Intersecting Circles, Line is a Secant to One and a Tangent to the Other

Steps

1. We choose the reference circle for the circular inversion. Its center is placed at the intersection of the given circles.
2. We apply circular inversion to the given objects. The circles are transformed into intersecting lines, and the given line becomes a circle.
3. We now look for circles tangent to the transformed objects – the circle and the two lines. To find them, we use homothety: specifically, four homotheties in which the given circle is the image of each solution circle. The centers of these homotheties are the tangency points between the circles. The original lines are transformed into parallel tangents to the circle.
4. The intersection of the original lines becomes the intersection points of the tangents. We draw lines connecting these points. Their intersections with the circle are the centers of the homotheties – that is, the points of tangency between the given circle and the solution circles.
5. We draw lines through the tangency points and the center of the given circle. These lines map the center of the given circle to the centers of the solution circles under the homotheties.
6. Since the solution circles must be tangent to the given lines, their centers lie on the angle bisectors defined by these lines. We find the centers as the intersections of the bisectors with the previously drawn lines.
7. We have found four solutions in the inverted image.
8. Two additional solutions are circles tangent to both the given circle and the given line at their common point of tangency. Their centers lie on the perpendicular to the line through this point.
9. The centers of these two circles are found at the intersections of this perpendicular with the angle bisectors.
10. We have now found the remaining two solutions.
11. These are the images, under circular inversion, of the solution circles of the original problem.
12. The problem has six solutions.

Steps

1. We choose the line of inversion. Its center is selected as the point of tangency of the given line and circle.
2. We apply circular inversion to the given objects. The circle passing through the point of tangency is transformed into a line. The given line is invariant.
3. We seek circles that are tangent to all the transformed objects. Their centers are found using loci defined by geometric conditions. We find four such circles. These are the images of four solution circles of the original problem.
4. The remaining two solutions are transformed into tangents to a transformed circle that is parallel to the given line.
5. We invert the images of the solution circles back using the same circular inversion.
6. The problem has six solutions.

Steps

1. The centers of all circles that are externally tangent to one of the given circles and internally tangent to the other lie on an ellipse. The foci of the ellipse are the centers of the two given circles, and it passes through their points of intersection.
2. The centers of all circles that are either externally tangent to both given circles or internally tangent to both lie on a hyperbola. The foci of the hyperbola are the centers of the two circles, and the hyperbola also passes through their points of intersection.
3. The centers of all circles tangent to the given circle and to a line that is tangent to it at a common point lie on the perpendicular to the line passing through the point of tangency.
4. The centers of all circles tangent to the given line and circle outside their point of tangency 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, offset by a distance equal to the radius of the circle.
5. The centers of the solution circles lie at the intersections of the ellipse, the perpendicular, the hyperbola, and the parabola. Some of these intersections do not correspond to valid solutions—this occurs when one curve contains centers of circles with internal tangency and another with external tangency (or vice versa).
6. The problem has six solutions.