Secant Line and Point on Circle

Steps

1. To find the solution, we use circular inversion with the given point as the center of the inverting circle.
2. Apply the inversion to the given line and the given circle. The given point is mapped to infinity.
3. In the inverted setting, the solutions appear as tangents to the image of the circle that are parallel to the image of the 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 a given line and passing through a given point lie on a parabola, where the given line is the directrix and the given point is the focus.
2. The centers of all circles tangent to a given circle at a given point lie on the line passing through that point and the center of the given circle.
3. The centers of the desired solution circles lie at the intersections of the parabola and the line.
4. The problem has two solutions.

Steps

1. The centers of the solution circles lie on the line passing through the center of the given circle and the given point. The reflection of the given line with respect to this axis of symmetry also becomes tangent to the solution circles.
2. We use a homothety centered at the intersection of the original line and its reflected image. Construct a circle whose center lies on the axis and which is tangent to both the given line p and its reflection p′. This circle is the image of the solution circles under the homothety.
3. The circle touches line p at point T′. Its intersections with the axis correspond to the images of the given point A.
4. The segments connecting point T′ with the images of A must be parallel to the segments connecting the original point A with the points of tangency of the solution circles.
5. The centers of the solution circles lie on perpendiculars to the given line that pass through the identified points of tangency.
6. The problem has two solutions.