Internal Tangency, Third Tangent to Inner and Crossing Outer Circle
Number of solutions: 4
GeoGebra construction
Steps
- We first focus on the pair of circles with external tangency. The centers of the solution circles lie on a straight line and a hyperbola. The line passes through the centers of both circles. The centers of the given circles are also the foci of the hyperbola. The hyperbola passes through the point of tangency of the two circles.
- Next, we determine the loci of centers of circles tangent to the pair of circles with internal tangency. The centers of the solution circles lie on a straight line and an ellipse. The line passes through the centers of both circles. The foci of the ellipse are the centers of the given circles, and the ellipse passes through the point of tangency of the two circles.
- The centers of the solution circles lie at the intersections of the identified lines, ellipse, and hyperbola. However, not every intersection of these curves is a center of a solution circle — only those for which the type of tangency (internal or external) to the shared circle matches on both curves.
- The problem has four solutions.
GeoGebra construction
Steps
- Choose the circle of inversion, placing its center at the tangency point of the two circles.
- Apply circular inversion to the given circles. The circle passing through the center of the inversion circle transforms into two parallel lines.
- 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.
- 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.
- The fourth solution appears as a tangent to the circle, parallel to both parallel lines.
- Map the four found solutions back using circular inversion.
- The problem has four solutions.