Rational parametrization of algebraic variety Suppose I have some algebraic variety (i. e. solution of system of polynomial equations). Sometimes I can find rational parametrization of it - for example, it case of circle defined by $x^2 + y^2 - 1 = 0$ I can parametrize all solutions $x = \frac{2t}{1+t^2}, y = \frac{1-t^2}{1+t^2}.$ Sometimes I can't.
So does algebraic geometry (or maybe other mathematics) provide general setting for decision whether or not given algebraic variety (not only curve or surface) has rational parametrisation? 
Or these varieties are exactly rational varieties? If so, it's not clear for me, so could anybody explain the link between this object and my question.
 A: Yes, your question is closely related to rational varieties. 
As the article you link to and Cass' comment explain, a rational variety $X$ of dimension $n$ is one that has a dense open subset that is isomorphic to an open subset of affine space $\mathbf A^n$. This isomorphism allows us to identify the coordinate functions on $X$ with rational functions on $\mathbf A^n$, just as in your example. So this gives a rational parametrisation of the points on $X$.
Conversely, if you have a rational parametrisation of the points of $X$, plugging in values of the parameters gives a dominant rational map $f:\mathbf A^m \dashrightarrow X$. (We don't demand that parametrisations be defined for all possible values of the parameters, just a dense open set of the possible values. Note also that our parametrising affine space is allowed to have bigger dimension than $X$.)
A variety $X$ for which one can find a map $f$ as described above is called unirational. It's an easy exercise to see that if there is a dominant rational map as above, then one can find a dominant map $\varphi :\mathbf A^n \dashrightarrow X$ from an affine space of the same dimension as $X$. Such a map must be generically finite, so it gives a finite-to-one parametrisation of the points of $X$.
However: unirational does not imply rational! It was known by around 1900 that a curve or surface that has a finite-to-one parametrisation must actually have a generically one-to-one (in other words birational) parametrisation. The question of whether this remains true in higher dimensions remained open until the early 1970s, when, remarkably, three completely independent proofs (by Artin–Mumford, Clemens–Griffiths, and Iskovskikh–Manin) showed that a smooth cubic hypersurface in $\mathbf P^4$ is unirational but not rational. 
In general, deciding whether varieties are rational or not remains one of the most fundamental and difficult issues in algebraic geometry.
On the other hand: although there are some cases in which deciding rationality is extremely difficult, the comment of Cass that varieties are "rarely" rational is quite accurate, because there are easliy-verified necessary conditions for rationality that rule out "most" candidates. 
For example, let $X$ be a smooth hypersurface in $\mathbf P^n$. Then the adjunction formula shows that if $X$ has degree at least $n+1$, then the canonical bundle $\omega_X = \wedge^{n-1} \Omega_X$ has nonzero global sections. This is impossible for a rational variety, and so we conclude that no smooth hypersurface of degree at least $n+1$ is rational.
