Problem of Apollonius

Draw a perpendicular bisector l parallel to and the same distance to both lines.
Draw a circle m with the center at A, and it's radius being the distance between bisector l and either one of the lines.
Mark the points of intersection of the circle m and the bisector l S_1, and S_2.
Finally, draw circles k_1, and k_2, with their centers at points S_1, and S₂ and their radii being the distance between each one's respective center and point A.

We solve for 2 parallel lines p₁ and p₂ and for the point A between them
Draw line s, for which |s p₁| = |s p₂|
Draw point M on the line s and construct circle c at the point M, with p₁, and p₂ as it's tangents.
Construct line g, which intersects points A and g || p₁.
Name the points of intersection A and g Q₁ and Q₂.
We draw the circle k₁, which is the displacement of the circle c by the vector Q₁ A. We draw a circle k₂, which is a displacement of the circle c by the vector Q₂ A. In this way, we have obtained the resulting circles, the centers of which are the displacement of the point M by the given vectors.

Given two parallel lines and a point lying between them.
We will solve the problem using circle inversion. First, we choose a reference circle, its center is the entered point, which will be displayed to infinity. This will make the solution circles appear as straight lines.
In the circle inversion, we display the specified straight lines. They are displayed as a circle passing through the specified point.
We construct the common tangents of both circles. These are images of solutions. There is a third common tangent passing through the specified point, but we are not interested in it.
We display the common tangents in a circle inversion, thereby obtaining both solutions.
The problem has two solutions.

Parallel Lines, Point In Between