One Circle Inside Another, Third Has Internal Tangency with Both

Steps

1. We choose the reference circle for the inversion. Its center is chosen at the point of tangency of the given circles.
2. We apply the circular inversion to the given objects. The circle passing through the point of tangency is transformed into a line.
3. The first solution circle is transformed into circle tangent to the images of the given circles. We find its centre using sets of points with given properties.
4. The second solution circle is transformed into tangent to a circle parallel to the images of the given circles.
5. We invert the images of the solution circles back using the same inversion.
6. The problem has two solutions.

Steps

1. The common points of tangency of the given circles will also be points of tangency with the solution circles. Therefore, the centers of the solution circles will lie on lines passing through the centers of the given circles and the points of tangency.
2. The common tangents of the touching circles will also be tangents of the solution circles.
3. We use two homotheties in which one of the given circles is mapped onto one of the solution circles. In these homotheties, the common tangent of two given circles is mapped onto a tangent to the third given circle that is parallel to it. The centers of the homotheties are the points of tangency between the third given circle and the solution circles.
4. Since the common tangent of two circles is mapped in each of the two homotheties onto a tangent to the third given circle that is parallel to it, the corresponding points of tangency are also mapped to one another. The centers of the homotheties lie on lines passing through these points of tangency. They lie at the intersections of these lines with the given circle.
5. We have now found all the points of tangency between the given circles and the solution circles. The centers of the solution circles lie on the perpendicular bisectors of the segments defined by these points of tangency.
6. The centers of the solution circles lie at the intersections of the found bisectors and the lines passing through the centers of the circles and the points of tangency.
7. The problem has two solutions.