Algebraic Geometry Santa Cruz 1995, Part 2: Summer Research by American Mathematical Society, János Kollár, Robert

By American Mathematical Society, János Kollár, Robert Lazarsfeld

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Additional resources for Algebraic Geometry Santa Cruz 1995, Part 2: Summer Research Institute on Algebraic Geometry, July 9-29, 1995, University of California, Santa Cruz

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The results of the previous section show that a circle intersects a line in two distinct points, just one point, or not at all. We can therefore conclude that any two distinct circles intersect in two distinct points, just one point, or not at all. In particular, the circles 28 Circles C C C D D D Fig. 2. Three ways in which circles can intersect intersect in a single point exactly when L touches C, D at the same point: in that case we say that C, D touch at that point. e. have the same centre.

Here is the promised pay-off. 3 Suppose that every point on the line L lies on a conic Q. Then Q = L L for some line L . In particular, that is the case when L meets Q in more than two points. ) Proof We can assume L it is not parallel to the y-axis. 13) holds. For every x there is a unique value of y for which L(x, y) = 0, and hence Q(x, y) = 0. That means that J (x) = 0 for all x: since a non-zero quadratic has ≤ 2 roots, that means J is identically zero, so Q = L L . 1 tells us that every point on L lies on Q, so we reach the same conclusion.

That reduces our problem to that of finding a practical criterion for the origin to be a centre for the general conic Q(x, y) = ax 2 + 2hx y + by 2 + 2gx + 2 f y + c. 2 The origin is a centre for a general conic ( ) if and only if the coefficients of the linear terms x, y are both zero. Proof The origin is a centre of Q if and only if the following conics coincide. e. the coefficients of the linear terms x, y are zero Q(x, y) = ax 2 + 2hx y + by 2 + 2gx + 2 f y + c Q(−x, −y) = ax 2 + 2hx y + by 2 − 2gx − 2 f y + c.

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