Two Internally Tangent Circles, Third Inside the Outer One

Number of solutions: 6

GeoGebra construction

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Steps

  1. First, we find the centers of the 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 to know at least one point on each hyperbola. We draw the line passing through the centers of the given circles and find its intersections with the circles. The midpoints of the segments between these intersections are the vertices of the hyperbolas.
  2. We construct the hyperbolas with foci at the centers of the given circles and passing through the found vertices.
  3. Next, we find all centers of circles tangent to another pair of the given circles. This time, the centers of the solution circles lie on a line and an ellipse. The line passes through the centers of the two circles. The foci of the ellipse are the centers of the given circles, and the ellipse passes through their common point of tangency.
  4. The centers of the solution circles lie at the intersections of the constructed hyperbolas, the ellipse, and the line. However, not every intersection of these curves is a center of a solution circle – only those for which the type of tangency (external or internal) to the common circle matches on both curves.
  5. The problem has six solutions.

GeoGebra construction

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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 the circular inversion to the given objects. The circle passing through the point of tangency is transformed into a line.
  3. The first four solution circles are transformed into circles tangent to the images of the given circles. We find them using sets of points with given properties.
  4. The remaining two solution circles are transformed into tangents to a circle parallel to the images of the given circles.
  5. We invert the images of the solution circles back using the same inversion.
  6. The problem has six solutions.