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

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.)

The radical centre has a defining property that follows directly from the equal-power definition of a radical axis: it is the unique point whose power is the same with respect to all three circles, since it lies on each pairwise radical axis (the locus of equal power for that pair) simultaneously. Consequently, if the radical centre lies outside all three circles, the tangent lengths from it to each of the three circles are all equal -- this common tangent length is exactly what is needed to construct the circle orthogonal to all three (see below).