Three Non-Intersecting Circles, Two Inside the Third

Number of solutions: 8

GeoGebra construction

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Steps

  1. We first find the centers of circles tangent to one pair of the given circles. The centers of the solution circles lie on a pair of hyperbolas. The foci of the hyperbolas are the centers of the given circles. To construct them, we need at least one point on each hyperbola. We draw a line passing through the centers of the given circles and find its intersections with the circles. The midpoints of the segments defined by these intersections are the vertices of the hyperbolas.
  2. We draw the hyperbolas with foci at the centers of the given circles and passing through the identified vertices.
  3. Next, we find all centers of circles tangent to another pair of given circles. This time, the centers of the solution circles lie on a pair of ellipses. The foci of the ellipses are the centers of the given circles. To construct them, we again need at least one point on each ellipse. We draw a line passing through the centers of the given circles and find its intersections with the circles. The midpoints of the segments defined by these intersections are the vertices of the ellipses.
  4. We draw the ellipses with foci at the centers of the given circles and passing through the identified vertices.
  5. The centers of the solution circles lie at the intersections of the constructed hyperbolas and ellipses. However, not every intersection of these curves is the center of a solution circle — only those for which the type of tangency (external or internal) to the common circle matches on both curves.
  6. The problem has eight solutions.

GeoGebra construction

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Steps

  1. To solve the problem, we use dilation in which the smallest circle is always reduced to a point. First, we look for the two solutions that are externally tangent to both of the smaller circles. If we imagine these two solutions and imagine shrinking the smallest given circle to its center, the solutions will expand. The remaining two circles must adjust their sizes to remain in contact with the solutions. The largest circle must be enlarged by the radius of the smallest circle, and the middle circle must be reduced by the radius of the given circle. In this way, we transform the problem into a circle–circle–point problem.
  2. We will solve the transformed problem using circle inversion. As the center of the inversion circle, we choose the center of the smallest circle. We can use the given circle itself as the inversion circle. The construction in the inverted image will be rather small, but we have access to zoom. We invert the two dilated circles.
  3. The center of the smallest circle is mapped to infinity in the inversion. Therefore, the solutions of the problem are mapped as common tangents of the transformed circles.
  4. We invert the found tangents back.
  5. We observe that the resulting circles are tangent to the dilated circles and pass through the center of the smallest given circle. They solve the transformed problem. However, after inverse dilation, only two of these circles correspond to solutions of the original problem.
  6. We dilate the selected circles by the radius of the smallest given circle. We have found the first two solutions to the problem.
  7. Next, we look for circles that are internally tangent to the smallest given circle and externally tangent to the middle circle. In this dilation, the middle circle will be enlarged and the largest one will be reduced. As mentioned above, the smallest circle is always reduced to a point.
  8. We invert the dilated circles.
  9. We find the common tangents of the transformed circles.
  10. We invert the tangents back.
  11. The resulting circles are solutions of the dilated problem.
  12. We dilate the found solutions back to obtain the solutions of the original problem.
  13. We now repeat the procedure for the case in which the two larger given circles are reduced in the dilation.
  14. We invert the dilated circles.
  15. We find the common tangents of the transformed circles.
  16. We invert the tangents back.
  17. The resulting circles are solutions of the dilated problem. After inverse dilation, however, only two of these circles correspond to solutions of the original problem.
  18. We dilate the selected solutions back to obtain the solutions of the original problem.
  19. Finally, we repeat the procedure for the case in which the two larger given circles are enlarged in the dilation.
  20. We invert the dilated circles.
  21. We find the common tangents of the transformed circles.
  22. We invert the tangents back.
  23. The resulting circles are solutions of the dilated problem. After inverse dilation, however, only two of these circles correspond to solutions of the original problem.
  24. We dilate the selected solutions back to obtain the solutions of the original problem.
  25. The problem has eight solutions.