Point on One of Two Intersecting Circles, Outside the Other

Number of solutions: 2

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 lying on one of them. We solve it, for instance, using loci of points with given properties. 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.

The solution circles will be tangent to both given circles. Their centers therefore lie on a hyperbola or an ellipse with foci at the centers of the given circles. The hyperbola and the ellipse pass 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 centers of the solution circles lie at the intersections of the hyperbola, the ellipse, and the line. However, not every intersection is the center of a solution circle — only those where the type of tangency (internal or external) to the common circle 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. 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 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 this given circle at the given point will also be a tangent of the solution circles.
There are two homotheties in which the second given circle is the image of the solution circles. The centers of these homotheties are the points of tangency between these circles. The tangent of 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 homothety centers, which are the intersections of these lines with the 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 defined by 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.

Circle Inversion