Answer
$x^{2}+y^{2}=3,\quad z=0$
Work Step by Step
(distance from $(0,0,1)$ to (x,y,z))=2
$\sqrt{x^{2}+y^{2}+(z-1)^{2}}=2$
(distance from $(0,0,-1)$ to (x,y,z))=2
$\sqrt{x^{2}+y^{2}+(z+1)^{2}}=2$
2=2, so we equate the LHS's
$\sqrt{x^{2}+y^{2}+(z-1)^{2}}=\sqrt{x^{2}+y^{2}+(z+1)^{2}}\quad$
square both sides
$ x^{2}+y^{2}+(z-1)^{2}=x^{2}+y^{2}+(z+1)^{2}\quad$ ... simplify
$(z-1)^{2}=(z+1)^{2}$
$z^{2}-2z+1=z^{2}+2z+1$
$-4z=0$
$z=0$
(this set of points is in the xy-plane)
Substitute $z=0$ in $x^{2}+y^{2}+(z-1)^{2}=4,$
$x^{2}+y^{2}+1=4,\quad z=0$
$x^{2}+y^{2}=3,\quad z=0$
(circle of radius $\sqrt{3}$ in the xy-plane, centered at the origin).
