Consider the family of circles $x^2+y^2-2x-2\lambda y-8=0$ passes through two fixed points A and B. then the distance between the points A and B is

$x^2+y^2-2x-2\lambda y-8=0$ 
 $(x^2+y^2-2x-8)-\lambda \left(2y\right)=0$
It is of the form  $S+\lambda L=0$ 
 All the circles passes through the point of intersection of the circle
 $x^2+y^2-2x-8=0$ and $y=0$

Solving we get $x^2-2x-8=0$ 
 $(x-4)(x+2)=0$$x=4$$x=-2$
$A=(4,0)$   $B=(-2,0)$  $\left|AB\right|=6$