Problem of Apollonius

First, we find the centers of circles tangent to one pair of the given circles. The centers of the solution circles lie on a pair of hyperbolas. The foci of the hyperbolas are the centers of the given circles. To construct them, we need to know at least one point on each hyperbola. We draw the line passing through the centers of the given circles and find its intersections with the circles. The midpoints of the segments between these intersection points are the vertices of the hyperbolas.
We construct the hyperbolas with foci at the centers of the given circles and passing through the identified vertices.
Next, we find the centers of solution circles tangent to a pair of given circles with internal tangency. The solution circles will touch the given 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 constructed hyperbolas and the line. However, not every intersection 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 a 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 where the third given circle is mapped onto the solution circles, the common tangent of the two circles is mapped onto tangents to this circle that are parallel to each other. The centers of these homotheties are the points of tangency between the third given circle and the solution circles.
Since the common tangent of the two circles is mapped to parallel tangents of the third given circle, the corresponding points of tangency are also mapped to one another. The centers of the homotheties therefore lie on the 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.

Two Circles Touching Internally, Third Outside