The intersections of two circles

This page describes a method for finding the points of intersection of two circles in a plane.

Given two circles A and B in a plane, let the centre of A be the vector \(C_A\) and its radius \(R_A\), and similarly let \(C_B\) and \(R_B\) be the centre and radius of B.

Let the distance from the centre of A to the centre of B be \(d\). Let the perpendicular distance from the line segment between the centres, which we'll call AB, to either intersection be \(g\), and the distance from the centre of A and B to the point P where this perpendicular meets the line segment AB be \(e\) and \(f\) respectively. From Pythagoras we have \(e^2+g^2=R_A^2\) and \(f^2+g^2=R_B^2\), so \(f^2-e^2=R_B^2-R_A^2\).

Now \(e^2+2ef+f^2=d^2\) so \(2ef+2f^2=d^2+R_A^2-R_B^2\), and since \(2ef+2f^2=2(e+f)f=2df\), \[ f=\frac{d^2+R_B^2-R_A^2}{2d}, \] and \(e=d-f=(d^2-R_B^2+R_A^2)/2d\) and \(g=\sqrt{R_B^2-f^2}\). We can then get the points of intersection by adding a vector of length \(\pm g\) perpendicular to AB to the vector P, which is given by \[ P = (f C_A + e C_B)/d. \]

★ I worked this out myself in the course of doing something involving drawing Venn diagrams. After I finished working it out myself, I searched for other solutions on the internet to see if someone had a better way, but the top solutions I found on search engines were actually much more complicated than the above calculations, so I've put this here as it may be useful for someone.