I am uncertain exactly what is wanted. If it is visualization you can use ContourPlot
:
fun1[_][x_, y_] := y^2 - 2 x^2 (x + 3);
fun2[c_][x_, y_] := c ((y + 1)^2 (y + 9) - x^2);
cp = ContourPlot[
Evaluate[(fun1[#][x, y] == fun2[#][x, y]) & /@ {0, 2}], {x, -10,
10}, {y, -10, 10}, MeshFunctions -> (fun2[2][#, #2] &),
Mesh -> {{0.}}, MeshStyle -> {PointSize[0.015], Red},
ContourStyle -> {Green, Purple}, PlotLegends -> "Expressions"]

The mesh function exploits when c=0 the second function is 0.
If an approximation of the points of intersection is desired you can extract them from graphic: e.g.
ind = Cases[cp, Point[x__] :> x, Infinity];
ans = cp[[1, 1, 1]][[First@ind]]
yielding:
{{2.63803, -8.8862}, {0.441098, -1.15861}, {0.335547, -0.880514}, \
{-0.376551, -0.864387}, {-0.539491, -1.19441}, {-2.71652, -2.02988}, \
{-2.9964, -0.00107654}, {2.64087, -8.88768}, {-2.7232, -2.03298}, \
{-2.99633, -0.00098642}, {-0.369225, -0.870458}, {0.327706, \
-0.882296}, {0.439565, -1.15777}, {-0.513967, -1.18753}}
I re-stress approximates.
If you want to "find":
red = N@Reduce[(fun1[#][x, y] == fun2[#][x, y]) & /@ {0, 2}, {x, y}]
yields:
(x == -3. || x == -2.72156 || x == -0.537205 || x == -0.378744 ||
x == 0.340628 || x == 0.441236 || x == 2.64504 ||
x == -2.8947 - 2.63671 I || x == -2.8947 + 2.63671 I) &&
y == 2.99448*10^-8 (-1.65309*10^7 + 162. x - 1.00263*10^8 x^2 -
3.40627*10^7 x^3 + 6.49485*10^6 x^4 + 4.1719*10^6 x^5 +
638632. x^6 - 19352. x^7 - 3200. x^8)
A way to extract this result (real solutions):
xval = Cases[red, x == r_ :> r, Infinity];
yval[u_] := First[Cases[red, y == r_ :> r]] /. x -> u;
calc = {#, yval@#} & /@ Cases[xval, _Real]
yielding:
{{-3., 0.}, {-2.72156, -2.03094}, {-0.537205, -1.19225}, {-0.378744, \
-0.867192}, {0.340628, -0.88046}, {0.441236, -1.15756}, {2.64504, \
-8.88754}}
Comparing with the approximations the original plot and the above real solutions are overlayed with an "x":
cp2 = Show[cp, ListPlot[calc, PlotMarkers -> {"\[Times]", 20}]]

If this is not what is desired, my apologies.
x^2
andSqrt[x]
might help. In the end you might have that $\endgroup$ – Öskå May 18 '14 at 20:42