Verified Solution • 2025–26 Session moderate

QJEEM23Ma8cdf: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8cdf Solution: Let be the circle in the complex plane with centre and radius . …

Let C\mathrm{C} be the circle in the complex plane with centre z0=12(1+3i)\mathrm{z}_{0}=\frac{1}{2}(1+3 i) and radius r=1r=1 . Let z1=1+i\mathrm{z}_{1}=1+\mathrm{i} and the complex number z2z_{2} be outside the circle CC such that ∣z1−z0∣∣z2−z0∣=1\left|z_{1}-z_{0}\right|\left|z_{2}-z_{0}\right|=1 . If z0,z1z_{0}, z_{1} and z2z_{2} are collinear, then the smaller value of ∣z2∣2\left|z_{2}\right|^{2} is equal to :

Multiple Choice Options

  • A 32\frac{3}{2}
  • B 52\frac{5}{2}
  • C 132\frac{13}{2}
  • D 72\frac{7}{2}

Step-by-Step Verified Solution

Given, z0=1+3i2,z1=(1+i)z_0=\frac{1+3 i}{2}, z_1=(1+i)

∣z1−z0∣=∣1−i2∣=12\left|z_1-z_0\right|=\left|\frac{1-i}{2}\right|=\frac{1}{\sqrt{2}}

amp;∣z1−z0∣∣z2−z0∣=1amp;⇒12∣z2−z0∣=1amp;⇒∣z2−z0∣=2\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}

amp;z2−z0z1−z0=∣z2−z0∣∣z1−z0∣(±1)=±2amp;z2=z0±2(z1−z0)\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}

z2=2z1−z0=32+12i⇒∣z2∣2=52z_2=2 z_1-z_0=\frac{3}{2}+\frac{1}{2} i \Rightarrow\left|z_2\right|^2=\frac{5}{2}

OR

z2=3z0−2z1=−12+52i⇒∣z2∣2=132z_2=3 z_0-2 z_1=\frac{-1}{2}+\frac{5}{2} i \Rightarrow\left|z_2\right|^2=\frac{13}{2}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring