Suppose you have two quadratic polynomials in two real variables, f(x, y) and g(x, y), and you want to know whether the two curves
f(x, y) = 0
g(x, y) = 0
intersect, and if they do, find where they intersect.
David Eberly has a set of notes on solving systems of polynomial equations that give the conditions you need. The conditions are quite complicated, but they are explicit and do not involve any advanced math. He gives 14 equations that combine to form the coefficients in a 4th degree polynomial that will tell you whether the curves intersect, and will get you started on finding the points of intersection.
There is a more elegant solution, but it requires much more background. It requires working in the complex projective plane ℙ² rather than the real plane ℝ². What does that mean? First, it means you work with complex numbers rather than real numbers. Second, it means adding “points at infinity.” This sounds awfully hand-wavy, but it can all be made rigorous using homogeneous coordinates with three complex numbers.
If you haven’t seen this before, you may be thinking “You want me to go from pairs of real numbers to triples of complex numbers? And this is supposed to make things easier?!” Yes, it actually does make things easier.
Once you move to ℙ², the question of whether the curves f(x, y) = 0 and g(x, y) = 0 intersect becomes trivial: yes, they intersect. Always. A special case of Bézout’s theorem says any two quadratic curves are either the same curve or they intersect in at 4 points, counting intersections with multiplicity.
Your original question was whether the curves intersect in ℝ², and now that question reduces to asking whether the 4 points guaranteed by Bézout’s theorem lie in ℝ². They might not be real, or they might be points at infinity.
Even if you go by David Eberly’s more elementary approach, you still face a similar question. You have a fourth degree polynomial, and you need to know whether it has real roots. How would you do that? You’d probably find all four complex roots first and see whether any are in fact real. Moving from the real numbers to the complex numbers makes it easier to find the roots of polynomials, and moving to complex projective space carries this idea further.
In the real plane ℝ² the zero set of a quadratic polynomial can be an ellipse, a parabola, or a hyperbola. (The circle is a special case of an ellipse.) But in ℙ² there is only one kind of conic section: there is a change of variables that turns the zero set of any quadratic polynomial into a circle. So now we’ve reduced the problem to finding where two circles intersect.
Finding the intersection of two conic sections illustrates the principles discussed in the preface to Algebraic Geometry: A Problem Solving Approach.
One way of doing mathematics is to ask bold questions about the concepts your are interested in studying. Usually this leads to fairly complicated answers having many special cases. An important advantage to his approach is that questions are natural and easy to understand. A disadvantage is that proofs are hard and often involve clever tricks, the origins of which are very hard to see.
A second approach is to spend time carefully defining the basic terms, with the aim that the eventual theorems and proofs are straightforward. Here the difficulty is in understanding how the definitions, which often initially seem somewhat arbitrary, ever came to be. The payoff is that the deep theorems are more natural, their insights more accessible, and the theory made more aesthetically pleasing.
- Fundamental theorem of algebra
- Working every problem in the book
- The Jerico-Masada approach to math
- Advantages of redundant coordinates
- What is an elliptic curve?
One thought on “Finding where two quadratic curves intersect”
I do beg your pardon; but “two circles”, while topologically correct, seems to obscure just how we get four points of intersection(for euclidean circles we expect two real intersections or two intersections on the imaginary radical axis and a tangency at infinity); but i believe a circle and a parabola should cover all possible cases.