Undo this transformation I have two variables $x$, $y$ and calculate the following:
$a = \frac{x}{\sqrt{x^2+y^2}}$, $b = \frac{y}{\sqrt{x^2+y^2}}$
Using $a$ and $b$ is there a way I can derive my original $x$ and $y$?
 A: No. You can't
For $x=1,y=0$ you get $a=1,b=0$ also for $x=2,y=0$ you still get $a=1,b=0$. So given $a,b$ there is no way to determine $x,y$.
Note however that $a/b$ gives you $x/y$ (when defined, otherwise it tells you whether $y$ is zero or not). This is the best you can do as the pairs $(x,y)$ and $(cx,cy)$ will give you the same $a$ and $b$.
A: This the classical normalization operation : 
$$\binom{a}{b} = \frac{1}{\sqrt{x^2+y^2}} \underbrace{\binom{x}{y}}_V=\frac{1}{\|V\|} V$$
transforming a vector into the proportional vector with unit norm ( = length) thus belonging to the unit circle.
It is like a projection onto a straight line (imagine the circle is unrolled). And, like a projection, there exists an infinity of vectors $(x,y)$ that have the same vector $(a,b)$ with unit length as their image. Thus this transformation has no inverse.
For example $\binom{0.6}{0.8}$ is the image of $\binom{3}{4}$, $\binom{6}{8}$, $\binom{9}{12}$,...
A: By squaring we get $$a^2=\frac{x^2}{x^2+y^2},b^2=\frac{y^2}{x^2+y^2}$$ and we get
$$a^2+b^2=\frac{x^2+y^2}{x^2+y^2}=1$$
and you can not compute $x$ or $y$.
