Intersecting Circles, Line Passes Through Common Intersection Point

Steps

1. Choose the circle of inversion so that its center lies at the common intersection point of the two circles and the line.
2. Apply circular inversion to the given objects. The two circles become lines, and the line remains fixed (it is self-inverse).
3. Now solve the problem for three intersecting lines. The centers of the solution circles lie on the angle bisectors formed by the lines.
4. This gives four solutions to the problem with three lines. In the chosen inversion, these are the images of the desired solution circles.
5. Invert the found circles back to obtain the solutions of the original problem.
6. The problem has four solutions.

Steps

1. The centers of all circles that are externally tangent to one given circle and internally tangent to the other lie on an ellipse. The foci of the ellipse are the centers of the given circles. The ellipse passes through the intersection points of the two circles.
2. The centers of all circles that are either externally tangent to both given circles or internally tangent to both lie on a hyperbola. The foci of the hyperbola are the centers of the given circles. The hyperbola passes through the intersection points of the two circles.
3. The centers of all circles tangent to a given circle and a line lie on a pair of parabolas. The focus of both parabolas is the center of the circle. The directrices are parallel to the given line. The distance between the directrices and the given line equals the radius of the circle.
4. The centers of circles tangent to all three given objects lie at the intersections of the ellipse, the hyperbola, and the parabolas. However, some of these intersections are not centers of solution circles – this occurs when one curve contains centers of circles internally tangent to the given circle, and another contains centers of circles externally tangent to it (or vice versa).
5. The problem has four solutions.