Number of solutions to a system of three quadratic equations in two variables I have a system composed of three quadratic equations,
$$\left\{ \begin{array}{rcl} a_{1}x^{2}+b_{1}xy+c_{1}y^{2}+d_{1}x+e_{1}y+f_{1}=0 \hfill\\ a_{2}x^{2}+b_{2}xy+c_{2}y^{2}+d_{2}x+e_{2}y+f_{2}=0\hfill\\ a_{3}x^{2}+b_{3}xy+c_{3}y^{2}+d_{3}x+e_{3}y+f_{3}=0\hfill\\ \end{array} \right.$$
Its coefficients are expressed by various parameters, and the coefficients are very complex. I know so far that this system has a solution, and I want to prove that it has no infinite set of solutions, what's the best way to think about it ? The shape of the solution to a single equation is not certain, it may be an oval, it may be hyperbolic or other shapes. At the same time, I also tried a series of simultaneous equation methods such as elimination, but because the coefficients are very complicated, they will become more complicated after addition, subtraction, multiplication and division.
 A: We assume of course that a preliminary check has been made to be sure that among the 3 equations
$$\left\{ \begin{array}{rcl} a_{1}x^{2}+b_{1}xy+c_{1}y^{2}+d_{1}x+e_{1}y+f_{1}=0 \hfill\\ a_{2}x^{2}+b_{2}xy+c_{2}y^{2}+d_{2}x+e_{2}y+f_{2}=0\hfill\\ a_{3}x^{2}+b_{3}xy+c_{3}y^{2}+d_{3}x+e_{3}y+f_{3}=0\hfill\\ \end{array} \right.$$
none is proportional to any other one.
Here is a "method of attack" of this issue, as you have asked it.
First of all : do you know that a quadratic expression
$$ax^2+bxy+cy^2+dx+ey+f=0\tag{1}$$
has an associated matrix which is:
$$M=\pmatrix{a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f}\tag{2}$$
(See here for details)
(Of course, as we can multiply (1) by any nonzero factor, matrix $M$ must be thought defined "up to a multiplicative factor").
If $\Delta:=det(M)\ne 0$, the corresponding curves are non-degenerated conic curves (reminder : ellipses if $\delta < 0$, parabolas if $\delta = 0$, hyperbolas if $\delta > 0$ where $\delta:=b^2-4ac$).
In the other case, we have a degenerate case : pairs of lines like in this example :
$$\underbrace{x^2-2xy+y^2-1=0}_{(x-y-1)(x+y+1)=0} \ \iff \ \underbrace{(x-y-1)=0}_L \ \text{or} \ \underbrace{(x+y+1)=0}_{L'}$$
If each of the 3 equations is non-degenerated, the associated curves cannot have an infinite number of common points (the maximal number of common 2 conic curves have a maximum of 4 common points).
If, on the contrary,

*

*only one or two of of them is degenerated, still no problem (for example, even if two of them are degenerated, generating 4 lines, these 4 lines will intersect the third non-degenerated conic curve in a finite number of points).


*the three of them are degenerated, for example the first one is $(L_1)(L'_1)=0$, $(L_2)(L'_2)=0$, $(L_3)(L'_3)=0$ with for example $L_1=L_2=L_3$, the infinity of points with equation $L_1=0$ will be solutions.
Remark: It is useful to know that (notations of (2)) :

*

*rank$(M)=3$: non-degenerate conic.


*rank$(M)=2$: degenerate conic into the union of 2 distinct lines.


*rank$(M)=1$: degenerate conic into the same line (taken twice).
(The rank can be calculated as $3$ minus the number of null eigenvalues).

Now that the objectives are clearer (so I think...), it remains to transform the cases above into an algorithmic form. Up to you...
