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

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4-1.Complex numbers

hard

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

A

$x +2 y -4=0$

B

$x^{2}+y+4=0$

C

$x^{2}+y-4=0$

D

$x^{2}-y+3=0$

(JEE MAIN-2022)

Solution

$|z-i|-|z+5 i|=0$

$|x+(y-1) i|=|x+(y+5) i|$

$x^{2}+(y-1)^{2}=x^{2}+(y+5)^{2}$

$(y-1)^{2}-(y+5)^{2}=0$

$(2 y+4)(-6)=0$

$y=-2$

$\therefore x^{2}+(-2)^{2}=4$

$x=0$

$Z \equiv(0,-2)$, check options

Standard 11

Mathematics

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