Two Externally Tangent Circles, Line Intersects One Of Them

Steps

1. We construct the inversion circle with its center at the point of tangency of the given circles.
2. Under the circular inversion, the given circles are mapped to parallel lines, while the given line is mapped to a circle.
3. The images of the two sought solutions are circles tangent to the transformed objects. Their centers are found using loci of points with the given properties.
4. We have constructed the images of the first two solutions.
5. The images of the remaining two solutions are lines tangent to the image of the given line and parallel to the images of the given circles.
6. We map the found images of all four solutions back using circular inversion.
7. The problem has four solutions.

Steps

1. We draw perpendiculars to the line passing through the centers of the circles, and we call their intersections with the specified line I and J, and we call their intersections with the specified circles G and H.
2. Construct a circle centered at point I with radius CG. We will call its intersections with the perpendicular K and L. We will construct a circle centered at point J and radius EH. Let's call its intersections with the perpendicular M and N.
3. At points K, L, M and N we construct parallels with the specified straight line
4. We construct a parabola with focus at point C, whose guiding line is a parallel passing through point K. We construct a parabola with the focus at point C, whose guiding line is the parallel passing through point L. Let's construct a parabola with focus at point E, whose guiding line is a parallel passing through point M. We will construct a parabola with the focus at point B, whose guiding line is the parallel passing through point N. We will mark their intersections as S₁, S₂, S₃ and S₄
5. From these found centers of the circles of the solution, we start a perpendicular to the specified straight line. The intersection of each with the given straight line forms the points of contact of the circle, which has its center at the point S, through which the perpendicular also passes.
6. We construct circles with centers at points S that intersect the point of contact.
7. We have 4 solutions