How to solve system of equations \(\displaystyle{A}_{{1}}{x}^{{2}}+{B}_{{1}}{x}{y}+{C}_{{1}}{y}^{{2}}+{D}_{{1}}{x}+{E}_{{1}}{y}+{F}_{{1}}={0}\)

You have the following system of quadratic equations:
${\displaystyle A_1{x}^2+B_{1}xy+C_1{y}^2+D_{1}x+E_{1}y+F_1=0}$
and ${\displaystyle A_2{x}^2+B_{2}xy+C_2{y}^2+D_{2}x+E_{2}y+F_2=0}$
First, transform each of the given quadratic equations into quadratic form notation as follows. Define $r=\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]$, then the first equation can be written as the quadratic form ${\displaystyle r^T{Q}_{1}r=0}$ where
$Q_1=\left[\begin{array}{ccccc}{A}_1& & \frac{B_1}{2}& & \frac{D_1}{2}\\ \frac{B_1}{2}& & C_1& & \frac{E_1}{2}\\ \frac{D_1}{2}& & \frac{E_1}{2}& & F_1\end{array}\right]$
Similarly, the second quadratic equation can be expressed as the quadratic form ${\displaystyle r^T{Q}_{2}r=0}$, with
$Q_2=\left[\begin{array}{ccccc}{A}_2& & \frac{B_2}{2}& & \frac{D_2}{2}\\ \frac{B_2}{2}& & C_2& & \frac{E_2}{2}\\ \frac{D_2}{2}& & \frac{E_2}{2}& & F_2\end{array}\right]$
Points of intersection of the two equations will satisfy ${\displaystyle r^T{Q}_{1}r=0}$
and ${\displaystyle r^T{Q}_{2}r=0}$
Therefore, the points of intersection will satisfy
${\displaystyle r^T(\alpha Q_1+\beta Q_2)r=0}$
Let ${\displaystyle Q^{\cdot }=\alpha Q_1+\beta Q_2}$, we want to make the determinant of Q∗ zero for reasons that will become clear promptly. And to simplify the expression for the determinant, we'll consider first the case where ${\displaystyle \alpha =0}$, if ${\displaystyle det\left(Q_2\right)=0}$, then we're done, otherwise, we can assume that ${\displaystyle \alpha \ne 0}$, and can therefore write
${\displaystyle Q^{\cdot }=\alpha (Q_1+d\frac{\beta }{\alpha }{Q}_2)}$
Taking ${\displaystyle \alpha =1}$ is as good as any non-zero value for α, this simplifies Q∗ to
${\displaystyle Q^{\cdot }=Q_1+\beta Q_2}$
Since Q1 and Q2 are ${\displaystyle 3\times 3}$ symmetric matrices, then det(Q∗) will be a cubic polynomial in β, which must have at least one real root. Once we substitute this root into Q∗ we have our rank deficient matrix Q∗. A Rank deficient square matrix has at least one of its eigenvalues equal to zero, and this leaves the other two eigenvalues as both positive , both negative, or one positive and one negative. Diagonalizing Q∗ we obtain
${\displaystyle Q\ast =RD{R}^T}$
where $D=\left[\begin{array}{ccccc}{D}_{11}& & 0& & 0\\ 0& & D_{22}& & 0\\ 0& & 0& & 0\end{array}\right]$
Remember that our equation which we are trying to solve is
${\displaystyle r^T{Q}^{\cdot }r=0}$
therefore, we want to solve ${\displaystyle r^{T}RD{R}^{T}r=0}$
Define ${\displaystyle w=R^{T}r}$, i.e. ${\displaystyle r=Rw}$, then we end up with
${\displaystyle w^{T}Dw=0}$
Recalling what D is, this is
${\displaystyle D_{11}{w}_1^2+D_{22}{w}_2^2=0}$
Now if both ${\displaystyle D_{11}\text{and}{D}_{22}}$ have the same sign, then the solution is ${\displaystyle w_1=w_2=0}$ while ${\displaystyle w_3}$ can be any real number. In this case, since ${\displaystyle r=Rw}$ must have its third coordinate equal to 1, then this case corresponds to a single point.
We're not done, we have to verify that point we obtained satisfies ${\displaystyle r^T{Q}_{1}r=0}$
If it does then this is our solution (a single point), otherwise there are no solutions.
Next, we'll consider the case where the eigenvalues have unequal signs, then we can choose that ${\displaystyle D_{11}>0\text{and}{D}_{22}<0}$. In this case, the equation in vector w becomes
${\displaystyle D_{11}{w}_1^2+D_{22}{w}_2^2=0}$
so that ${\displaystyle w_2=\pm \sqrt{-d\frac{D_{11}}{D_{22}}}{w}_1}$
while ${\displaystyle w_3}$ can be any real number. Thus in this case
${\displaystyle w=t(0,0,1)+s(1,\pm \sqrt{-d\frac{D_{11}}{D_{22}}},0)}$
That is w lies in two possible planes spanned by the above two vectors for each of the two possible signs. Now
${\displaystyle r=Rw}$
Again, a plane in the r space for each choice of the sign in the above equation. Intersecting each plane (in x,y,z of vector r ) with the plane ${\displaystyle z=1}$ (because we want the third coordinate of r to be 1 ), we obtain exactly two lines of intersection.
Each line is of the form
$r=r_0+\lambda d_0,\lambda \in \mathrm{ℝ}$
For each of these lines, we now enforce
${\displaystyle r^T{Q}_{1}r=0}$
This will ensure that ${\displaystyle r^T{Q}_{2}r=0}$ as well, because we know that r satisfies ${\displaystyle r^T(Q_1+\beta Q_2)r=0}$. This will give at most two solutions for the resulting quadratic equation in ${\displaystyle \lambda }$. Thus we can have at most four solutions satisfying our system.