# Why can two non-overlapping circles intersect in at most two points, while two non-overlapping ellipses can intersect at four?

When reading about why no Venn diagram for four sets can be formed by intersecting four circles, I found that the author claimed that any two distinct circles can intersect in at most two points, while any two distinct ellipses can intersect in at most four.

Why is this? I can easily see from examples that it's intuitively obvious, but is there a geometric reason for it?

Thanks!

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You know the fact that any circle can be chartereised by any three points it passes through, i.e, If a circle passes through through 3 points then it is unique.Therefore two distinct circles can intersect at atmost 2 points.

In case of ellipses, they are characterised by five points.

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Surely you mean five points for ellipses? If an ellipse was characterized by four points, then two distinct ellipses could not have four points in common. –  Rahul Oct 5 '12 at 5:33

With the right viewpoint, any two distinct circles intersect in exactly four points, indeed any two distinct nondegenerate conics intersect in exactly four points. Like the circle of radius $2$ and the hyperbola $xy=1$. The relevant fact is the Theorem of Bézout, which says that under certain conditions of nondegeneracy, two plane curves, of degrees $m$ and $n$ respectively, intersect in $mn$ points, as long as you count multiplicity, set the points in the projective plane, and look for coordinates of intersection-points in an algebraically closed field. The equation of the circle $(x-1)^2 +y^2=1$, for instance, becomes $X^2-2XZ+Z^2+Y^2=Z^2$ upon homogenization, and it has the two distinct (in projective plane) points $(1,i,0)$ and $(1,-i,0)$. These are the two “imaginary points at infinity” carried by the circle, and they are common to all circles.

Two concentric circles intersect at these two points, and are tangent at both, so each point must be counted with multiplicity $2$.

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