JEE MAIN - Mathematics (2022 - 26th July Evening Shift - No. 2)

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

$$x+2 y-4=0$$

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

$$x+2 y+4=0$$

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

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

So, $$\mathrm{z}$$ lies on $${ \bot ^r}$$ bisector of $$(0,1)$$ and $$(0, - 5)$$

i.e., line $$y = - 2$$

as $$|z| = 2$$

$$ \Rightarrow z = - 2i$$

$$x = 0$$ and $$y = - 2$$

so, $$x + 2y + 4 = 0$$