Show that the common chord of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-4\mathrm{y}+9=0$ and ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}-6\mathrm{y}+23=0$ is the diameter of the second circle and also find its length

Given circles are
$\begin{array}{l}\mathrm{S}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-4\mathrm{y}+9=0.....\left(1\right)\\ {\mathrm{S}}^{\text{'}}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}-6\mathrm{y}+23=0....\left(2\right)\end{array}$
Now common chord of (1) & (2) is $\mathrm{S}-{\mathrm{S}}^{\mathrm{\prime }}=0$

$\begin{array}{l}\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-4\mathrm{y}+9\right)-\left({\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}-6\mathrm{y}+23\right)=0\\ \Rightarrow 2\mathrm{x}+2\mathrm{y}-14=0\\ \Rightarrow \mathrm{x}+\mathrm{y}-7= 0.....\left(3\right)\end{array}$
from (2) $\Rightarrow$g = -4, f = -3, c = 23
Centre = (4,3)
$\begin{array}{l}\mathrm{r}=\sqrt{(-4{)}^2+(-3{)}^2-23}\\ =\sqrt{16+9-23}\\ r=\sqrt{2}\end{array}$
Now sub centre = (4,3) in (3)
$\begin{array}{l}\Rightarrow 4+3=7= 0\\ \Rightarrow 0=0\end{array}$
$\therefore$Common chord is diameter of second circle.
Now length of common chord (AB) = length of diameter of second circle.
                                                               $\begin{array}{l}=2\mathrm{r}\\ =2\sqrt{2}\end{array}$