Consider a circle bundle (I) of circles f.e. generated by two given circles c_{1} and c_{2}. Consider also an arbitrary circle c(C,r). Find the member circles d, d' of family (I) which are tangent to the given circle c.

The solution is found at once by observing that the radical axis of all pairs of circles (c_{3}, c) passes through the same point O, lying on the (common radical) axis of bundle (I). O is the radical center of the three circles c_{1}, c_{2} and c. To construct the tangent members draw tangents from O to c: OA_{1}, OA_{2}. The circle O(|OA_{1}|) belongs to bundle (II) consisting of all circles orthogonal to the circles of bundle (I). The circles d, d' have their centers on the intersection points B_{1}, B_{2} of CA_{1}, CA_{2} with the line joining the centers of c_{1}, c_{2}.
The construction can be applied to study Steiner chains of circles (see SteinerChain.html ).