Diverging Lines, One Line Is Tangent To A Circle

Steps

1. The centers of the solution circles must lie on the angle bisectors formed by the intersecting lines.
2. Two of the solution circles will be tangent to the given circle at the point where one of the given lines is also tangent to it. Their centers will lie on the perpendicular to the line passing through this point of tangency.
3. The centers of these two solution circles are located at the intersections of the perpendicular and the angle bisectors.
4. The remaining two centers must lie on a parabola that contains the centers of all circles tangent to the given circle and its tangent line (other than at their common point of tangency). The focus of this parabola is the center of the given circle. The directrix is a line parallel to the given tangent, with the distance between the two parallel lines equal to the radius of the given circle.
5. The centers of the remaining two solution circles are located at the intersections of the angle bisector and the parabola.
6. The problem has four solutions.

Steps

1. We first search for two solution circles that are tangent to the given circle and its tangent line at their common point of tangency. If we dilate the given circle to its center, both given lines shift by the radius of the circle.
2. Through this dilation, the problem is transformed into one involving two intersecting lines and a point that lies on one of them. We solve this using loci of points with specific properties. The centers of circles tangent to the dilated lines lie on the angle bisectors defined by those lines.
3. The centers of circles tangent to the dilated line at point S lie on the perpendicular passing through that point.
4. The centers of the solution circles lie at the intersections of the angle bisectors and the perpendicular.
5. This gives us the first two solutions.
6. Next, we look for two more solution circles that lie on the same side of the given tangent line as the given circle and are externally tangent to it. If we dilate the given circle to its center, both given lines again shift by the radius, but this time in the opposite direction.
7. The centers of the desired circles will lie on the angle bisector between the lines.
8. We continue using homothety. Draw an arbitrary circle centered on the bisector that is tangent to the dilated lines. This circle is homothetic to the solution circles of the dilated problem. The center of the homothety is the intersection point of the dilated lines.
9. In this homothety, point S is mapped to the points of tangency. Therefore, the tangency points must lie on a line passing through point S and the center of the homothety.
10. Connect the images of point S with the images of the centers of the circles.
11. Since homothety preserves parallelism, the centers of the solution circles lie on lines parallel to these connections, passing through point S. The centers are found at the intersections of these parallels with the angle bisector.
12. This gives us the remaining two solutions.
13. The problem has four solutions.

Steps

1. We have two divergent lines and a circle with one point of touch
2. We draw perpendicular lines on a assigned straight line passing through the center of a assigned circle and mark four points of intersection.
3. We draw four parallel lines to the assigned lines going through points of intersection from the previous step. We mark four new points of intersection of these lines.
4. We draw an angle bisector between the assigned lines. Then we draw a line going through point of intersection K and point of intersection of the assigned lines. We mark two points of intersection for these new lines with the assigned circle.
5. We draw two lines going through points of intersection from the previous step. We mark the points of intersection (S1 and S2) of these lines with the angle bisector.
6. We draw two circles defined by a centre and a point. With the centre in points S1 and S2 and points in O and N. Centres of both circles are located on the same line as in step 4.
7. We draw a perpendicular line on the angle bisector of the assigned lines which is passing through the point of intersection of the assigned lines. We use the perpendicular line to the assigned line that passes through the center of the assigned circle. It will intersect this new line at point S3. We draw a circle with centre S3 defined by point B.
8. We draw the intersection of the perpendicular to the assigned line, which passes through the centre of the assigned circle, and the angle bisector of the assigned lines. We draw the circle centred at point S4 and defined by point B.
9. The problem has are four solutions.