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 z2 be outside the circle C such that \[\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1\]. If z0,z1 and z2 are collinear, then the smaller value of \[\left|z_{2}\right|^{2}\] is equal to :
Solutions · 1
Official solution
Given, \[
z_0=\frac{1+3 i}{2}, z_1=(1+i)
\]
\[
\left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}}
\]
\[
∣z1−z0∣∣z2−z0∣=1⇒21∣z2−z0∣=1⇒∣z2−z0∣=2
\]
\[
z1−z0z2−z0=∣z1−z0∣∣z2−z0∣(±1)=±2z2=z0±2(z1−z0)
\]
\[
z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2}
\]
OR
\[
z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2}
\]