Points Lie Outside A Centre Equidistant From It

Steps

0. We have a circle k and points A, B
1. Choose a circle e with its center on k and an arbitrary radius
2. Perform a circular inversion of A, B, and k with respect to e
3. Now, solve the BBp problem with points on the same half-plane: draw a line g passing through points A' and B', name the intersection gp as E, and construct a perpendicular line h to line g passing through point B' and another perpendicular line i passing through point A'
4. Place point I at the midpoint between B' and E, which is the center of the circle f with a radius equal to the segment BI
5. The intersections of circle f and line i are points C and D
6. Draw a circle d with center at E and a radius equal to the line segment ED
7. The intersections of line k' and circle d are points G and H, which are the tangent points to the resulting circles
8. Using three points, we can draw the resulting, but still inverted, circles k1' defined by points A'B'G and k2' defined by points A'B'K
9. To obtain the finalised solution, perform another circular inversion on k1' with respect to e and k2' with respect to e, which yields the solutions k1 and k2

Steps

1. Draw segment bisector of the line AB. The centers of the resulting circles will lie on it. Mark the points of intersection of the bisector and the circle C, and D.
2. Construct segment bisectors of the segments AC, and AD. The centers of the circles will lie in the intersections of these bisectors with the bisector of the segment AB.
3. Draw circles with a radius from their center to point A.