Intersecting Circles, Line is a Secant of One

Steps

We choose the reference circle for the circular inversion. Its center is placed at the intersection point of the given circles. 1. We choose the reference circle for the circular inversion. Its center is placed at the intersection point of the given circles. 2. We apply circular inversion to the given objects. The circles are transformed into intersecting lines, and the given line is transformed into a circle. 3. Now we look for circles tangent to the transformed objects – a circle and two lines. To find them, we use homothety, specifically four homotheties in which the given circle is the image of each solution circle. The centers of these homotheties are the points of tangency between the given circle and the solution circles. The given lines are transformed into parallel tangents to the circle. 4. The intersection point of the original lines is transformed into the intersection points of the tangents. We draw lines connecting these points. The intersections of these lines with the circle are the centers of the homotheties – i.e., the tangency points between the given circle and the solution circles. 5. We draw lines passing through the found points of tangency and the center of the given circle. These lines represent, in the homotheties, the images of the center of the given circle mapped to the centers of the solution circles. 6. Since the solution circles must be tangent to the given lines, their centers lie on the angle bisectors of the two lines. We find the centers of the solution circles as intersections of these bisectors with the previously constructed lines. 7. We have found four solutions to the problem in the inverted image. 8. These are the images of the original problem’s solutions under circular inversion. 9. The problem has four solutions.

Steps

1. The centers of all circles that are externally tangent to one of the given circles and internally tangent to the other lie on an ellipse. The foci of the ellipse are the centers of the given circles, and it 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 again the centers of the given circles, and it also passes through their points of intersection.
3. The centers of all circles tangent to the given line and to the circle it intersects lie on a pair of parabolas. The focus of each parabola is the center of the circle. The directrices are lines parallel to the given line, and their distance from it is equal to the radius of the given circle.
4. The centers of the solution circles—tangent to all three given objects—are found at the intersections of the ellipse and hyperbola with the parabolas. Some of these intersection points do not correspond to valid solutions. This occurs when one curve contains centers of circles with internal tangency, and the other with external tangency (or vice versa).
5. The problem has four solutions.