Circles with Internal Tangency, Line Is Secant to Both

Steps

1. We choose the reference circle for the circular inversion. Its center is chosen at the point of tangency of the two given circles.
2. We apply circular inversion to the given objects. The circles are transformed into a pair of parallel lines, and the line is transformed into a circle.
3. We look for circles tangent to the transformed parallel lines and the circle. The centers of these circles are found using loci of points with given properties. Four circles satisfy the conditions.
4. Two of the solution circles to the original problem pass through the point of tangency of the given circles. In the inversion, these two solutions appear as a pair of tangents to the circle that is the image of the original line. The tangents are parallel to the lines that are images of the original circles.
5. We map the found images of the solution circles back using circular inversion.
6. The problem has six solutions.

Steps

1. All centers of circles tangent to both given circles at their common point of tangency lie on the line passing through the centers of the two given circles and their common tangent point.
2. The common tangent of the two given circles is also tangent to two of the solution circles.
3. The centers of these two solution circles lie on the angle bisectors defined by the common tangent and the given line.
4. The first two centers of the solution circles are found as the intersections of the line through given centers with the respective angle bisectors.
5. The centers of all circles tangent to both given circles outside their common point lie on an ellipse. The foci of the ellipse are the centers of the given circles. The ellipse passes through their common point of tangency.
6. The centers of all circles tangent to the given line and one of the circles lie on a pair of parabolas. The common focus of the parabolas is the center of that circle. The directrices are lines parallel to the given line, at a distance equal to the radius of the circle.
7. The centers of the remaining four solution circles lie at the intersections of the ellipse and the two parabolas.
8. The problem has a total of six solutions.