1. Given the real number pair $(x, y)$ satisfies the equation $(x-2)^{2}+y^{2}=3$, let the minimum and maximum values of $\frac{y}{x}$ be $m$ and $n$ respectively. Then $m+n=$
$\begin{array}{l}\text { II. 1.0. } \\ \text { Let } y=t x \text {. Then }\left(1+t^{2}\right) x^{2}-4 x+1=0 \text {. } \\ \text { By } \Delta=(-4)^{2}-4\left(1+t^{2}\right) \geqslant 0 \\ \Rightarrow-\sqrt{3} \leqslant t \leqslant \sqrt{3} \\ \Rightarrow m=-\sqrt{3}, n=\sqrt{3} \\ \Rightarrow m+n=0 \text {. }\end{array}$