Q.Find the radical centre of the circles $x^2 + y^2 - 4x - 6y + 5 = 0$, $x^2 + y^2 - 2x - 4y - 1 = 0$, $x^2 + y^2 - 6x - 2y = 0$.

Given three circles whose centres are not collinear, consider the three radical axes obtained by pairing them up: the radical axis of circles $1$ and $2$, of circles $2$ and $3$, and of circles $3$ and $1$. Writing $S_1,S_2,S_3$ for the three circles' equations, these three radical axes are $L_{12}=S_1-S_2$, $L_{23}=S_2-S_3$, $L_{31}=S_3-S_1$, and a short algebraic observation makes their relationship immediate:

Since the three lines' equations sum identically to zero, any point satisfying two of them automatically satisfies the third -- so the three radical axes are concurrent: they meet at a single point, called the radical centre of the three circles. (When the three centres are collinear, the three radical axes are parallel rather than concurrent, since a line perpendicular to a common direction shared by all three centre-pairs cannot meet the others in a single common point unless it coincides with them.) …