Let C be the circle in the complex plane with centre z_0=(1)/(2)(1+3i) and radius r=1.

Question

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

Accepted Answer

Given, ${z}_{0}=\frac{1+3i}{2},{z}_{1}=(1+i)$

$|{z}_{1}-{z}_{0}|=|\frac{1-i}{2}|=\frac{1}{\sqrt{2}}$

$\begin{gathered} |{z}_{1}-{z}_{0}||{z}_{2}-{z}_{0}|=1 \ \Rightarrow \frac{1}{\sqrt{2}}|{z}_{2}-{z}_{0}|=1 \ \Rightarrow |{z}_{2}-{z}_{0}|=\sqrt{2} \end{gathered}$

$\begin{gathered} \frac{{z}_{2}-{z}_{0}}{{z}_{1}-{z}_{0}}=\frac{|{z}_{2}-{z}_{0}|}{|{z}_{1}-{z}_{0}|}(\pm 1)=\pm 2 \ {z}_{2}={z}_{0}\pm 2({z}_{1}-{z}_{0}) \end{gathered}$

${z}_{2}=2{z}_{1}-{z}_{0}=\frac{3}{2}+\frac{1}{2}i\Rightarrow {|{z}_{2}|}^{2}=\frac{5}{2}$

OR

${z}_{2}=3{z}_{0}-2{z}_{1}=\frac{-1}{2}+\frac{5}{2}i\Rightarrow {|{z}_{2}|}^{2}=\frac{13}{2}$

Solutions · 1