Consider the family of circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{\lambda y}-8=0$ passing through two fixed points A and B. Then the distance between the points A and B is ______

${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{\lambda y}-8=0$
or $\left({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-8\right)-2\mathrm{\lambda y}=0$
which is of the form of$\mathrm{S}+\mathrm{\lambda L}=0$
All the circles pass through the point of intersection of the circle
${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-8=0$ and $\mathrm{y}=0$
Solving, we get
$\begin{array}{r}{\mathrm{x}}^2-2\mathrm{x}+8=0\text{ or }(\mathrm{x}-4)(\mathrm{x}+2)=0\\ \therefore \mathrm{A}\equiv (4,0)\text{ and }\mathrm{B}\equiv (-2,0)\end{array}$
So, the distance between A and 'B is 6'