac{\cos ϕ}{\cos θ} = \frac{\sqrt{1 - y^{2}}}{\sqrt{1 - x^{2}}} \\ \Rightarrow \sqrt{1 - x^{2}} y_{1} = \sqrt{1 - y^{2}} \end{array}$