Suppose $\displaystyle x,y \in R^3 $such that $\displaystyle |x-y| \geq 0.$

1. Show the set of points $\displaystyle z \in R^3$ such that $\displaystyle |z-y|=|z-x|$ is the plane

$\displaystyle u + \frac{x+y}2 : u \dot (x-y)=0$

2. Determine the $\displaystyle r \geq 0$ such that the set of points $\displaystyle |z-y|=|z-x|=r$ is a circle.