Find the equation and length of the common chord of the following circle.
${\mathrm{x}}^2+{\mathrm{y}}^2-5\mathrm{x}-6\mathrm{y}+4=0, {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2=0$

Given circles are
$\begin{array}{l}\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-5\mathrm{x}-6\mathrm{y}+4=0......\left(1\right)\\ {\mathrm{S}}^{\text{'}}={\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2=0......\text{ (2) }\end{array}$
Now equation of common chord of (1) & (2) is $\mathrm{S}-{\mathrm{S}}^{\text{'}}=0$
$\begin{array}{l}\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-5\mathrm{x}-6\mathrm{y}+4\right)-\left({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\right)=0\\ \Rightarrow -3\mathrm{x}-6\mathrm{y}+6=0\\ \Rightarrow \mathrm{x}+2\mathrm{y}-2=0......\left(3\right)\end{array}$
from (2) g =-1, f=0, c=-2
Centre $\left(\mathrm{c}\right)=(1,0), \mathrm{r} =\sqrt{1+0+2}=\sqrt{3}.$
Now d = perpendicular distance $(1,0),\mathrm{to} \left(3\right)$
$\begin{array}{l}\mathrm{d}=\frac{\left|1\right(1)+2(0)-2\mid }{\sqrt{5}}\\ \mathrm{d}=\frac{1}{\sqrt{5}}\end{array}$
Now length of chord (AB) = $2\sqrt{{\mathrm{r}}^2-{\mathrm{d}}^2}$
                                             $\begin{array}{l}=2\sqrt{3-\frac{1}{5}}\\ =2\sqrt{\frac{14}{5}}\end{array}$