Circles with External Tangency, Third Touches One and Intersects the Other

Number of solutions: 4

GeoGebra construction

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Steps

  1. We start by focusing on one pair of tangent circles. The centers of the solution circles lie on a straight line and a hyperbola. The line passes through the centers of the two given circles. The foci of the hyperbola are the centers of the given circles, and the hyperbola also passes through their point of tangency.
  2. Similarly, we find the loci of centers of circles tangent to the second pair of tangent circles. Again, the centers lie on a straight line and a hyperbola. The line passes through the centers of the given circles. The foci of the hyperbola are the centers of the circles, and the hyperbola passes through their point of tangency.
  3. The centers of the solution circles lie at the intersections of the identified hyperbolas and lines. However, not every such intersection corresponds to a valid solution — only those where the type of tangency (external or internal) to the shared circle is consistent on both curves.
  4. The problem has four solutions.

GeoGebra construction

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Steps

  1. Choose the circle of inversion, placing its center at the tangency point of the two circles.
  2. Apply circular inversion to the given circles. The circle passing through the center of the inversion circle transforms into two parallel lines.
  3. The center of the first solution in the inverted setting is found at the intersection of the axis of the strip between the parallel lines and the perpendicular to the parallels passing through the tangency point and the center of the transformed circle.
  4. The centers of the next two inverted solution circles lie on the same axis as the previous circle and also on a circle whose radius equals the sum of the radius of the transformed circle and half the distance between the parallel lines.
  5. The fourth solution appears as a tangent to the circle, parallel to both parallel lines.
  6. Map the four found solutions back using circular inversion.
  7. The problem has four solutions.