Tangent Line and Point on Circle

Steps

1. To find the solution, we use circular inversion with the center of the inverting circle at the given point.
2. Apply the inversion to the given line and circle. The given point maps to infinity.
3. In the inverted setting, the solution appears as a tangent to the image of the circle that is parallel to the image of the line.
4. Apply the inverse transformation to the found tangent.
5. The problem has a unique solution.

Steps

1. The set of all centers of circles that are tangent to a given line and externally tangent to a given circle lies on a parabola. The focus of this parabola is the center of the given circle. Construct the directrix of the parabola—it is parallel to the given line, and its distance from the point of tangency T is equal to the distance from T to the center of the given circle.
2. Construct the parabola according to the previous step.
3. All centers of circles tangent to the given circle at a fixed point lie on the line passing through that point and the center of the given circle.
4. The center of the desired circle lies at the intersection of the constructed line and the parabola.
5. The problem has a unique solution.

Steps

1. To solve the problem, we use a similarity transformation (specifically, a homothety) in which the given circle is the image of the solution circle, and the center of the homothety is the given point. Under this transformation, the given line maps to a tangent to the given circle that is parallel to the original line.
2. The identified point of tangency on the given circle is the image of the point of tangency on the solution circle. Therefore, both tangency points must lie on a straight line with the center of the homothety.
3. The point of tangency and the given point both lie on the solution circle. Hence, the center of the desired circle must lie on the perpendicular bisector of the segment connecting these two points.
4. The given line is tangent to the solution circle. Therefore, its center must lie on the perpendicular to the line passing through the point of tangency.
5. The center of the solution circle lies at the intersection of the bisector and the perpendicular.
6. The problem has a unique solution.