If the locus of z ∈ C , such that Re( z-1 2 z+i )+Re( z -1 2 z -i )=2 is a circle of radius r and center (a, b) then (15 a b)/(r^2) is equal to :

Correct answer: C. 18 aligned & Re((z-1)/(2 z+i))+Re( z -1 2 z -i )=2 & Here, (z-1)/(2 z+i)=( z -1 2 z -i )=2 & =Re((z-1)/(2 z+i))+Re( z-1 2 z+i )=2 & =2 Re((z-1)/(2 z+1))=2 Re((z-1)/(2 z+i))=1 aligned Let z=x+i y aligned & Re(((x-1)+i y)/(2 x+i(2 y+1)))=1 Re[(((x-1)+i y)(2 x-i(y+1))/((2 x+i(2 y+1)(2 x-i(2 y+1)))]=1 & (2 x(x-1)+y(2 y+1))/(4 x^2+(2 y+1)^2)=1 & 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y & 2 x^2+2 y^2+3 y+2 x+1=0 & x^2+y^2+x+(3)/(2) y+(1)/(2)=0 & centre =((-1)/(2), (-3)/(4)), r= (1)/(4)+(9)/(16)-(1)/(2) = 5 4 & a=(-1)/(2), b=(-3)/(4), r^2=(5)/(16) & 15 (a b)/(r^2)=15 ×((-1)/(2)) ×((-3)/(4)) × (16)/(5)=18 aligned

JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Complex Number

If the locus of z ∈ C , such that Re( z-1 2 z+i )+Re( z -1 2 z -i...

If the locus of

z \in \mathrm{C}

, such that

\operatorname{Re}\left(\frac{z-1}{2 z+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-\mathrm{i}}\right)=2

is a circle of radius

r

and center

(a, b)

then

\frac{15 a b}{r^2}

is equal to :

Options

Correct answer

C. 18

Step-by-step solution

\begin{aligned}& \operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\frac{\bar{z}-1}{2 \bar{z}-i}\right)=2 \\ & \text{ Here, } \frac{z-1}{2 z+i}=\left(\frac{\overline{\bar{z}-1}}{2 \bar{z}-i}\right)=2 \\ & =\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)+\operatorname{Re}\left(\frac{\mathrm{z}-1}{2 z+i}\right)=2 \\ & =2 \operatorname{Re}\left(\frac{z-1}{2 z+1}\right)=2 \Rightarrow \operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1\end{aligned}

Let

z=x+i y

\begin{aligned}& \operatorname{Re}\left(\frac{(x-1)+i y}{2 x+i(2 y+1)}\right)=1 \Rightarrow \operatorname{Re}\left[\frac{((x-1)+i y)(2 x-i(y+1)}{(2 x+i(2 y+1)(2 x-i(2 y+1))}\right]=1 \\ & \Rightarrow \frac{2 x(x-1)+y(2 y+1)}{4 x^2+(2 y+1)^2}=1 \\ & \Rightarrow 2 x^2-2 x+2 y^2+y=4 x^2+4 y^2+1+4 y \\ & \Rightarrow 2 x^2+2 y^2+3 y+2 x+1=0 \\ & \Rightarrow x^2+y^2+x+\frac{3}{2} y+\frac{1}{2}=0 \\ & \text{ centre }=\left(\frac{-1}{2}, \frac{-3}{4}\right), r=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4} \\ & a=\frac{-1}{2}, b=\frac{-3}{4}, r^2=\frac{5}{16} \\ & 15 \frac{a b}{r^2}=15 \times\left(\frac{-1}{2}\right) \times\left(\frac{-3}{4}\right) \times \frac{16}{5}=18\end{aligned}

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