Intersecting Circles, Line Passing Through Common Point as Secant to One and Tangent to the Other

Steps

1. Choose the circle of inversion such that its center lies at the intersection point of the two circles, which also lies on the given line.
2. Apply circular inversion to the given objects. The given line remains unchanged (self-inverse), and the two circles are transformed into straight lines.
3. Solve the problem for three lines, two of which are parallel. The centers of the solution circles lie at the intersections of the angle bisectors formed by the lines.
4. In the inverted configuration, two solutions are found. These correspond to the images of the solution circles of the original problem.
5. Invert the found circles back to obtain the actual solutions.
6. The problem has two 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 in cases where 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).
5. The problem has two solutions.