Given curves are $x^2+y^2=4$ and $y^2=3 x$ Put the value of $y^2$.
$$
\begin{aligned}
& \Rightarrow x^2+3 x-4=0 \\
& \Rightarrow x^2+4 x-x-4=0 \\
& x(x+4)-1(x+4)=0 \\
& (x-1)(x+4)=0 \\
& x-1=0, x+4=0 \\
& x=1, x=-4
\end{aligned}
$$
Put in $y^2=3 \mathrm{x}$, then, $y^2=3, y= \pm \sqrt{3}$. $(x, y) \rightarrow(1, \sqrt{3}) \&(1, \sqrt{3})$
Now, Differentiate both curves w.r.t $x$
$$
\begin{aligned}
& \Rightarrow x^2+y^2=4 \\
& 2 x+2 y \cdot \frac{d y}{d x}=0 \\
& y \frac{d y}{d x}=-x \\
& \frac{d y}{d x}=-\frac{x}{y} \\
& \Rightarrow y=\sqrt{3 x} \\
& \frac{d y}{d x}=\frac{\sqrt{3}}{2 \sqrt{x}}
\end{aligned}
$$
Put the point $(1, \sqrt{3})$ in their differentiation.
$$
\begin{aligned}
& m_1=\frac{d y}{d x}=\frac{-1}{\sqrt{3}}, m_2=\frac{\sqrt{3}}{2} \\
& \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{2}}{1-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2}}\right|=\left|\frac{\frac{-2-3}{2 \sqrt{3}}}{\frac{1}{2}}\right| \\
& \tan \theta=\left|\frac{-5}{\sqrt{3}}\right|=\frac{5}{\sqrt{3}}
\end{aligned}
$$
So, $\frac{5}{\sqrt{3}}$ is the correct option.