Verified Solution • 2025–26 Session moderate

QJEEM23Mab4ee: JEE Main 2023 Mathematics - Class 11 JEEM23Mab4ee Solution: If the center and radius of the circle are respectively and ,…

If the center and radius of the circle ∣z−2z−3∣=2\left| {{{z - 2} \over {z - 3}}} \right| = 2 are respectively (α,β)(\alpha,\beta) and γ\gamma , then 3(α+β+γ)3(\alpha+\beta+\gamma) is equal to :

Multiple Choice Options

  • A 12
  • B 10
  • C 11
  • D 9

Step-by-Step Verified Solution

∣z−2z−3∣=2\left| {{{z - 2} \over {z - 3}}} \right| = 2

amp;(x−2)2+y2=2(x−3)2+y2amp;⇒x2+y2−4x+4=4x2+4y2−24x+36amp;⇒3x2+3y2−20x+32=0amp;⇒x2+y2−203x+323=0amp;⇒(α,β)=(103,0)amp;γ=1009−323=49=23amp;3(α+β+γ)=3(103+23)amp;=12\begin{aligned}& \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\& \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\& \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\& \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\& \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\& \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\& 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\& =12\end{aligned}

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