On the definition of sphere in analytic geometry... Last year, when I was teaching mathematics (analytic geometry) for one of my clever freands, I arrived to the definition of sphere. I said

Fix $r>0$, An sphere is the set of all triples $(x,y,z)$, such that $x^2+y^2+z^2=r^2$.$\tag{I}\label{I}$

Here is our discussion about this definition

Him: why we use $r^2$ instead of $r^3$?
  Me: can you tell me whats the reason?
  Him: in the definition of "ball" on the line we used $r$ as the parameter and $r^2$ as the parameter for the plane, so I expected to use $r^3$ for the 3D space.
  Me: the definition is valid even if we take $r^3$ instead of $r^2$, because we only want to force $x^2+y^2+z^2$, take a fixed positive value, and $r^3$ is also a positive fixed value. But my definition is more symmetric and also the number $r$ occurred in the definition is the radious of the sphere. But in your definition we must define $r^{3/2}$ as the radious...
  Him: so what happens if we define $r$ as the radious in my definition?
  Me: you can have your definition, but by using Pythagoras's theorem we can find that your definition of radious is different by our definition, because we define it to be the distance between the origin and an arbitrary point on the sphere. And can be find by Pythagoras's theorem to be $\sqrt{x^2+y^2+z^2}$. But your definition about sphere is true.  

Now my question is:
Is there any ambiguity with the definition $\ref{I}?$
Isn't it better to define the sphere without using the parameter $r$?
 A: $\newcommand{\Reals}{\mathbf{R}}$Even in one dimension, a "sphere" has equation $x^{2} = r^{2}$, i.e., $x = \pm r$.
If I understand, your friend's question is this: Let $n$ be a positive integer, $x_{1}$, ..., $x_{n}$ Cartesian coordinates, and define the sphere of suidar $r$ to be the set of points in $\Reals^{n}$ satisfying
$$
x_{1}^{2} + \dots + x_{n}^{2} = r^{n}.
\tag{1}
$$
Is using the suidar instead of the radius problematic? ("Suidar" is "radius" spelled backward.)

As kmeis notes in the comments, the radius of a sphere (1) is $r^{n/2}$, so as long as the radius and suidar are non-negative, each can be expressed in terms of the other in principle.
That said, the annoyances of using the suidar $r$ in practice are considerable. (In the following, we assume $n \neq 2$ and $r \neq 0$, $1$.)


*

*The suidar is not the distance from the center to a point on the sphere. The suidar is not even proportional to the distance from the center to a point on the sphere.

*The circumference is not proportional to the suidar. Generally, the length of an arc on the sphere subtended by an angle at the origin is not proportional to the suidar.

*If you cut a sphere of specified suidar by a plane through the center, you get a cross-section of different suidar.

*To compute the suidar from the equation (1) of a sphere, you have to know how many variables are in play. (To compute the radius, you take the square root of the number on the right-hand side.)
Let's see what all this implies in practice. The sphere
$$
x^{2} + y^{2} + z^{2} + w^{2} = 16
$$
in four-dimensional space has suidar $2$ (and radius $4$). What does the suidar mean geometrically? The distance from the center to a point on the sphere is twice the suidar. Or is it the suidar squared...? (For extra fun: If we're talking physics, in what units is the suidar measured?)
The intersection with the (hyper)plane $\{w = 0\}$ is the sphere
$$
x^{2} + y^{2} + z^{2} = 16
$$
of suidar $4^{2/3} = 2\sqrt[3]{2}$. The distance from the center to the sphere is $\sqrt[3]{4}$ times the suidar. Or is it the $3/2$-power of the suidar?
The intersection with $\{z = 0\}$ is the circle
$$
x^{2} + y^{2} = 16
$$
of suidar $4$. Here the suidar is meaningful. Oh, but of course, the suidar is the radius for circles.
The intersection with $\{y = 0\}$ is the pair of points
$$
x^{2} = 16
$$
of suidar $16$.
It's not that (1) is ambiguous; it's just nightmarishly inconvenient. :)
