Two Circles With Contact, Third Circle Inside One Of Them

Number of solutions: 2

We first find the centers of circles tangent to one pair of the given circles. The centers of the solution circles will lie on a pair of ellipses. The foci of the ellipses are the centers of the given circles. To construct them, we need to know at least one point on each ellipse. We draw the line passing through the centers of the given circles and find its intersections with the circles. The midpoints of the segments defined by the obtained intersection points are the vertices of the ellipses.
We draw the ellipses with foci at the centers of the given circles and passing through the identified vertices.
Next, we find the centers of the solution circles tangent to the two given circles that are internally tangent to each other. The solution circles will also be tangent to these two circles at their common point of tangency. The centers of such circles lie on the line passing through the centers of the given circles.
The centers of the solution circles lie at the intersections of the found ellipses and the line. However, not every intersection point is the center of a solution circle — only those for which the type of tangency (external or internal) to the common circle matches on both curves.
The problem has two solutions.

We choose the reference circle for the inversion. Its center is chosen at the point of tangency of the given circles.
We apply the circular inversion to the given objects. The circles through the point of tangency are transformed into lines.
The solution circles are transformed into tangents to a circle parallel to the images of the given circles.
We invert the found tangents back using the same inversion.
The problem has two solutions.

The common point of tangency of the given circles will also be the point of tangency with the solution circles. Therefore, the centers of the solution circles will lie on the line passing through the centers of the given circles and the point of tangency.
The common tangent of the two touching circles will also be a tangent of the solution circles.
In the homotheties in which the outer given circle is mapped onto the solution circles, the common tangent of the two circles is mapped to tangents of this circle that are parallel to each other. The centers of these homotheties are the points of tangency between the outer given circle and the solution circles.
Since the common tangent of the two circles is mapped onto parallel tangents of the third given circle, the points of tangency of the tangents and the circles are also mapped onto each other. Therefore, the centers of the homotheties lie on lines passing through these points of tangency. They lie at the intersections of these lines with the given circle.
We have now found all points of tangency between the given circles and the solution circles. The centers of the solution circles lie on the perpendicular bisectors of the segments defined by these points of tangency.
The centers of the solution circles lie at the intersections of the found bisectors and the line passing through the centers of the circles and the point of tangency.
The problem has two solutions.

Loci of Points Satisfying Certain Properties