Substitution on $\mathbb{R}$ as isomorphisms

Question

I was trying to figure out different ways to find the shortest distance between a curve such as $(x+a)^2+(y+b)^2 = r ^2$ and a line, say $y=mx+b$. One method I tried was to do a substitution $u = x+a$ and $v = y+b$, since I prefer having a more cumbersome linear expression as opposed to a more complicated circle.

Now, proving that this mapping preserves the structure of the plane was the main difficulty. I thought it would be as simple as just finding a isomorphism between them, and tried to use the definition of a holomorphism between two rings, and here is what I got:

Since $f(x,y) = (x+a,y+b)$
$$f(x_1 +x_2,y_1 + y_2) = ((x_1+x_2)+a),(y_1+y_2)+b)\neq ((x_1+a)+(x_2+a),(y_1+b) +(y_2+b)) = f(x_1,y_1) + f(x_2,y_2)$$

Which I think serves as a proof that there can't be a holomorphism between the rings, and hence no isomorphism.

But intuitively, shouldn't all structures be preserved? I mean, the distance between the circle and the line will remain the same?

My guess is that I might be wrong in assuming that it's a mapping between to rings, but I can't see why that wouldn't be the case.

I'm not at all interested in an alternative way to solve the problem, my question is:

Why isn't the mapping given above a isomorphism, alternatively, where did my reasoning fail, can I motivate the mapping in some other way?

Answer 1

What you need is that $f$ preserves distances. This means that $f$ is isometric.

It is irrelevant whether $f$ is an isomorphism between rings, groups, or real vector spaces. In this case all those different isomorphisms mean essentially the same thing, but in general you should be careful as a number of different things can be called isomorphisms. In fact, an isometry is an isomorphism between metric spaces, so it all depends on the structure you are looking at.

The kind of isomorphism you mean (addition-preserving) always needs to fix the origin: $f(0)=0$. But this doesn't work since $f$ is a shift. The mapping $f$ you have is a good way to simplify the problem, and $f$ preserving addition wouldn't really help. The simple explicit formula is helpful.