Equivalence Class of Circles Centred at the Origin: $ \{ (c, d) : c^2 + d^2 = r \}$?

Question

I have a question regarding this equivalence relation ($R = \{ ((a, b), (c, d)) : a^2 + b^2 = c^2 + d^2 \}$) and its equivalence classes.

As said in the answers to the aforementioned question, this is the set containing all of the equivalence classes:

$\{ (c, d) : c^2 + d^2 = r \}$

And we are told that it is the equivalence class of all circles entered at the origin. However, the equation of a circle, of course, is $c^2 + d^2 = r^2$.

Evidently, there is something that I am not understanding here. I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

First: $\{(c,d) : c^2 + d^2 = r\}$ is not the set containing all of the equivalence classes. It is one way to write one of the equivalence classes, the equivalence class of a point $(a,b)$ having the property that $a^2 + b^2 = r$.

Second: It's not the equivalence class of all circles - in any situation, "the equivalence class of all $X$'s" would need to include some $X$'s! This set has no circles in it, only points. What is true - and what is stated in the answer you're reading - is that the set $\{(c,d) : c^2 + d^2 = r\}$ is a circle, and all of the equivalence classes of this relation are circles.

Third: Yes, the equation of a circle is usually $c^2 + d^2 = r^2$. Except that it's not - it's usually $x^2 + y^2 = r^2$. The difference between $x$ and $c$ didn't bother you, because it's just a name; variables can be named whatever you like. The thing is, $r$ is just a variable, too. It doesn't have to mean "radius". And in this case, it doesn't; in the set you're looking at, the author of the original answer used $r$ to mean $a^2 + b^2$, which is the square of the distance between $(a,b)$ and the origin. If it helps, you could use $s$ instead (so $s = a^2 + b^2$), rewrite the set as $\{(c,d):c^2 + d^2 = s\}$, and then say $r = \sqrt{s}$ (so $s = r^2$ and we're looking at $c^2 + d^2 = r^2$).

Answer 2

$(x_1,y_1)$ is a point on a cirlce a radius, $r$ units from the origin.

That point satisfies the equation $x_1^2+y_2^2 = r^2$

And $(x_2,y_2)$ is lies in the same cirlce. $x_2^2 + y_2^2 = r^2$

And setting $r^2 = r^2$ in our system of equations.

$x_1^2 + y_1^2 = x_2^2 + y_2^2$

For any $r,$ if two points $(x,y)$ and $(c,d)$ lie on the same circle centered at the origin $x^2+y^2 = c^2+d^2$