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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 $1$?

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The other two intersections are on the complex plane. – JChau Feb 17 '14 at 6:02
Could you please elaborate? – Grobber Feb 17 '14 at 6:05
For example by providing these complex points. Mathematica cannot find them. – Andrej Bauer Feb 17 '14 at 6:06
Note that solitions at $\infty$ also count, and that you have to take into account multiplicities of zeros (although that does not seem to be an issue here). – Andrej Bauer Feb 17 '14 at 6:07
Yeah I have considered infinite points. – Grobber Feb 17 '14 at 6:09
up vote 1 down vote accepted

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.

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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.

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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.

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The missing two points are the

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".

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