If x + y = 3 is the equation of the chord AB of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-8=0$ find the  equation of the circle having $\overline{AB}$  as diameter.

$\text{ Given circle }\mathrm{s}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-8=0.....\left(1\right)$
$\text{ Given line }\mathrm{x}+\mathrm{y}-3=0.....\left(2\right)$
$\text{ Now equation of circla Passing through (1) \& (2) is }\mathrm{S}+\mathrm{\lambda L}=0$
$\begin{array}{l}\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-8+\mathrm{\lambda }(\mathrm{x}+\mathrm{y}-3)=0-\left(3\right)\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-8+\mathrm{\lambda x}+\mathrm{\lambda y}-3\mathrm{\lambda }=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+(\mathrm{\lambda }-2)\mathrm{x}+(\mathrm{\lambda }+4)\mathrm{y}-8-3\mathrm{\lambda }=0.\end{array}$
$\text{ where }\mathrm{g}=\frac{\mathrm{\lambda }-2}{2},\mathrm{f}=\frac{\mathrm{\lambda }+4}{2},\mathrm{c}=-8-3\mathrm{\lambda }\text{. }$
$\text{ centre }=\left(\frac{2-\mathrm{\lambda }}{2},-\frac{\mathrm{\lambda }-4}{2}\right)$
$\mathrm{since} \overline{\mathrm{AB}} \mathrm{is} \mathrm{diameter} \mathrm{then} \mathrm{centre} \mathrm{lies} \mathrm{on} \left(2\right)$
$\begin{array}{l}\frac{2-\mathrm{\lambda }}{2}+\left(\frac{-\mathrm{\lambda }-4}{2}\right)-3=0\\ \Rightarrow \frac{2-\mathrm{\lambda }-\mathrm{\lambda }-4-6}{2}=0\\ \Rightarrow -2\mathrm{\lambda }-8=0\\ \Rightarrow -2\mathrm{\lambda }=8\\ \Rightarrow \mathrm{\lambda }=-4\end{array}$
$\therefore \text{ from (3) Equation of circle is }$
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-8-4(\mathrm{x}+\mathrm{y}-3)=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+4\mathrm{y}-8-4\mathrm{x}-4\mathrm{y}+12=0.\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+4=0.\end{array}$