Two Tangent Circles Of The Same Size, Point On The Circle

Steps

1. point A is chosen as the centre of a circle inversion, the radius of the circle is chosen to intersect both of the given circles, we construct circle k
2. we display circles k₁ and k₂ in a circle inversion, circle k₁' and line k₂' are created
3. in the point of tangency of k₁' and k₂' is located point C, we run circle o to line k₂' passing through point C
4. In the point of intersection of k₁', point C is found
5. we create a parallel line k₄' to line k₂ passing through point C
6. we display line k₄' in a circle inversion through circle k, the final circle is the solution of the problem

Steps

1. We draw lines passing through point A and the centers of the given circles, i.e., points C and B.
2. We mark the intersections of the line with the circle that does not contain point A from the given conditions. In this case, it is circle b.
3. We find the midpoints between the intersections F, E, and point A, which will then lie on the sought hyperbolas.
4. Using two foci and a point lying on the hyperbola, we construct the hyperbola. We choose points A and B as the foci, and one of the midpoints G or H as the point lying on the hyperbola.
5. For the center of circle C and point A, there already exists a set of points in the form of line g. It intersects with the hyperbola at two points: point C and a new intersection, which we label as S.
6. We draw the final circle k around center S. Point C is already the center of circle c, which follows from the given conditions and is therefore not a solution.
7. The problem has a unique solution.