11. Let real numbers $x, y$ satisfy the equation $(x+2)^{2}+y^{2}=1$. Then the maximum value of $\frac{y}{x}$ is $\qquad$
$=11 \cdot \frac{\sqrt{3}}{3}$.
According to the problem, to maximize the value of $\frac{y}{x}$, it is necessary to draw a tangent from the origin to the circle
$$
(x+2)^{2}+y^{2}=1
$$

such that the slope of the tangent is maximized, at which point the maximum value is $\frac{\sqrt{3}}{3}$.