Answer
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=1$
Work Step by Step
The difference of $r$ and $r_0$ can be expressed as a new vector $r_{diff}=⟨x-x_0,y-y_0,z-z_0⟩$. We are told that $|r_{diff}|=1$. Therefore, $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}=1$. By squaring both sides:
$$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=1$$
The points that fulfill the equation above form a sphere of radius $1$ and center at $(x_0, y_0, z_0)$.
