Non-Tangent Circles, Line Is Secant to Both

Steps

1. We choose the reference circle for the circular inversion. Its center is chosen at the intersection of one of the given circles with the line.
2. We apply circular inversion to the given objects. A circle passing through the center of inversion is mapped to a line. The given line is invariant under the inversion.
3. In the image, we solve the problem of finding circles tangent to two intersecting lines and a circle. The centers of the sought circles lie on the angle bisectors.
4. We proceed using homothety. We consider four different homotheties. In each of them, the given circle is the image of one of the solution circles. The centers of these homotheties are always the points of tangency between the given circle and the solution circle. The lines the solution circles are to be tangent to are thus mapped to parallel tangents to the circle.
5. The intersections of the lines and their images must be collinear with the centers of homothety. We draw the lines connecting them. The intersections of these lines with the given circle are therefore the centers of the considered homotheties.
6. In these homotheties, the centers of the solution circles are the images of the center of the given circle. They must therefore lie on lines passing through both the center of the circle and the centers of homothety. The intersections of these lines with the angle bisectors give the desired centers.
7. We have found four solutions to the inverted problem. These are the images of the original problem’s solutions under circular inversion.
8. We map the found circles back using circular inversion.
9. The problem has four solutions.

Steps

1. The centers of all circles tangent to a given circle and its secant lie on a pair of parabolas. Their focus is the center of the given circle. The directrices are lines parallel to the given line. The distance between the given line and the directrix equals the radius of the given circle.
2. A second pair of parabolas forms the locus of centers of circles tangent to the second given circle and the line. The focus of these parabolas is the center of the second circle, and the distance between the directrices and the given line again equals the radius of the given circle.
3. The centers of the solution circles lie at the intersections of these parabolas.
4. The problem has four solutions.