11. The trajectory equation of a moving point $z_{1}$ in the complex plane is $\left|z_{1}-z_{0}\right|=\left|z_{1}\right|, z_{0}$ is a fixed point, $z_{0} \neq 0$. Another moving point $z$ satisfies $z_{1} z=-1$, find the trajectory of point $z$, and indicate its shape and position on the complex plane.
Notice that the moving point $z_{1}$ is equidistant from the fixed point $z_{0}$ and $O$, then the trajectory of the moving point $z_{1}$ is the perpendicular bisector of the line segment $O z_{0}$. Let $z_{0}=a+b \mathrm{i}(a, b \in \mathbf{R})$, so the point $z_{1}=u+v \mathrm{i}(u, v \in \mathbf{R})$ lies on the line $2(a u+b v)=a^{2}+b^{2} \quad(*)$. Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$, then by $z_{1}=-\frac{1}{z}=-\frac{1}{x+y \mathrm{i}}=-\frac{x-y \mathrm{i}}{x^{2}+y^{2}} \Rightarrow\left\{\begin{array}{l}u=-\frac{x}{x^{2}+y^{2}}, \\ v=\frac{y}{x^{2}+y^{2}} .\end{array}\right.$
Substituting into $(*)$ we get $2(-a x+b y)=\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \Rightarrow x^{2}+y^{2}+\frac{2 a x}{a^{2}+b^{2}}-\frac{2 b y}{a^{2}+b^{2}}=0$. Also, $z_{1} z=-1 \Rightarrow z \neq 0$, so the trajectory of the point $z$ is a circle in the complex plane with center $\left(-\frac{a}{a^{2}+b^{2}}, \frac{b}{a^{2}+b^{2}}\right)$ and radius $r=\frac{1}{\sqrt{a^{2}+b^{2}}}$, excluding the origin.