Circles with Internal Tangency and Secant Line to the Outer Circle

Steps

1. The problem will be solved in three phases. In each phase, we apply a dilation to transform the problem into a configuration that can be solved using circular inversion. First, we look for a solution circle that touches both given circles at their point of tangency and lies inside the larger circle. During the dilation that contracts the smaller circle to a point, the line is shifted toward the center of the smaller circle, and the larger circle is reduced by the radius of the smaller one. This transforms the original problem into one involving a point, a line, and a circle, where the point lies on the circle.
2. We solve the transformed problem using circular inversion. As the inverting circle, we choose the smaller of the two given circles, although any circle centered at the center of the smaller one would suffice. We apply the inversion to the dilated line and the larger circle. The center of the smaller circle is mapped to infinity.
3. We find the parallel tangents from the circle (the image of the line) to the line (the image of the larger circle). These two parallels are the images of the solutions to the dilated problem. However, only one of them corresponds to the solution of the original problem.
4. From the two parallels, we select the one that corresponds to the desired solution and apply the inverse transformation to it.
5. The resulting circle is then dilated by the radius of the smaller given circle, yielding the first solution.
6. – 10. In steps 6 to 10, we find a solution circle that touches both given circles at their common point of tangency and lies outside both of them. The steps mirror those in steps 1 to 5. The only difference is that in the initial dilation, the line is shifted in the opposite direction.
7. We now find circles that are internally tangent to the larger circle and externally tangent to the smaller one. We begin by applying a dilation. The line is shifted away from the given circles, and the larger circle is enlarged by the radius of the smaller one.
8. The dilated objects are then mapped using circular inversion. As the inverting circle, we again use the smaller of the given circles.
9. The solutions in the inverted setting correspond to the common tangents of the two inverted circles.
10. These tangents are transformed back using the inverse inversion.
11. The resulting circles are then dilated by the radius of the smaller given circle. This yields the remaining two solutions.
12. The problem has four solutions in total.

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.