JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q15

← JEE Main 2022 · 28 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 15 of 90

Let $S_1=\left\{z_1\in\mathbf C:|z_1-3|=\dfrac12\right\}$ and $S_2=\left\{z_2\in\mathbf C:\big|z_2-|z_2+1|\big|=\big|z_2+|z_2-1|\big|\right\}$ . Then, for $z_1\in S_1$ and $z_2\in S_2$ , the least value of $|z_2-z_1|$ is :

\dfrac12

\dfrac32

\dfrac52

Answer and worked solution

Official answer: C

$S_2$ . Squaring $|z-a|=|z+b|$ with $a=|z+1|$ , $b=|z-1|$ gives $4x=2x(a+b)$ . So $x=0$ , or $a+b=2$ (the segment $[-1,1]$ of the real axis).

Nearest point to the circle $|z-3|=\frac12$ is $z=1$ : distance $2-\frac12=\frac32$ .

(C)

Topics:Geometry in the complex plane Modulus and argument

← Previous question Next question →

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 15