CIRCLE • CIRCLE • CIRCLE

Two Intersecting Circles, Third Circle Outside Both Of Them

Number of solutions: 4

GeoGebra construction

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Steps

  1. Three circles k1, k2 and k3 are given. Two of them intersect.
  2. We will solve the problem using dilation. Consider a solution that is tangent fron the outside with circles k1 and k2 and tangent from inside to circle k3. If this solution in the dilation is reduced to pass through the center of S3, both circles k1 and k2 will increase in radius by the radius of circle k3. The same is true for a solution that has inner contact with circles k1 and k2 and outer contact with circle k3.
  3. By dilation, we changed the kkk problem to the Bkk problem, where the point is the point S3 and the circles are the images of the circles k1 and k2. We will solve this problem using circular inversion. We will use k3 with center S3 as the base circle (we can use any circle with center S3, so why not use k3).
  4. The image of the dilated solutions are the common tangents of the circles k1'' and "k2''.
  5. We display the found tangents back in the circular inversion. The obtained circles are the dilated images of the searched solutions. The centers S4 and S7 are already the centers of the searched solutions.
  6. We enlarge and reduce the images of circles k4 and k7, respectively, in the dilation by the radius of circle k3. This gives the first two solutions to the problem.
  7. We have found two solutions to the problem. Now we repeat the procedure to find the remaining two solutions, one of which has an outer tangent to all the given circles and the other has an inner tangent to all of them.
  8. Whereas in the previous case the circles k1 and k2 were both increasing in size, now they will be decreasing in size.
  9. The dilated images of circles k1 and k2 are shown in circular inversion with the main circle k3.
  10. Find the common tangents of the circles shown.
  11. Display the found tangents back in the circular inversion. Their images are also the dilated images of the solutions we are looking for.
  12. Reduce the circle k5 ' by dilating it by the radius of the circle k3, and enlarge the circle k6 '. This gives us the remaining two solutions.
  13. In total, we have found four solutions to the problem.