Three Circles With No Common Points

Steps

1. To solve the problem, we use dilation by continuously shrinking the smallest circle to a point. We begin by seeking the two solutions where the solution circles are externally tangent to both of the two smaller given circles. If we visualize these two solutions and imagine shrinking the smallest circle to its center, the solution circles will expand. The remaining two given circles must change size accordingly to maintain contact. The largest circle must be enlarged by the radius of the smallest circle, and the middle circle must be enlarged by the radius of the smallest circle as well. This transforms the original problem into a circle–circle–point configuration.
2. We solve the transformed problem using circular inversion. The center of inversion is the center of the smallest circle. The inverting circle can be the smallest given circle itself. While the resulting inverted construction may appear small, we can use zoom to handle it comfortably. We now invert the two dilated circles.
3. The center of the smallest circle maps to infinity under the inversion. Thus, the solutions of the problem correspond to the common tangents of the two inverted circles.
4. The found tangents are inverted back to their original form.
5. The resulting circles are tangent to the dilated circles and pass through the center of the smallest given circle. These are the solutions to the transformed problem. However, only two of them will correspond to solutions of the original problem after reversing the dilation.
6. We dilate the selected circles by the radius of the smallest given circle. This gives us the first two solutions of the original problem.
7. Next, we search for solution circles that are internally tangent to the smallest given circle and externally tangent to the middle circle. In this case, the middle circle is enlarged, and the largest circle is reduced in size. As stated earlier, the smallest circle is always reduced to a point. 8.–12. Repeat steps 2 through 6. This yields two more solutions. 13.–24. Repeat the same process for the remaining two combinations of enlarging and reducing the middle and largest given circles.
8. The problem has a total of eight solutions.