Thomas' Calculus 13th Edition

$x^{2}+y^{2}=3,\quad z=0$
(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).