Let y=y(x) be a function of x satisfying $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$ where k is a constant and $y\left(\frac{1}{2}\right)$=$-\frac{1}{4}$, Then $\frac{dy}{dx}$ at x=$\frac{1}{2}$is equal to:

$$\begin{gathered} \mathrm{y}\sqrt{1-{\mathrm{x}}^2}+\mathrm{x}\sqrt{1-{\mathrm{y}}^2}=\mathrm{k} \\ \frac{\mathrm{dy}}{\mathrm{dx}}\sqrt{1-{\mathrm{x}}^2}-\frac{\mathrm{xy}}{\sqrt{1-{\mathrm{x}}^2}}+\sqrt{1-{\mathrm{y}}^2}-\frac{\mathrm{xy}}{\sqrt{1-{\mathrm{y}}^2}}\frac{\mathrm{dy}}{\mathrm{dx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{\sqrt{3}}{2}\right)-\frac{{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{-1}{4}}\right)}{{\displaystyle \frac{\sqrt{3}}{2}}}+\frac{\sqrt{15}}{4}-\frac{{\displaystyle \frac{1}{2}}\left({\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}{{\displaystyle \frac{\sqrt{15}}{4}}}\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{\sqrt{3}}{2}+\frac{1}{2\sqrt{15}}\right)=-\left(\frac{\sqrt{15}}{4}+\frac{1}{4\sqrt{3}}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{\sqrt{45}+1}{2\sqrt{15}}\right)=-\frac{\left(\sqrt{45}+1\right)}{4\sqrt{3}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\sqrt{5}}{2} \end{gathered}$$