10. Solution: (1) The vertex of the parabola is $M(1,0)$, and the center of the circle $C(-1,0)$. Since the circle $C$ is tangent to the $y$-axis, the equation of the circle $C$ is $(x+1)^{2}+y^{2}=1$.
(2) The directrix of the parabola is $x=0$. Let $N\left(x_{0}, y_{0}\right)$, then $y_{0}^{2}=4\left(x_{0}-1\right)$. The two tangent lines intersect the $y$-axis at $A(0, a)$ and $B(0, b)$. The equation of $AN$ is $y=\frac{y_{0}-a}{x_{0}} x+a$, which can be rewritten as $\left(y_{0}-a\right) x-x_{0} y+a x_{0}=0$. Since the line is tangent to the circle, we have $\frac{\left|a-y_{0}+a x_{0}\right|}{\sqrt{\left(y_{0}-a\right)^{2}+x_{0}^{2}}}=1$, which simplifies to $\left(x_{0}+2\right) a^{2}-2 y_{0} a-x_{0}=0$.
Similarly, we get $\left(x_{0}+2\right) b^{2}-2 y_{0} b-x_{0}=0$.
Therefore, $a$ and $b$ are the roots of the equation $\left(x_{0}+2\right) t^{2}-2 y_{0} t-x_{0}=0$, so $|A B|=|a-b|=\sqrt{(a+b)^{2}-4 a b}$. Since $a+b=\frac{2 y_{0}}{x_{0}+2}$ and $a b=-\frac{x_{0}}{x_{0}+2}$,
we have $|A B|=\sqrt{\left(\frac{2 y_{0}}{x_{0}+2}\right)^{2}+4 \frac{x_{0}}{x_{0}+2}}=\sqrt[2]{-12\left(\frac{1}{x_{0}+2}-\frac{1}{12}\right)^{2}+\frac{13}{-12}}$.
Since $x_{0} \geqslant 1$, then $\frac{1}{x_{0}+2} \in\left(0, \frac{1}{3}\right]$. When $\frac{1}{x_{0}+2}=\frac{1}{12}$, $|A B|_{\text {max }}=\frac{\sqrt{39}}{3}$. When $\frac{1}{x_{0}+2}=\frac{1}{3}$,
$$
|A B|_{\text {min }}=\frac{2 \sqrt{3}}{3} \text {. }
$$
10. Given the parabola $y^{2}=4(x-1), M$ is its vertex.
(1) The center of circle $C$ is symmetric to the vertex $M$ of the parabola with respect to the $y$-axis, and circle $C$ is tangent to the $y$-axis. Find the equation of circle $C$;
(2) Through any point $N$ on the parabola, draw two tangent lines to circle $C$. These two tangent lines intersect the directrix of the parabola at points $A, B$. Find the range of $|A B|$.