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

Given, $z = x + iy$

Also, \left|z\right| = 2$</p><p>So,${x}^{2} + {y}^{2} = 4 . . . \left(1\right)$</p><p>Now, \left|z - i\right| = \left|z + 5i\right|$

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

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

$⇒ y = - 2 . . . \left(2\right)$

So, $x = 0$ by solving equation \left(1\right) \text{ and } \left(2\right)$</p><p>Hence, only$x + 2y + 4 = 0 is true, as point only satisfy this equation.