Secant Line, Point on Line Outside Circle

Steps

1. The centers of the solution circles must lie on the perpendicular to the given line passing through the given point.
2. When we dilate the given circle down to its center, the given line along with the given point shifts by the radius of the circle. This transforms the problem into one involving a line and two points.
3. The center of the circle passing through the center of the given circle and the image of the given point lies on the perpendicular bisector between them.
4. The center of the solution in both the dilated and original problem lies at the intersection of the bisector and the perpendicular.
5. To find the second solution, we again dilate the circle to its center, but this time we shift the line and point into the opposite half-plane.
6. The center of the desired circle lies on the bisector between the image of the given point and the center of the circle.
7. The center of the solution in both the dilated and original problem lies at the intersection of the bisector and the perpendicular.
8. The problem has two solutions.

Steps

1. Choose the circle of inversion so that the given point is its center.
2. Apply circular inversion to the given circle. The given line remains invariant under this inversion, and the given point maps to infinity.
3. The images of the solution circles are tangents to the image of the circle that are parallel to the given line.
4. Invert the found tangents back to obtain the solution circles.
5. The problem has two solutions.

Steps

1. The centers of all circles tangent to both a given circle and a given line lie on a pair of parabolas. The focus of both parabolas is the center of the given circle. Their directrices are parallel to the given line and are located at a distance from it equal to the radius of the given circle.
2. The centers of all circles tangent to the given line at the given point lie on the perpendicular to the line passing through the given point.
3. The centers of the desired solution circles are located at the intersections of the perpendicular line and the parabolas.
4. The problem has four solutions.