Two Externally Tangent Circles, Different Sizes, Point On The 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.

The centers of circles that are externally tangent to both given circles lie on a hyperbola. Its foci are the centers of the given circles. The hyperbola also passes through the intersection points of the two given circles.
The centers of circles tangent to the given circle and passing through the given point lie on the line passing through the center of the circle and the given point.
The center of the solution circle lies at the intersection of the hyperbola and the line. Not every intersection point is the center of a solution circle — only the one for which the type of tangency (external or internal) matches on both the hyperbola and the line.
The problem has one solution.

choose point A as the center of a circular inversion, and choose the radius of the circle so that it passes through both given circles, then draw circle k
display circles k₁ and k₂ in the circular inversion, circles k₁ and k₂ are created
point C is the point of tangency of k₁' and k₂', construct line o from point C perpendicular to line k₂
point F is the intersection of circle k₁ and the perpendicular line o
construct line k₄' that is parallel to the line k₂ and passes through point F
display line k₄' to its inverse image under the circular inversion with respect to circle k, that circle is the solution to the problem

The solution circle will be tangent to one of the given circles at the given point lying on it. 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 circle.
There is a homothety in which the second given circle is the image of the solution circle. The center of this homothety is the point of tangency between these two circles. The tangent of the solution circle is mapped to a parallel tangent of the given circle.
In the homothety, the point of tangency is mapped to another point of tangency. The line passing through both tangency points also passes through the center of the homothety, which is the intersection of this line with the given circle.
The identified center of homothety is also the tangency point between the given circle and the solution circle. Thus, we have two points lying on the desired circle. Its center must therefore lie on the perpendicular bisector of the segment defined by these two points.
The center of the solution circle lies at the intersection of this bisector and the line from step 1.
The problem has one solution.

Circle Inversion