7. If real numbers $x, y$ satisfy $(x-2)^{2}+y^{2}=3$, then the maximum value of $\frac{y}{x}$ is
7. $\sqrt{3}$.
$y^{2}=3-(x-2)^{2} \geqslant 0$, we get $2-\sqrt{3} \leqslant x \leqslant 2+\sqrt{3}$. The given equation can be transformed into $\left(\frac{y}{x}\right)^{2}=-\frac{1}{x^{2}}+\frac{4}{x}-1=-\left(\frac{1}{x}-2\right)^{2}+3$, when $\frac{1}{x}=2$, i.e., $x=\frac{1}{2} \in[2-\sqrt{3}, 2+\sqrt{3}]$, $\left(\frac{y}{x}\right)_{\max }^{2}=3$,
$$
\left(\frac{y}{x}\right)_{\max }=\sqrt{3} .
$$