If $\left|\mathrm{z}\right|=2\sqrt{2}$and $\arg \left(\frac{\mathrm{z}-(5+\mathrm{i})}{\mathrm{z}-(1+\mathrm{i})}\right)=\frac{\mathrm{\pi }}{4}$Then Z is

$\left|\mathrm{z}\right|=2\sqrt{2}\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}=2\sqrt{2}\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=8;$

${\mathrm{Tan}}^{-1}\left(\frac{\mathrm{y}-1}{\mathrm{x}-5}\right)-{\mathrm{Tan}}^{-1}\left(\frac{\mathrm{y}-1}{\mathrm{x}-1}\right)=\frac{\mathrm{\pi }}{4}$
$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-6\mathrm{y}+10=0,$

locus of $\mathrm{z}$ lies on  $8-6x-6y+10=0\Rightarrow x+y=3$(common chord of given two circles)

$\Rightarrow \mathrm{a},\mathrm{b}$ satisfices  both the circles