Is there any easy way to calculate the multiplicity of intersection of this two algebric curves?

Question

I've been given this two algebric curves in $\mathbb{P}^{2}_{\mathbb{C}}$: $$\mathcal{C}=\{[x,y,z]\in \mathbb{P}^{2}_{\mathbb{C}}|\hspace{0.5em} P(x,y,z)= y^{2}z-x(x-2z)(x+z)=0\}$$ $$\mathcal{D}=\{[x,y,z]\in \mathbb{P}^{2}_{\mathbb{C}}|\hspace{0.5em}Q(x,y,z)=y^{2}+x^{2}-2xz=0\}$$ Searching for their intersection gave me: $$\mathcal{C}\cap\mathcal{D}=\{[0,0,1],[2,0,1],[-2,2i\sqrt{2},1],[-2,-2i\sqrt{2},1] \}$$ For Bezout's theorem we know that $$\sum_{p\in \mathcal{C}\cap\mathcal{D}}\mathcal{I}_{p}(\mathcal{C},\mathcal{D})=3*2=6$$ For every $p$ in $\mathcal{C}\cap\mathcal{D}$ find $\mathcal{I}_{p}(\mathcal{C},\mathcal{D})$. I'm not quite sure about how to proceed, I know that if $\mathcal{C}$ and$\mathcal{D}$ have no common factors then for example we can take $p=[0,0,1]$, $P(x,y,z-1)$, $Q(x,y,z-1)$, calculate the resultant $\mathcal{R}_{P,Q}(y,z)$ and then searching for the bigger $k$ in $\mathbb{N}$ such that $(\alpha y-\beta z)^{k}$ divedes $\mathcal{R}_{P,Q}(y,z)$. This should be working but it's pretty complicate and requires at least 3 determinants of $5\times 5$ matrixes so I'd like to know if there is any easier strategy.

Answer 1

On the affine piece $z=1$, the curves are of the form $y^2=C(x-x_0)+\text{higher order}$ with $C\neq 0$, $x_0=0\text{ or }2$. So both curves have the same tangent line at $[x,y,z]=[0,0,1]$ or $[x,y,z]=[2,0,1]$, hence local intersection multiplicity $>1$ there.