Both Points Outside A Circle

Steps

1. Construct a circle for circular inversion. Choose the point A as its centre and the distance between the assigned points (the line segment AB) as its radius.
2. Invert the assigned circle into the newly constructed one with the help of circular inversion.
3. After constructing the new circle with c. inversion, construct its tangents which pass through the point B.
4. Construct two circles with the circular inversion of tangents of the circle constructed in the first step. The two circles, k1 and k2, are the solution.

Steps

1. First, we determine the locus of centers of circles that are tangent to a given circle and pass through one of the given points. This locus is a hyperbola with foci at the given point and the center of the circle. To construct the hyperbola, we need at least one additional point on it. For example, we can use the center of a circle that is tangent to the given circle at point C. Let point E be the midpoint of segment AC.
2. Construct the hyperbola with foci at points A and S and passing through point E.
3. Construct the perpendicular bisector of the segment defined by the two given points. All centers of circles passing through both points lie on this line.
4. Find the intersections of the hyperbola with the perpendicular bisector. The intersection points are the centers of the solution circles.
5. The problem has two solutions.

Steps

0. We have a circle k with center S and points A, B
1. Draw a line f passing through points A and B and the perpendicular bisector of segment AB, g
2. Draw a circle c with its center on g having an arbitrary radius
3. Mark the intersections of c and k as E and F, and label the line EF as h
4. Mark the intersection of f and h as G
5. From G, draw tangents to k, which we label as i and j
6. From the intersections of i and j with k, labeled as H and I, respectively, draw lines HS, labeled as l, and IS, labeled as m
7. Mark the intersection of g with m as J and the intersection of g with l as K
8. Construct the finalised solution by creating k1, with center at J passing through A, and k2, with center at K passing through A