Intersection points of a circle and a parabola On solving to find the intersection points of a circle ($x^2+y^2=2$) and a parabola ($x^2=y$), we get a quadratic equation $$y^2+y-2=0$$ which gives two values  $y=1$ and $y=-2$.
With $y=1$ we get two points of intersection $(1,1)$ and $(1,-1)$.
But $y=-2$ doesn't lie neither on the circle nor on the parabola.
So why does the negative solution exist?
Thank you!
 A: $y=-2$ doesn't lie on the parabola. That much is true. But , we are looking at this only in the real plane. For $y=-2$ , we will get complex solutions, which we do not see on the real plane.
To find the complex solutions:
$$x^2+y^2=2$$
$$x^2+4=2$$
$$x^2=-2$$
$$x=\pm i\sqrt2 $$
A: According to a wonderful theorem of Bézout, two conic sections will ordinarily intersect in four points, counting multiplicity. But you have to allow for points with complex coordinates as well, and these will not be seen  in the real picture that we can draw.
Your two values of $x$ are $-2\pm\sqrt6$. If my hand computation is correct, the points with negative $x$-coordinate are $\bigl(-2-\sqrt6,\pm\sqrt{-9-4\sqrt6}\,\bigr)$, definitely complex.
I suppose you may ask, in the case of two circles, where the remaining two intersections are; and worse, where the four intersections of concentric circles are. There you have to go back and get the full statement of Bézout: that we must look not only in the finite plane, but in the projective plane too. But that is going beyond what I want to write about this late in the evening.
