TITLE: Roots of bivariate polynomials
QUESTION [6 upvotes]: A bivariate polynomial of degree $m+n$ is,
$ p(x,y)  = \sum_{k=1}^n\sum_{j=1}^m a_{jk}x^ky^j$
where $a_{mn}\neq0$ and $a_{jk}\in\mathbb{R}$ for $1\leq j\leq m$, $1\leq k\leq n$. 
Univariate polynomial root-finding is ill-conditioned because a small change in coefficients cause characteristic changes in the roots. For example, $x^2 + 2x+1$ has a repeated root at $x=-1$ but $x^2+ (2-\epsilon)x+1$ has two complex roots for any small $\epsilon>0$.  I was wondering about examples of things that can go wrong with small perturbations of the coefficients in the bivariate case. 

Is there an example of a bivariate polynomial $p(x,y)$ such that $p$ has real roots lying along a line/curve but one small perturbation of the coefficients causes there now to be real roots only at isolated point(s)?
An example of a bivariate polynomial with real roots lying along a line but after a small perturbation having no real roots is, of course, $p(x,y) = x^2+2x+1$.

Thank you.

REPLY [2 votes]: I'm not sure I understood the meaning of "real roots lying along a line/curve"; if roots are real numbers, how can they lie along a curve? Anyway, the following example might fit the intended meaning of the question:
The polynomial $p(x,y)=(x^2+2x+1)((x-1)^2+y^2)$ has roots along the line $x=-1$. The perturbed polynomial $(x^2+(2-\epsilon)x+1)((x-1)^2+y^2)$ has a sole root at $(1,0)$.