Invariance of circles and lines, a geometric proof

Inversion of lines

Up to conjugation, the inversion transformation moves a point on a ray through the origin of distance r to a point on the same ray with a distance of 1/r. This immediately shows that the image of a line through the origin is another line through the origin.

Suppose we have a line that does not pass through the origin O, and a circle that passes through O such that the (extension of the) diameter OB is perpendicular to the line at C (diagram above). Draw any line from O to a point D on the line, and let A be the intersection with the circle. The angle OAB is a right angle since it is based on the diameter OB. This means that the triangles OAB and OCD have the same angles, so they are proportional. As a result OA/OB = OC/OD, and after rearranging OA*OD = OB*OC = constant for all D. If we choose the dimensions of the circle such that OB*OC is 1 then this circle is the inversion image of the line.

Inversion of circles

Suppose we have two circles, with the line PQ connecting their centers crossing a common tangent AB at the origin O. We draw a line OCDEF intersecting both circles (diagram above). The angles OAP and OBQ are right angles (between a tangent and a radius). Since the angles AOP and BOQ are the same, the triangles AOP and BOQ have the same angles, so they are proportional. Thus OA/OB = OP/OQ = AP/BQ. Now look at the triangles POC and QOE. By the sine theorem OP/sin(angle OCP) = CP/sin(angle POC), so sin(angle OCP) = (OP/CP)sin(angle POC). Similarly, sin(angle OEQ) = (OQ/EQ)sin(angle QOE). Since the angles POC and QOE are the same, AP = CP and EQ = BQ we get OP/CP = OQ/EQ (using the previous proportion), so sin(angle OCP) = sin(angle OEQ). Both angle OCP and OEQ are obtuse so they are equal. This again means that the triangles OCP and OEQ are proprtional, and OP/OQ = OC/OE.

Now look at the triangles OBE and OFB (diagram below). The angles OBE and OFB are equal, one is between a tangent and a chord and the other is based on the chord. Since the angle BOE is the same as FOB again the triangles are proportional and OE/OB = OB/OF.

Summing up the equalities we have established we get that OC/OE = OP/OQ = OA/OB and OE/OB=OB/OF. Substituting OE we get OC*OF = OA*OB = constant for all lines OCF. Choosing B so that OA*OB = 1 finds the inversion image of the circle centered at P. This shows the inversion image of any circle not containing the origin.