Find the radical centre of the following circles. ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+5=0, {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-1=0, {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-2\mathrm{y}=0$

Given circles are.
$\begin{array}{l}\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+5=0......\left(1\right)\\ {\mathrm{S}}^{\text{'}}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-1=0......\left(2\right)\\ {\mathrm{S}}^{\text{'}\text{'}}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-2\mathrm{y}=0.....\left(3\right)\end{array}$
Radical axis of (1) & (2) is $\mathrm{s}-{\mathrm{s}}^{\mathrm{\prime }}=0$
$\begin{array}{l}\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+5\right)-\left({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-1)=0\right.\\ \Rightarrow -2\mathrm{x}-2\mathrm{y}+6=0\\ \Rightarrow \mathrm{x}+\mathrm{y}-3=0.....\text{ (4) }\end{array}$
Radical axis of (2) & (3) is $\mathrm{S}-{\mathrm{S}}^{11}=0$
$\begin{array}{l}\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-1\right)-\left({\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-2\mathrm{y})=0\right.\\ \Rightarrow 4\mathrm{x}-2\mathrm{y}-1=0-\text{ (5) }\end{array}$
Now point of intersection of (4) & (5) is $\left(\frac{7}{6},\frac{11}{6}\right)$
$\therefore$ Radical Centre = $\left(\frac{7}{6},\frac{11}{6}\right)$