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

If you have an irrational tangent point between your curves, it must have at least another conjugate, and since tangent points are double points, by Bézout's theorem, the product of the degrees of the curves must be at least $4$.
Since dtldarek gave an example of a degree $4$ curve and a degree $1$ curve, let me give an example between two degree $2$ curves :
The circle $x^2+y^2= 1$ and the ellipse $17x² + 8y² + 12x = 4$ are tangent at $(-2/3, \pm \sqrt {5}/3)$

