circle. Naturally there is no inversion for a point . Draw a circle (in red). Assume that the red circle intersects the black circle at and if the circles do not intersect in two points see below for an alternative construction. if the circles intersect in only one point, , it is possible to invert simply by doubling the length of (quadrupling the length of ). Reflect the circle center across the line : Construct two new circles and (in light blue). The light blue circles intersect at and at another point . Point is the desired inverse of in the black circle. Point is such that the radius of is to as is to the radius; or . In the event that the above construction fails (that is, the red circle and the black circle do not intersect in two points), find a point on the line so that the length of line segment is a positive integral multiple, say , of the length of and is greater than (this is possible by Archimede's axiom). Find the inverse of in circle as above (the red and black circles must now intersect in two points). The point is now obtained by extending so that = . Determining the center of a circle through three points Given three non-collinear points , and , find the center of the circle they determine. Construct point , the inverse of in the circle . Reflect in the line to the point . is the inverse of in the circle . Intersection of two non-parallel lines (construction #3) Given non-parallel lines and , find their point of intersection, . Select circle of arbitrary radius whose center does not lie on either line. Invert points and in circle to points and respectively. The line is inverted to the circle passing through , and . Find the center of this circle. Invert points and in circle to points and respectively. The line is inverted to the circle passing through , and . Find the center of this circle. Let be the intersection of circles and . is the inverse of in the circle . Intersection of a line and a circle (construction #4) The compass-only construction of the intersection points of a line and a circle breaks into two cases depending upon whether the center of the circle is or is not collinear with the line. Circle