This intersection point has equal power distance to all the circles in the pencil containing A and B.
The limiting points themselves can be found at this distance on either side of the intersection point, on the line through the two circle centers.
From this fact it is straightforward to construct the limiting points algebraically or by compass and straightedge.
[4] An explicit formula expressing the limiting points as the solution to a quadratic equation in the coordinates of the circle centers and their radii is given by Weisstein.
An inversion centered at one limiting point maps the other limiting point to the common center of the concentric circles.