Suppose three circles are given for which the centers are not collinear. Each pair of circles determines a radical axis, and these three radical axes are concurrent.

[Construction of Assumptions] Construct three circles named for their center point, [math]A,B,C[/math], that are not collinear. Note that each pair of circles creates a radical axis. [math]A[/math] and [math]C[/math] overlap and the radical axis is the line through those intersection points, [math]G,H[/math]. [math]C[/math] and [math]E[/math] also overlap and form a similar radical axis through their intersection points [math]J,K[/math]. To create the third radical axis, construct a line segment between points [math]A,E[/math]. The perpendicular bisector of this segment is the radical axis between these non-intersecting circles by definition. [br][br]Proof: Assume there are three noncollinear circles and three radical axes as described above. Now consider a triangle that is created by connecting [math]A,E,M[/math] where [math]M[/math] is the point of concurrency of the radical axes created by [math]A,C[/math] and [math]C,E[/math]. Notice, [math]\overline{AN}\cong\overline{EN}[/math] and [math]\angle ANM\cong\angle ENM=90^\circ[/math] by the definition of a perpendicular bisector. By Common Notion 1, we know that equal things are equal, so [math]\overline{MN}\cong\overline{MN}[/math]. By Proposition 4 (SAS), we can conclude that [math]\triangle ANM\cong\triangle ENM[/math]. Therefore, the radical axis of [math]A[/math] and [math]E[/math] must be concurrent with the other two radical axes at point [math]M[/math] by construction. [math]\diamondsuit[/math]