There is an extensive theory of the real and complex roots of the chromatic polynomial of a graph, a substantial fraction of this being due to the connections between the chromatic polynomial and a particular limit of the partition function of the $q$-state Potts model.
More generally we can consider the full Tutte polynomial $T(x,y)$ which is now a bivariate polynomial, and now precisely equivalent to the partition function of the $q$-state Potts model. Much work has been done on the values that this function takes and the theoretical difficulty of computing the values both for individual points and curves on the $xy$-plane.
An obvious thing that one might try in order to get a different angle on the roots of the chromatic, flow etc. polynomials is to consider the level curves of the Tutte polynomial in the $xy$-plane, particularly the curves corresponding to $T(x,y) = 0$.
If we do some computation, say for the Petersen graph, then a superficially interesting plot emerges, especially the zero curve.
As the Petersen graph has no $1$-colourings and $2$-colourings, then it must be the case that $T(0,0) = 0$ and $T(-1,0) = 0$ and indeed we see this. My question is therefore:
Question: Are there connections between the nature and/or location of the zero curves of the Tutte polynomial in the $xy$-plane and the structure of the graph?
This seems such an obvious thing to do that I am sure it must have been done before, so perhaps the dearth of published results means "no, you can't tell anything from these plots except the obvious", but I thought I would check here.
I also tried to find some literature on "zeros of bivariate polynomials" but I got lost in a maze of results about numerical computation of these curves.
ADDITION
As requested, here are the commands used to create this plot (in Sage):
x,y = var('x','y')
tp = graphs.PetersenGraph().tutte_polynomial()
xmin = -4
ymin = -4
delta = 5
tplot = contour_plot(tp,(x,xmin,xmin+delta),(y,ymin,ymin+delta),plot_points=400,colorbar="true",cmap="coolwarm")
show(tplot,gridlines=true)