Given, $z=x+iy$

Also, $\left|z\right|=2$

So, $x^2+y^2=4 ...\left(1\right)$

Now, $\left|z-i\right|=\left|z+5i\right|$

$\Rightarrow {\left|z-i\right|}^2={\left|z+5i\right|}^2$

$\Rightarrow x^2+{\left(y-1\right)}^2=x^2+{\left(y+5\right)}^2$

$\Rightarrow y=-2 ...\left(2\right)$

So, $x=0$ by solving equation $\left(1\right) \& \left(2\right)$

Hence, only $x+2y+4=0$ is true, as point only satisfy this equation.