3Ne * Given: When the positive number $m$ takes different values, the equation $x^{2}+y^{2}-(4 m+2) x-2 m y+$ $4 m^{2}+4 m+1=0$ represents different circles. Find: The common tangent line (i.e., the line that is tangent to all these circles) of these circles.
Parse the circle equation as $(x-2 m-1)^{2}+(y-m)^{2}=m^{2}$. Therefore, the coordinates of the center of the circle are $(2 m+1, m)$. The radius is $m$. The line $y=k x+b$ is a common tangent to these circles if and only if for all positive numbers $m$, the following holds: $\frac{|k(2 m+1)-m+b|}{\sqrt{k^{2}+1}}=m$. Squaring both sides and simplifying, we get
$$
\left(3 k^{2}-4 k\right) m^{2}+2(k+b)(2 k-1) m+(k+b)^{2}=0,
$$

The common tangents, their equations are: $y=0$ (the $x$-axis) or $y=\frac{4}{3} x-\frac{4}{3}$.