Diverging Lines With Their Intersect On A Circle, One Line Is Tangential To The Circle

Steps

1. The centers of the solution circles must lie on the angle bisectors formed by the intersecting lines.
2. When the given circle is dilated down to its center, the given lines shift by the radius of the circle.
3. This transforms the problem into one involving two intersecting lines and a point. We solve it (for example, using homothety).
4. One of the solutions of the transformed problem, after dilating back, shrinks into the point of tangency. The other provides the first solution to the original problem.
5. Now we repeat the entire procedure, but this time the secant line shifts to the opposite side during dilation.
6. Again, we solve the problem for two lines and a point.
7. As before, one of the solutions of the transformed problem shrinks into the point of tangency after dilating back. The other gives the second solution to the original problem.
8. The problem has two solutions.

Steps

1. Choose the circle of inversion, placing its center at the intersection point of the lines and the circle.
2. Apply circular inversion to the given objects. The lines remain invariant, and the circle is mapped to a line. Through inversion, the problem is transformed into finding circles tangent to three lines.
3. Solve the problem for one non-parallel line and two parallel lines. Use the method of loci of points with given properties.
4. Invert the found circles back using circular inversion.
5. The problem has two solutions.

Steps

1. The centers of the solution circles must lie on the angle bisectors formed by the intersecting lines.
2. The centers of the solution circles must also lie on a parabola that contains the centers of all circles tangent to the given circle and its tangent line (except at their common point of tangency). The focus of this parabola is the center of the circle. The directrix is a line parallel to the given tangent, and the distance between the two parallel lines equals the radius of the given circle.
3. The centers of the solution circles are located at the intersections of the angle bisector and the parabola.
4. The problem has four solutions.

Steps

0. To construct the solution we will use the homogenous dilation, in which the solution is shown into the assigned circle. The points where the assigned circle and the circle from the solution meet are the center of the homogenous dilation. The assigned lines are the tangents of the circle from the solution, which is why they will be shown as the tangents in the homogenous dilation. The point of intersection of the lines will be shown into the point of intersection. The center of the homogenous dilation, which also is the tangent point of the circles has to be on the connecting line of these intersections. That way we can find the tangent point.
1. At first we will display the assigned lines, which are displayed as the tangents of the assignedcircle. Homogenous dilation preserves the lines being parallel.
2. At first we will display the assigned lines, which are displayed as the tangents of the assignedcircle. Homogenous dilation preserves the lines being parallel.
3. The tangential point has to be on the connecting line of intersection of the assigned lines and its image.
4. The tangential point is also the center of the homogenous dilation. It has to be on the same line as the center of the assigned circle and the center of the circle from the solution.
5. The center fo the circle from the solution also has to be on the angle bisector of the assigned lines.
6. We will repeat the entire previous process once more. One of the lines will be shown on the opposite side of the assigned circle.
7. We will repeat the entire previous process once more. One of the lines will be shown on the opposite side of the assigned circle.
8. The second tangent point is at the line of intersection of the lines and the intersection of their images.
9. The tangent point is also the center of the homogenous dilation. It has to be on the same line as the center of the assigned circle and the center of the circle from the solution.
10. The center of the circle from the solution also has to be on the angle bisector of the assigned lines.
11. We have the centers of both of the circles from the solution. The radius of the circles is given by the distance to the tangent points.
12. The exercise has two solutions.