1. ## Real analysis

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.

2. Have you done nothing on this yourself? Given two points, x and y, in 3 dimensional space, there exist a line segment between them. Imagine the plane perpendicular to that line, passing through its midpoint. What can you say about that plane?

Imagine the circle in that plane, center at the midpoint of the original line segment, radius R. What can you say about the distances from any point on that circle to x and y?

3. I can visualize a general answer in my head, but I'm having difficulty actually showing it.