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

$1 / \sqrt{2}$

The equations of the curves are

$x^2+y^2-4 x=0$     ....(i)       and,        $x^2+y^2-8=0$      ......(ii)

Now,

$x^2+y^2-4 x=0$

$\Rightarrow 2 x+2 y \frac{d y}{d x}-4=0 \Rightarrow \frac{d y}{d x}=\frac{2-x}{y} \Rightarrow m_1=\left(\frac{d y}{d x}\right)_{(2,2)}=0$

and, $x^2+y^2-8=0$

$\Rightarrow 2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y} \Rightarrow m_2=\left(\frac{d y}{d x}\right)_{(2,2)}=-1$

∴  $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$\Rightarrow \tan \theta=\frac{0-(-1)}{1+0 \times-1}=1 \Rightarrow \theta=\frac{\pi}{4}$

Clearly, $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$