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

$\begin{array}{l}|\mathrm{z}-\mathrm{i}|-|\mathrm{z}+5\mathrm{i}|=0\\ \Rightarrow |\mathrm{x}+(\mathrm{y}-1\left)\mathrm{i}\right|=|\mathrm{x}+(\mathrm{y}+5\left)\mathrm{i}\right|\\ {\mathrm{x}}^2+(\mathrm{y}-1{)}^2={\mathrm{x}}^2+(\mathrm{y}+5{)}^2\\ (\mathrm{y}-1{)}^2-(\mathrm{y}+5{)}^2=0\\ (2\mathrm{y}+4)(-6)=0\\ \mathrm{y}=-2\\ \therefore {\mathrm{x}}^2+(-2{)}^2=4\\ \mathrm{x}=0\\ \mathrm{Z}\equiv (0,-2),\text{ check options }\end{array}$