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.

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...
