Find the centre and radius of the circle $\sqrt{1+{\mathrm{m}}^2}\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)-2\mathrm{cx}-2\mathrm{mcy}=0$

Equation of the given circle is
${\mathrm{x}}^2+{\mathrm{y}}^2-\frac{2\mathrm{c}}{\sqrt{1+{\mathrm{m}}^2}}\mathrm{x}-\frac{2\mathrm{mc}}{\sqrt{1+{\mathrm{m}}^2}}\mathrm{y}=0$
comparing with ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0,$
we get $2\mathrm{g}=\frac{-2\mathrm{c}}{\sqrt{1+{\mathrm{m}}^2}},2\mathrm{f}=\frac{-2\mathrm{mc}}{\sqrt{1+{\mathrm{m}}^2}},\mathrm{c}=0$
$\therefore \text{ Centre }=(-\mathrm{g},-\mathrm{f})$
$=\left(\frac{\mathrm{c}}{\sqrt{1+{\mathrm{m}}^2}},\frac{\mathrm{mc}}{\sqrt{1+{\mathrm{m}}^2}}\right)$
$\text{ Radius }=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}=\sqrt{\frac{{\mathrm{c}}^2}{1+{\mathrm{m}}^2}+\frac{{\mathrm{m}}^2{\mathrm{c}}^2}{1+{\mathrm{m}}^2}-0}=\sqrt{{\mathrm{c}}^2}=\left|\mathrm{c}\right|$