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

Given circles are
$\mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-4\mathrm{y}+9=0 ...\left(1\right)$
${\mathrm{S}}^{\mathrm{\prime }}={\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}-6\mathrm{y}+23=0 ...\left(2\right)$
The equation of the common chord of (1) & (2) is $\mathrm{S}-{\mathrm{S}}^{\mathrm{\prime }}=0\Rightarrow 2\mathrm{x}+2\mathrm{y}-14=0\Rightarrow \mathrm{x}+\mathrm{y}-7=0 ...\left(3\right)$
Centre of (2), C = (4, 3)
Clearly C satisfies (3) ; (3) is a diameter of (2)
Length of common chord = Diameter of (2)
$=2\sqrt{16+9-23}=2\sqrt{2}$