Find the equation and length of the common chord of the two circles $\mathrm{s}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}+5\mathrm{y}+4=0$ and ${\mathrm{s}}^{\mathrm{\prime }}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2+5\mathrm{x}+3\mathrm{y}+4=0$

Common chord is radical axis
$\therefore$Equation of common chord is $\mathrm{s}-{\mathrm{s}}^{\mathrm{\prime }}=0$
$\Rightarrow -2\mathrm{x}+2\mathrm{y}=0\Rightarrow \mathrm{x}-\mathrm{y}=0$

Centre of 1^st circle  ${\mathrm{c}}_1=\left(\frac{-3}{2},\frac{-5}{2}\right)$
$\text{ Radius }{\mathrm{r}}_1=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}=\sqrt{\frac{9}{4}+\frac{25}{4}-4}$

$=\sqrt{\frac{9+25-16}{4}}=\sqrt{\frac{18}{4}}=\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}$
Length of chord $\mathrm{AB}=2\sqrt{{\mathrm{r}}_1^2-{\mathrm{d}}^2}$
$\mathrm{d}=\perp \text{ distance from }{\mathrm{c}}_1\left(-\frac{3}{2},-\frac{5}{2}\right)\text{ to }\mathrm{x}-\mathrm{y}=0$
$=\frac{\left|-\frac{3}{2}+\frac{5}{2}\right|}{\sqrt{1^2+(1{)}^2}}=\frac{1}{\sqrt{2}}$
$\therefore$Length of chord $\mathrm{AB}=2\sqrt{{\mathrm{r}}_1^2-{\mathrm{d}}^2}$
$=2\sqrt{\frac{9}{2}-\frac{1}{2}}=2\sqrt{\frac{8}{2}}=2\sqrt{4}=4$