CIRCLE • CIRCLE • LINE

Two Separate Circles with Common Tangent on Same Side

Number of solutions: 4

GeoGebra construction

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Steps

  1. Choose the circle of inversion with its center at the tangency point of the given circle and the line.
  2. Apply circular inversion to the given objects. The circle passing through the center of the inversion transforms into a straight line. The given line is self-inverse.
  3. Find the circles tangent to the transformed objects. To locate their centers, use loci of points defined by specific geometric properties.
  4. The solution that passes through the tangency point (the center of the inversion circle) is transformed into a tangent to the image of the circle, parallel to the given line.
  5. Invert the found images of the solution circles back to obtain the original solutions.
  6. The problem has four solutions.

GeoGebra construction

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Steps

  1. The centers of all circles that are tangent to the first given circle and the given line at their point of tangency lie on the perpendicular to the line passing through the common tangency point.
  2. The centers of all circles that are tangent to the first given circle and the line at a point other than their point of tangency lie on a parabola. The focus of this parabola is the center of the circle, and the directrix is a line parallel to the given line. The distance between the circle and the directrix is equal to the radius of the circle.
  3. The centers of all circles that are tangent to the second given circle and the given line at their point of tangency lie on the perpendicular to the line passing through their common tangency point.
  4. The centers of all circles that are tangent to the second given circle and the line at a point other than their point of tangency lie on a parabola. The focus is the center of the circle, and the directrix is a line parallel to the given line. The distance between the circle and the directrix is equal to the radius of the circle.
  5. Find the intersections of the respective perpendiculars and parabolas. These points are the centers of the solution circles.
  6. The problem has four solutions.