Can two integer polynomials touch in an irrational point? We define an integer polynomial as polynomial that has only integer coefficients. Here I am only interested in polynomials in two variables.
Example:


*

*$P = 5x^4 + 7 x^3y^4 + 4y$


Note that each polynomial P defines a curve by considering the set of points where it evaluates to zero. We will speak about this curve.
Example: 
The circle can be described by


*

*$x^2 + y^2 -1 = 0$


We say two polynomials $P,Q$ are touching in point $(a,b)$ if $P(a,b) = Q(a,b) = 0$ and the tangent at $(a,b)$ is the same. Or more geometrically, the curves of $P$ and $Q$ are not crossing.

(The Figure was created with IPE - drawing editor.)
We also need a further technical condition. For this let $D$ be a ''small enough'' disk around $(a,b)$. Then $Q$ and $P$ define two regions indicated green and yellow. Those regions must be interior disjoint.  Without this condition for $P = y-x^3$ and $Q=y$ the point $(0,0)$ would be a touching point as well. See also the right side of the figure.
(I know that I am not totally precise here, but I don't want to be too formal, so that I can reach a wide audience.)
(Thanks for the comment from Jeppe Stig Nielsen.)
Example:


*

*$P = y - x^2$  (Parabola)

*$Q = y$  ($x$-axis)


They touch at the origin $(0,0)$.
My question: 
Does there exist two integer polynomials $P,Q$ that touch in an irrational point $(a,b)$? 
(It would be fine for me if either $a$ or $b$ is irrational)
Many thanks for answers and comments.
Till

Here's a general way to find such examples where both curves are of the form $y=f(x)$.  Notice that $y=f(x)$ and $y=g(x)$ meet at a given value of $x$ iff that value of $x$ is a root of the polynomial $h(x)=f(x)-g(x)$, and they have the same tangent line iff that value is a root of $h(x)$ of multiplicity greater than $1$.
So this means that to find an example, we just need a polynomial $h(x)$ with integer coefficients that has a double root at some irrational value of $x$ (we can then take $g(x)$ to be any polynomial with integer coefficients at all, and $f(x)=h(x)+g(x)$).  This is easy to do: just take any polynomial $p(x)$ with integer coefficients and an irrational root, and let $h(x)=p(x)^2$.
