A circumference as intersection of a quadric with a plane

Question

I have to prove that the following intersection, is a circumference: $$ \mathcal{C} = \begin{cases} x^2+y^2+xy-3x-4y+z+\frac{5}{2}=0 \\ x+y+z=2 \end{cases} $$ The method I would like to use is to find the improper points, and observe that they fulfill the relationship $x^2+y^2+z^2=0$. Is this a right procedure? For improper points: $$ \begin{cases} x^2+y^2+xy=0 \\ z=2-x-y \\ t=0 \end{cases} \Rightarrow \begin{cases} x=(\frac{-1 \pm i\sqrt{3}}{2})y \\ z=2-(\frac{1 \pm i\sqrt{3}}{2})y \\ t=0 \end{cases} $$ $P_{\infty} = (\frac{-1 \pm i\sqrt{3}}{2},1,\frac{3 \mp i\sqrt{3}}{2},0)$ and $(\frac{-1 \pm i\sqrt{3}}{2})^2+1+(\frac{3 \mp i\sqrt{3}}{2})^2$ is not zero.

Answer 1

Your idea is fine, but you didn't get the points at infinity right. The correct equations are

$$ \begin{cases} x^2+y^2+xy=0 \\ z=-x-y \\ t=0 \end{cases}$$

which give

$$ P_{\infty} = \left(\frac{-1 \pm i\sqrt{3}}{2},1,\frac{1 \mp i\sqrt{3}}{2},0\right). $$