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We have two conics $Q_1,Q_2$ on $\mathbb{P}_2$ over some algebraically closed field. Also $Q_1$ and $Q_2$ are supposed to be smooth.
I've just discovered Bezout's theorem, which states that two algebraic curves with degrees $m$ and $k$ have exactly $km$ intersection points (of course some of them can be the same). So, I'd like to build some examples when two conics have exactly $1,2,3$ and $4$ distinct intersection points. All cases but not the $1$st are quite easy. For example, $q_1(x,y,z)=x^2+y^2-z^2$ and $q_2(x,y,x)=x^2+2y^2-z^2$ are two polynomials which define smooth conics on $\mathbb{P}_2$ and they have exactly $2$ intersection points $(1:0:1)$ and $(1:0:-1)$. When we need three or four points, there are similar examples. But what about $1$ point? I cannot build an example. Is it possible? I'd like to note that conics are smooth.

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up vote 6 down vote accepted

The conics $x^2-yz=0$ and $x^2+z^2-yz=0$ are smooth (in all characteristics) and intersect only in one point, namely $[0:1:0]$.

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Take any two pairs of intersecting lines with different slope: $x^2+a_1y^2=0$ and $x^2+a_2y^2=0$ for $a_1 \neq a_2$. They only cross at the origin $(0:0:1)$

Also, take the double y axis $x^2=0$ and the double x axis $y^2=0$ (or any pair of double-lines with different slopes), which again only cross at the origin.

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Dear rewritten, your conics are not smooth. – Georges Elencwajg Jan 8 '14 at 21:40
You are completely right. – rewritten Jan 8 '14 at 21:41

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