Circles with Internal Tangency, External Line Tangent

Steps

1. Construct the circle of inversion. Choose the point of tangency T as its center.
2. Apply the inversion to the given objects. In the chosen inversion, the given line remains invariant. One of the given circles is transformed into a line parallel to it.
3. Identify the images of the solutions. The first is a circle located in the strip between the two parallel lines. The second is a tangent to the image of the original circle, which is parallel to both lines. The image of the point of tangency with the remaining two objects lies at infinity.
4. Apply the inverse transformation to the identified circle and line.
5. The problem has two solutions.

Steps

1. The centers of circles tangent to the given circles at their common point of tangency lie on the line connecting the centers of the given circles.
2. The tangent passing through the point of tangency of the circles is also a tangent to the solution circle.
3. The center of the solution circle lies on the angle bisector defined by this tangent and the given line.
4. The center of the first solution circle lies at the intersection of this bisector and the line connecting the centers of the given circles.
5. We have found the first solution.
6. The second solution circle is tangent to the given circle and the given line at their common point of tangency. Its center therefore lies on the perpendicular to the given line passing through this point.
7. The centers of all circles tangent to the given circles outside their common point of tangency lie on an ellipse. The foci of the ellipse are the centers of the given circles. The ellipse passes through their point of tangency.
8. The center of the second solution circle lies at one of the intersections of the perpendicular line and the ellipse.
9. The problem has two solutions.

Steps

1. We construct the line on which lie the centers of all circles tangent to the given circles at their common point of tangency. This line passes through the centers of the given circles.
2. We construct a tangent to the inner circle that is parallel to the given line and determine its point of tangency. This tangent represents the image of the given line under two homotheties, whose centers are the points of tangency between the inner circle and the solution circles. The inner given circle is the image of the solution circles.
3. To find the first solution, we use the first homothety. We construct a line passing through the newly found point of tangency and the point of tangency of the given circles, which is the center of this homothety. The intersection of this line with the given line is the point of tangency between the given line and the solution circle.
4. Now we know two tangency points of the circle we are looking for. Its center must therefore lie on the perpendicular bisector of the segment connecting these two points.
5. The center of the circle lies at the intersection of this bisector and the line connecting the centers of the given circles.
6. We have found the first solution.
7. The second solution circle is tangent to the outer given circle and the given line at their common point. Its center therefore lies on the perpendicular to the given line passing through this point.
8. The previously found point of tangency is now the image of the tangency point between the given circle and the line in the second homothety. The center of this homothety thus lies on the line connecting these two points. This center is also the point of tangency between the given circle and the solution circle.
9. The center of the solution circle again lies on the perpendicular bisector of the segment connecting its points of tangency.
10. The center lies at the intersection of this bisector and the perpendicular to the given line.
11. The problem has two solutions.