Given, curves are : $x^2+y^2=4 x$ ...(i)
and$x^2+y^2=8$ ...(ii)
$\because \frac{d y}{d x}$ is the slope of tangent drawn to curve
Slope of the tangent drawn to curve (i) is
$m_1=\frac{d y}{d x} \quad$ [from curve (i)]
On differentiating Eq. (i) w.r.t. $x$
$\frac{d}{d x}\left(x^2+y^2\right)=\frac{d}{d x}(4 x)$
$\Rightarrow \quad 2 x+2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}$
$\left.\therefore \quad m_1\right|_{a t(2,2)}=\frac{4-2 \times 2}{2 \times 2}=0$
On differentiating Eq. (ii) w.r.t. $x$,
$\frac{d}{d x}\left(x^2+y^2\right)=\frac{d}{d x}(8)$
$\begin{array}{ll}\Rightarrow & 2 x+2 y \frac{d y}{d x}=0 \\ \Rightarrow & \frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y}\end{array}$
$\left.m_2\right|_{(2,2)}=\frac{-2}{2}=-1$
$\therefore \quad \tan \theta=\left|\frac{m_2-m_1}{1+m_1 \cdot m_2}\right|$
$=\left|\frac{-1-0}{1+0}\right|=1$
$\Rightarrow \quad \theta=\tan ^{-1}(\mathrm{l})=\frac{\pi}{4}$
$\because \quad \theta=\sin ^{-1}(a) \quad$ [Given]
$\Rightarrow \quad \sin ^{-1}(a)=\frac{\pi}{4}$
$\Rightarrow \quad a=\sin \frac{\pi}{4} \Rightarrow a=\frac{1}{\sqrt{2}}$