If \(\left|\frac{\mathrm{z}}{1+\mathrm{i}}\right|=2\), where \(\mathrm{z}=x+\mathrm{i} y, \mathrm{i}=\sqrt{-1}\) represents a circle, then centre ' \(C\) ' and radius ' \(r\) ' of the circle are

\(\begin{aligned} & \text { Given, }\left|\frac{\mathrm{z}}{1+\mathrm{i}}\right| 2, \mathrm{z}=x+\mathrm{i} y \\ & \Rightarrow\left|\frac{x+\mathrm{i} y}{1+\mathrm{i}}\right|=2 \\ & \Rightarrow\left|\frac{(x+\mathrm{i} y)(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})}\right|=2 \\ & \Rightarrow\left|\frac{x-x \mathrm{i}+\mathrm{i} y-\mathrm{i}^2 y}{1^2-(\mathrm{i})^2}\right|=2 \\ & \Rightarrow\left|\frac{(x+y)+\mathrm{i}(y-x)}{2}\right|=2 \\ & \Rightarrow|(x+y)+(y-x) \mathrm{i}|=4 \\ & \Rightarrow \sqrt{(x+y)^2+(y-x)^2}=4 \\ & \Rightarrow(x+y)^2+(y-x)^2=16\end{aligned}\)
\(\begin{aligned}
& \Rightarrow x^2+2 x y+y^2+y^2-2 x y+x^2=16 \\
& \Rightarrow 2 x^2+2 y^2=16 \\
& \Rightarrow x^2+y^2=8,
\end{aligned}\)
which is a circle with centre \((0,0)\) and radius is \(\sqrt{8}\) i.e \(2 \sqrt{2}\)