The slope of the tangent to the curve $\mathrm{y}=\sqrt{4-{\mathrm{x}}^2}$ at the point where the ordinate and the abscissa are equal is

Here ,y> 0. Putting y = x in $\mathrm{y}=\sqrt{4-{\mathrm{x}}^2}\text{, we get }\mathrm{x}=\sqrt{2},-\sqrt{2}\text{. }$
So, the point is $(\sqrt{2},\sqrt{2})$
Differentiating y² + x² = 4 w.r.t. x, we get

$\begin{array}{r}2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{x}=0\text{ or }\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}\\ \therefore \text{ At }(\sqrt{2},\sqrt{2}),\frac{\mathrm{dy}}{\mathrm{dx}}=-1\end{array}$