Answer
$x^{2}+y^{2}=3,\quad z=0$
Work Step by Step
The distance from $(0,0,1)$ to$ (x,y,z)=2$ is:
$\sqrt{x^{2}+y^{2}+(z-1)^{2}}=2$
The distance from $(0,0,-1)$ to $(x,y,z)=2$ is:
$\sqrt{x^{2}+y^{2}+(z+1)^{2}}=2$
The distances are equal, 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 $x^{2}+y^{2}+(z-1)^{2}=4,$
$x^{2}+y^{2}+1=4,\quad z=0$
$x^{2}+y^{2}=3,\quad z=0$
We get a circle of radius $\sqrt{3}$ in the xy-plane, centered at the origin.
