Two circles intersect in two points Take for example two circles $$\begin{cases}x^2+y^2=1\\x^2+y^2-x-y=0\end{cases}$$ These two circles intersect in two points namely $(0,1)$ and $(1,0)$. But by Bezout's theorem they must intersect four points. Which points am I missing?
When I take the projective curves $$\begin{cases}X^2+Y^2=Z^2\\X^2+Y^2-XZ-YZ=0\end{cases}$$
I still get two points of intersection. Does these points have multiplicities more than $1$?
 A: If you eliminate $Y$ from both the equations you get
$$-2\,X\,Z^2\,\left(Z-X\right)=0$$
Thus there is a pair of roots at $Z=0$ (points at infinity).
Set $Z=0$ and you get $X^2+Y^2=0$. Setting $X=1$, you get
the two other roots $$(1, i, 0)$$ and $$(1, -i, 0)$$ I had a typo in the earlier answer.
A: Quote from Example section of the Wikipedia entry on Bézout's Theorem (see here):

Two circles never intersect in more than two points in the plane, while Bézout's theorem predicts four. The discrepancy comes from the fact that every circle passes through the same two complex points on the line at infinity. Writing the circle
$(x-a)^2+(y-b)^2 = r^2$
in homogeneous coordinates, we get
$(x-az)^2+(y-bz)^2 - r^2z^2 = 0$,
from which it is clear that the two points $(1:i:0)$ and $(1:-i:0)$ lie on every circle. When two circles don't meet at all in the real plane (for example because they are concentric) they meet at exactly these two points on the line at infinity with an intersection multiplicity of two.

So the four intersection points of your two circles are exactly the two points you already gave and the two points at infinity that are contained in every circle.
A: For Beozout's theorem to work we must work in the complex projective plane, as you noted. Mathematica's Reduce transforms the system
$$x^2 + y^2 = z^2 \ \ \text{and}\ \ x^2 + y^2 - x z - y z = 0$$
to
$$(y\neq 0\land ((x=0\land y=z)\lor (z=0\land (y=-i x\lor y=i x))))\lor (x\neq
   0\land ((y=0\land x=z)\lor (z=0\land (y=-i x\lor y=i x))))$$
We can read off four kinds of solutions from this:


*

*$x = 0$, $y = z \neq 0$, or in homogenous coordinates $0 : 1 : 1$

*$y = 0$, $x = z \neq 0$, or in homogenous coordinates $1 : 0 : 1$

*$y = i x \neq 0$, $z = 0$, or in homogenous coordinates $1 : i : 0$

*$y = - i x \neq 0$, $z = $, or in homogenous coordinates $1 : (-i) : 0$


You have found the first two. The other two are on the "complex line at infinity" so they're a bit easier to miss.
A: The missing two points are the
http://en.wikipedia.org/wiki/Circular_points_at_infinity
From the complex projective point of view, a circle is the same as any other conic, and 5 points are needed specify a conic.  The reason 3 points in the plane are enough to define a circle is that circles are the conics that pass through the 2 "circular points at infinity".  
