Circles with Internal Tangency and Secant Line to the Outer Circle

Number of solutions: 4

GeoGebra construction

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Steps

  1. We solve the problem in three phases. Each time, we modify the original problem using dilation to transform it into a form that can be solved with circle inversion. First, we look for a solution that is tangent to both circles at their common point of tangency and lies inside the larger circle. When we dilate the smaller circle to a point, the given line shifts toward the center of the smaller circle, and the larger circle shrinks by the radius of the smaller one. This transforms the original problem into a point–line–circle problem, where the point lies on the circle.
  2. We solve the modified problem using circle inversion. We choose the smaller of the two original circles as the inversion circle, though any circle centered at the center of the smaller circle would work. We invert the dilated images of the line and the larger circle. The center of the smaller circle maps to infinity.
  3. We find the lines parallel to the image of the line that are tangent to the image of the circle. These two parallels are the images of the solutions to the transformed problem. Only one of them corresponds to a solution of the original problem.
  4. We select the correct line and invert it back.
  5. We dilate the resulting circle by the radius of the smaller circle. This gives us the first solution.
  6. Next, we look for a circle that is tangent to both original circles at their common tangency point and lies outside both circles. The steps are analogous to Steps 1 through 5, with the only difference being that in the initial dilation, we shift the line in the opposite direction.
  7. We invert the dilated objects.
  8. We find the lines tangent to the image of the circle and parallel to the image of the line.
  9. We select the correct tangent and invert it back.
  10. We dilate the resulting circle by the radius of the smaller circle. This gives us the second solution.
  11. Now we look for circles that are internally tangent to the larger circle and externally tangent to the smaller circle. We first perform a dilation. The line moves away from the two circles, and the larger of the two circles is enlarged by the radius of the smaller one.
  12. We invert the dilated objects. Again, we use the smaller of the original circles as the inversion circle.
  13. The solutions of the problem appear in the inversion as common tangents of the two circles.
  14. We invert the tangents back.
  15. We dilate the resulting circles by the radius of the smaller given circle. This gives us the remaining two solutions.
  16. The problem has four solutions.

GeoGebra construction

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Steps

  1. First, construct the loci of centers of circles tangent to both given circles. The set of centers of all circles tangent to both circles at their common point of tangency is the line passing through the centers of the given circles.
  2. The set of centers of circles that are externally tangent to the smaller circle and internally tangent to the larger circle is an ellipse with foci at the centers of these circles. The ellipse passes through the common point of tangency.
  3. Now, construct the loci of centers of circles tangent to the smaller circle and the given line. The centers of circles internally tangent to the given circle and also tangent to the line lie on a parabola. The focus of the parabola is the center of the circle. The directrix is a line parallel to the given line, shifted by the radius of the circle. This shift is based on a dilation in which the given circle is reduced to a point.
  4. The centers of circles externally tangent to the given circle and also tangent to the line lie on another parabola. The focus is again the center of the circle, and the directrix is a line parallel to the given line, shifted in the opposite half-plane compared to the previous step.
  5. We now find the centers of the solution circles as the intersections of the previously constructed loci. The first center lies at the intersection of the line and one of the parabolas. Of the two possible intersection points, we select the one where both loci imply an internal tangency with the smaller circle.
  6. The second center lies at the intersection of the line and the other parabola. Here, we select the intersection point where both loci imply an external tangency with the smaller circle.
  7. The remaining two centers are found as the intersections of the parabola corresponding to external tangency with the smaller circle and the ellipse.
  8. The problem has four solutions in total.