Two Circles With Inner Contact, Point On The Smaller Circle

Number of solutions: 1

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.

Find the centres of the given circles
The first set of points is the line h containing the given point A and the centre of the smaller circle S2
Draw a line g containing the given point A and the center of the larger circle S1
Find the intersections with the line g and the larger circle and name it D
Draw a point F in the middle of the line AD
Construct an ellipse with foci at point A and S1
Find the intersection of the ellipse and the line h that is not point S2 and name it G
Draw a circle k with centre G passing through point A

Given are the circles a and b such that a is internally tangent to b. Also given is the point A, which lies on a 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 inner 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 outer 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 outer 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.

Circle Inversion