If ${\sin }^{-1}\left(\mathrm{a}\right)$ is the acute angle between the curves ${\mathrm{x}}^2+{\mathrm{y}}^2=4\mathrm{x}$ and ${\mathrm{x}}^2+{\mathrm{y}}^2=8$ at the point (2,2), then a =

$\begin{array}{rcl}{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}& =& 0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{(2\mathrm{x}-4)}{2\mathrm{y}}\\ {\mathrm{m}}_1& =& {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(2, 2)}=0\\ {\mathrm{x}}^2+{\mathrm{y}}^2-8& =& 0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}}{\mathrm{y}}\\ & \Rightarrow & {\mathrm{m}}_2={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(2, 2)}=\frac{-2}{2}=-1\\ \mathrm{tan\theta }& =& \left|\frac{0-(-1)}{1+0}\right|=1 \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\\ {\sin }^{-1}\mathrm{\theta }& =& \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \Rightarrow \mathrm{\theta }=\frac{1}{\sqrt{2}}=a\end{array}$