Two Circles Inside Each Other, Line Passes Through Their Annulus

Steps

1. The centers of circles tangent to both given circles lie on a pair of ellipses. The foci of the ellipses are the centers of the given circles. To draw them, we need at least one point on each ellipse. We draw a line passing through the centers of the given circles and find the midpoints of the segments whose endpoints are the intersections of this line with the given circles. These points are the centers of circles tangent to both given circles.
2. We construct ellipses with foci at the centers of the given circles that pass through these points.
3. The centers of circles tangent to the given line and the inner circle lie on the perpendicular to the given line passing through their point of tangency, and on a parabola. The focus of the parabola is the center of the given circle; the directrix is parallel to the given line, and its distance from the given line equals the radius of the circle.
4. The centers of the solution circles lie at the intersections of the ellipses, the perpendicular line, and the parabola.
5. The problem has four solutions.

Steps

2. Prepare a dilation to reduce the circle k₁ centered at C to C itself. That is, we will move the given line p₁ to and from point C by the radius of circle k₁, and we will also increase and decrease the radius of the given circle k₂ by the radius of circle k₁.
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 subtracted the radius of the given circle k₁ for both the circle and the line.
5. We will use the circle inversion. For convenience, we will choose the specified circle k₁ 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 meet these criteria.
7. We invert back the tangents through the circle k₁. 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
9. Since the given line is supposed to be the tangent of 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.
10. Construct the resultant circles k1 and k2, which have centres at S1 and S2 and pass through the points T1 and T2.
11. We now look at the case where for a circle and a line we add the radius k₁ of the given circle.
12. • 17) Repeat steps 5 - 10.