Problem of Apollonius

The solution circles will be tangent to the outer given circle at the specified point. Therefore, their centers lie on the line passing through the given point and the center of the given circle.
Now we use dilation. We apply two different dilations. In both cases, the inner given circle is contracted to its center. The outer given circle is either enlarged or contracted by the radius of the given circle. The given point moves along the line passing through the center of the given circle during these dilations.
The solution circles of the dilated problems will pass through the image of the given point and the center of the inner circle, to which the inner circle was contracted. Their centers therefore lie on the perpendicular bisectors of the segments defined by these points.
The centers of the solution circles of the dilated problem lie at the intersections of these bisectors with the line passing through the center of the outer given circle and the given point. These centers are also the centers of the solution circles of the original problem.
The found solutions are then dilated back.
The problem has two solutions.

The solution circles will be tangent to the outer given circle at the specified point. Therefore, their centers lie on the line passing through the given point and the center of the given circle.
We find the centers of circles that are tangent to the inner given circle and pass through the given point. These centers lie on a hyperbola. The foci of the hyperbola are the center of the given circle and the given point. To construct it, we need to know at least one of its points. We draw the line through the center of the given circle and the given point and find its intersections with the circle. The midpoints of the segments defined by these intersections and the given point are the vertices of the hyperbola.
We construct the hyperbola with foci at the center of the given circle and the given point, passing through the identified vertices.
The centers of the solution circles lie at the intersections of the line and the hyperbola.
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. One of them transforms into another circle, and the second one into a line.
In the chosen inversion, the solution circles are mapped to tangents to the image of the circle that are parallel to the line, which is the image of the second circle.
Invert the found tangents back to obtain the solution circles.
The problem has two solutions.

The solution circle will be tangent to the outer given circle at the given point. The centers of all such circles lie on the line passing through the given point and the center of the given circle.
The tangent to the given circle at the given point will also be a tangent of the solution circles.
There are two homotheties in which the inner given circle is the image of the solution circles. The centers of these homotheties are the points of tangency between the given and solution circles. The tangent to the solution circle is mapped to a parallel tangent of the given circle.
In the homotheties, the point of tangency is mapped to the points of tangency. The lines passing through these points also pass through the centers of the homotheties, which are the intersections of these lines with the inner given circle.
The identified homothety centers are also the tangency points between the given circle and the solution circles. Thus, for each solution circle, we have two of its points. Their centers lie on the perpendicular bisectors of the segments joining these points.
The centers of the solution circles lie at the intersections of these bisectors and the line from step 1.
The problem has two solutions.

One Circle Inside Another, Point on Outer Circle