Problem of Apollonius

We choose the reference circle for the inversion. We take the point of tangency of the two given circles as its center.
We apply a circular inversion to the given objects. Since the point of tangency lies on both circles, they are mapped to lines.
We find a circle that is tangent to the images of the given circles and passes through the image of the given point. To locate its center, we may use sets of points with defined properties. This circle is the image of the solution circle.
We invert the found image of the solution circle back using the original inversion.
The problem has one solution.

Imagine a circle that is a solution to the problem. If we enlarge this circle by the radius of the smaller of the two given circles, the smaller given circle shrinks to a point, and the larger one increases its radius by the same amount. The given point is adjusted so that it remains on the enlarged larger circle and stays tangent to the enlarged solution circle.
The center of the solution circle must lie on the perpendicular bisector of the segment A'S₂.
At the same time, it must lie on the line defined by the center of the larger circle and the given point. Therefore, we find the center of the solution circle at the intersection of this line and the bisector from the previous step.
The problem has one solution.

The centers of all circles tangent to the outer given circle at the given point lie on the line passing through the given point and the center of the circle.
We find the centers of circles tangent to the inner given circle and passing 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 passing through the center of the given circle and the given point and find its intersection with the circle. The midpoint of the segment defined by this intersection and the given point is a vertex of the desired hyperbola.
We construct the hyperbola with foci at the center of the given circle and the given point, passing through the identified vertex.
The center of the solution circle lies at the intersection of the line and the hyperbola.
The problem has one solution.

Given are the circles a and b such that a is internally tangent to b. Also given is the point A, which lies on b and is not the tangent point of the two circles.
Draw any circle c with A as its origin.
Invert this circle. Point A' is now at infinity. The inverted forms of a and b shall be named a' and b' respectively.
Now, construct the line d that is tangent to b' and parallel to a'. This line must touch a' and go through A' in infinity.
Invert d around c. The resulting circle is the solution.

The solution circle will be tangent to the outer given circle at the specified 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 specified point will also be a tangent of the solution circle.
There exists a dilation in which the inner given circle is the image of the solution circle. The center of this dilation is the point of tangency between these two circles. The tangent of the solution circle maps to a parallel tangent of the given circle.
In this dilation, the point of tangency maps to the point of tangency. The line passing through both tangency points also passes through the center of the dilation, which is the intersection point of this line with the inner given circle.
The found center of dilation is also the point of tangency between the given circle and the solution circle. This gives us two points lying on the sought circle. Its center must therefore lie on the perpendicular bisector of the segment between these two points.
The center of the solution circle lies at the intersection of the perpendicular bisector and the line from step 1.
The problem has one solution.

Two Circles With Inner Contact, Point On The Larger Circle