Let $\mathrm{C}$ be the circle in the complex plane with centre $\mathrm{z}_{0}=\frac{1}{2}(1+3 i)$ and radius $r=1$. Let $\mathrm{z}_{1}=1+\mathrm{i}$ and the complex number $z_{2}$ be outside the circle $C$ such that $\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1$. If $z_{0}, z_{1}$ and $z_{2}$ are collinear, then the smaller value of $\left|z_{2}\right|^{2}$ is equal to :

Given, $z_0=\frac{1+3 i}{2}, z_1=(1+i)$ <br/><br/>$\left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}}$ <br/><br/>$$ \begin{aligned} & \left|z_1-z_0\right|\left|z_2-z_0\right|=1 \\\\ & \Rightarrow \frac{1}{\sqrt{2}}\left|z_2-z_0\right|=1 \\\\ & \Rightarrow\left|z_2-z_0\right|=\sqrt{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \frac{z_2-z_0}{z_1-z_0}=\frac{\left|z_2-z_0\right|}{\left|z_1-z_0\right|}( \pm 1)= \pm 2 \\\\ & z_2=z_0 \pm 2\left(z_1-z_0\right) \end{aligned} $$ <br/><br/>$$ z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2} $$ <br/><br/>OR <br/><br/>$$ z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2} $$