Intersecting Circles, Line Through Intersection Point and Shared Interior

Steps

1. We choose the reference circle for the circular inversion. Its center is chosen as the common intersection point of all three given objects.
2. We apply circular inversion to the given objects. The circles are transformed into lines. The given line is invariant under the inversion.
3. In the image, we find circles tangent to three intersecting lines. Their centers lie on the angle bisectors of these lines.
4. The found circles are the images of the solution circles of the original problem. We map them back using circular inversion.
5. The problem has four solutions.

Steps

1. The centers of all circles that are externally tangent to one given circle and internally tangent to another lie on an ellipse. The foci of the ellipse are the centers of the given circles. The ellipse passes through the points where the two circles intersect.
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 points where the two circles intersect.
3. The centers of all circles tangent to a given circle and a line that intersects it lie on a pair of parabolas. The focus of each parabola is the center of the given circle. The directrix is a line parallel to the given line. The distance between the given line and its parallel is equal to the radius of the given circle.
4. The second parabola, on which lie the centers of circles tangent to the given circle and its secant, has the same focus as the first. Its directrix lies at the same distance, but on the opposite side relative to the given line.
5. The centers of circles tangent to all three given objects lie at the intersections of the ellipse, the hyperbola, and both parabolas. However, some of these intersections are not centers of solution circles. This occurs when one curve contains centers of circles that are internally tangent to the given circle, while another contains those that are externally tangent (or vice versa).
6. The problem has four solutions.