JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q2

MathematicsSingle correct+4 / -1Question 2 of 90

If $z=x+iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$ , then

x+2y-4=0

x^2+y-4=0

x+2y+4=0

x^2-y+3=0

Answer and worked solution

Official answer: C

Locus. $|z-i|=|z+5i|$ is the perpendicular bisector of $i$ and $-5i$ , the line $y=-2$ . With $|z|=2$ this gives $z=-2i$ , the point $(0,-2)$ .

Check. $x+2y+4=0-4+4=0$ ; the other three fail.

(C)

Topics:Geometry in the complex plane Modulus and argument

JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 2