Problem of Apollonius

We choose the reference circle for the inversion. Its center is the intersection point of the two given circles.
We apply a circular inversion to the given objects. Since both circles pass through the center of the inversion, they are mapped to lines.
In the image, we solve the Apollonius problem for two intersecting lines and a point. We solve it, for instance, using homothety. We find two solutions, which are the images of the solution circles of the original problem.
We invert the found circles back using the original inversion.
The problem has two solutions.

We find the centers of circles that are tangent to one of the given circles and pass through a given point. The centers of these circles lie on an ellipse. The foci of the ellipse are the center of the given circle and the given point. To construct the ellipse, we need at least one of its points. We draw a line passing through the center of the given circle and the given point and find its intersections with the circle. The midpoints of the segments between the intersection points and the given point are the vertices of the ellipse.
We construct the ellipse with foci at the center of the given circle and the given point, passing through the identified vertices.
The centers of the circles that are tangent to both given circles and may be solutions to the problem lie on a hyperbola. Its foci are the centers of the given circles. The hyperbola passes through the intersection points of the two circles.
The centers of the solution circles lie at the intersections of the ellipse and the hyperbola. However, not every intersection is the center of a solution circle – only those for which the type of tangency (external or internal) matches on both curves.
The problem has two solutions.

Choose the circle of inversion so that its center is at the given point.
Apply the inversion to both given circles.
In the chosen inversion, the solution circles are mapped to the common tangents of the images of the two circles.
Invert the found tangents back to obtain the solution circles.
The problem has two solutions.

Point Inside Two Intersecting Circles