$\textbf{1}\ w=\frac{5-4z}{4z-2}\\[2ex] 4wz-2w=\ 5-4z\ \ \\[2ex] 4wz+4z=5+2w\\[2ex] $

$\displaystyle z(4w+4)=5+2w\\[2ex] \displaystyle z=\frac{5+2w}{4w+4}\\[2ex] \displaystyle \left\vert{}z\right\vert{}=\frac{\left\vert{}5+2w\right\vert{}}{\left\vert{}4w+4\right\vert{}}\\[2ex] $

$\displaystyle\ \therefore{}1=\frac{\left\vert{}5+2(u+iv)\right\vert{}}{\left\vert{}4(u+iv)+4\right\vert{}}\\[3ex] \displaystyle \therefore{}\left\vert{}(5+2u)+2iv\right\vert{}=\left\vert{}\left(4u+4\right)+4iv\right\vert{}\\[3ex] \displaystyle \therefore{}\sqrt{{(5+2u)}^2+{(2v)}^2}=\sqrt{{(4+4u)}^2+{(4v)}^2}\\[3ex] \displaystyle \therefore{}25+20u+4u^2+4v^2=16+32u+16u^2+16v^2\\[3ex] \displaystyle \therefore{}12u^2+12v^2+12u-9=0\\[2ex]\therefore{}u^2+v^2+u-\frac{3}{4}=0 \\[2ex] $

$large Comparing this with the standard equation of circle: \\[2ex] \displaystyle u^2+v^2+2gu+2hv+c=0 \\[2ex] \displaystyle g=\frac{1}{2},h=0,\ \ c=\ -\frac{3}{4} \\[2ex] \displaystyle \therefore{}Center\equiv{}\left(-g,\ -h\right)\equiv{}(-\frac{1}{2}\ ,0) \\[2ex] \displaystyle And\ radius=\sqrt{g^2+h^2-c}=\sqrt{\frac{1}{4}+0-(-\frac{3}{4})}\ \ =\ 1 \\[2ex] $