One Circle Inside Another, Line Is Secant to Both

Steps

1. We choose the reference circle for the circular inversion. Its center is chosen at the intersection point of the given circle and the line.
2. We apply circular inversion to the given objects. A circle passing through the center of the reference circle is mapped to a line. The given circle is invariant under the inversion.
3. In the image, we solve the problem for two lines and a circle. The centers of the solution circles lie on angle bisectors.
4. We proceed using homothety. We consider four different homotheties. In each of them, the given circle is the image of one of the solution circles. The centers of these homotheties are always the points of tangency between the given circle and the solution circle. The lines that the solution circles should be tangent to are therefore mapped to parallel tangents of the circle.
5. The intersections of the lines and their images must be collinear with the centers of homothety. We draw the lines connecting them. The intersections of these lines with the given circle are thus the centers of the considered homotheties.
6. In these homotheties, the centers of the solution circles are the images of the center of the given circle. They must therefore lie on lines passing through both the center of the circle and the centers of the homotheties. The intersections of these lines with the angle bisectors determine the centers of the sought circles.
7. We have found four solutions to the inverted problem. These solutions are the images of the original problem's solutions under circular inversion.
8. We map the found circles back using circular inversion.
9. The problem has four solutions.

Steps

1. The solution circles will have internal tangency with the outer given circle and external tangency with the inner given circle. The centers of such circles lie on an ellipse. The foci of the ellipse are the centers of the given circles. To draw the ellipse, we need at least one point that lies on it. We draw a line through the centers of the given circles and find the midpoint of the segment whose endpoints are the intersections of this line with the given circles. This point is the center of one of the circles tangent to both given circles.
2. We construct the ellipse with foci at the centers of the given circles that passes through this point.
3. The centers of the circles tangent to the given line and the inner circle lie on a pair of parabolas. The focus of both parabolas is the center of the circle, and the directrices are lines parallel to the given line. The distance between the directrices and the given line equals the radius of the circle.
4. The centers of the solution circles lie at the intersections of the ellipse and the two parabolas.
5. The problem has four solutions.

Steps

2. Prepare a dilation to reduce the circle d centered at C to C itself. That is, we move the given line g to and from point C by the radius of circle d, and we also increase the radius of the given circle c by the radius of circle d.
3. Construct dilated circles and lines. In this way we transform the problem of two circles and a line into the problem of a circle, a line and a point.
4. First, we start with the case where we have added the radius of the given circle d to the line.
5. We will use the circle inversion. For convenience, we will choose the specified circle d as the circle over which to do the circular inversion. We do the circular inversion of the dilated line and circle.
6. Since the circle over which we are doing the circular inversion has its center at point C, we send point C to infinity. But since we are looking for something to touch point C and the inverted circle and line, this something must pass through infinity. The tangents of the circles created by inverting the dilated circle and line satisfy this criteria.
7. We invert back the tangents through the circle d. This gives two resulting circles for the dilated problem.
8. Since the dilation does not move the centers of the circles, the centers of these circles in the dilation are also the centers of the resultant circles in the original problem. Since the given line is supposed to be tangent to the resultant circles, the tangent point of these circles lies on the perpendicular to this line passing through their centers respectively. We thus form the tangent points T1 and T2.
9. Construct the resultant circles k1 and k2, which have centres at S1 and S2 and pass through the points T1 and T2.
10. Now we look at the case where we subtract the radius of the given circle d from the line.
11. • 17) Repeat steps 5 - 9.