Both Points Inside A Circle

Steps

1. Construct a circle for the inversion: select the centre of it at one of the points from the assignment, then select the radius, so that the second point lies on the circle.
2. Project the circle from the assignment through the circle from step 1 using a circular inversion.
3. Construct tangents from the point that lies on the circle (from step 1) and the projected circle (these two lines automatically passes through the last projected point, because the point is at infinity).
4. Project these tangents over the circle from step 1 using a circular inversion.
5. The projected tangents appear to be circles - the two solutions of the assignment.

Steps

1. Construct the axis of points A and B.
2. Draw a line through points A and S.
3. Name the intersections of the line with the given circle k E and F.
4. Find the centres of the line segments EA and FA.
5. These points are the extreme points of the major axis of the ellipse p, the focuses being points A and S.
6. Find the intersections of the ellipse and the axis from point 1.
7. These intersections are the centers of the resulting circles.

Steps

1) All centers of circles passing through the given points lie on the perpendicular bisector of segment AB. 2) Construct an arbitrary circle passing through the given points. 3) Draw the line through the intersection points of the constructed and the given circle. Also draw the line through points A and B. Let P be the intersection of these two lines. Since this point lies on the line through the intersection points of the two circles, it has equal power with respect to both circles. As it also lies on line AB, it has the same power with respect to all circles passing through A and B. 4) The power of point P is the same with respect to the given circle and both of the solution circles. Therefore, the tangents drawn from point P to the given circle are also tangents to the solution circles. 5) The centers of the desired circles lie at the intersections of the perpendicular bisector of segment AB and the perpendiculars to the tangents at the points of tangency. 6) The problem has two solutions.