Two Externally Tangent Circles, Third Intersects 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 circular inversion to the given circles. The circles that pass through the center of the reference circle are transformed into lines.
3. We find circles tangent to the images of the given circles. Their centers are found using sets of points with given properties. The four resulting circles are the images of four solution circles.
4. The remaining two solutions appear as tangents to the image of a circle that is parallel to the lines representing the images of the remaining two given circles.
5. We invert the obtained circles and tangents back using the same inversion.
6. The problem has six solutions.

Steps

1. We first focus on one pair of the given circles. The centers of circles that touch both given circles either externally or internally lie on a hyperbola. The foci of this hyperbola are the centers of the given circles, and the hyperbola passes through their common point of tangency.
2. The centers of circles that touch both given circles at their common tangency point lie on the line passing through their centers.
3. Next, we find the loci of centers of circles touching the second pair of circles. The centers of circles that touch both of these circles either externally or internally again lie on a hyperbola. The foci of this hyperbola are the centers of the given circles, and it passes through their points of intersection.
4. The centers of circles that touch one of the circles externally and the other internally lie on an ellipse. The foci of this ellipse are again the centers of the given circles, and the ellipse passes through their intersection points.
5. The centers of the solution circles lie at the intersections of the constructed hyperbolas, the line, and the ellipse. However, not every intersection corresponds to a valid solution — only those where the type of tangency (external or internal) to the shared circle matches across all involved curves.
6. The problem has six solutions.