Two Non-Tangent Circles, Third Intersecting Both

Steps

1. Choose the circle of inversion so that its center is at one of the intersection points of the given circles.
2. Apply circular inversion to the given circles. Since the center of the inversion lies at the intersection of the circles, two of the circles are transformed into lines.
3. Now we solve the problem in the inverted setting with two lines and a circle. To simplify further, we apply a dilation, shrinking the circle to its center. The lines are then shifted by the radius of the circle.
4. This dilation reduces the problem to one with two lines and a point. We can solve this, for example, using circular inversion. As the inversion circle, we use the circle already available. After applying the inversion, we reduce the problem to finding common tangents of two circles, which we then solve. The found tangents are the images of two of the solutions, which now need to be mapped back.
5. First, we invert them back using the second circular inversion. This gives us the solutions for the two lines and point configuration.
6. We then dilate this solution back to the two lines and circle configuration.
7. Finally, we invert the obtained circles in the original inversion to get the first two solution circles.
8. Now we return to the configuration of the given objects in the first inversion. We again apply dilation, but this time we shift the secant line into the opposite half-plane. The dilation transforms the problem again into one with two lines and a point.
9. We solve the two lines and point problem using circular inversion. We place the center of the inversion at the point. This maps the problem to finding common tangents of two circles, which we solve.
10. We invert the found tangents back in the circular inversion, giving us the solutions for the two lines and point problem.
11. We then dilate the obtained circles back to the two lines and circle configuration.
12. Finally, we invert the dilated circles in the first circular inversion to obtain the remaining two solutions.
13. The problem has four solutions.