Let $u=u(x,y)=xy+x$ and $v=v(x,y)=xy+y$.
Denote $f_a=u-a$, $g_b=v-b$, where $a,b \in \mathbb{C}$.
I am looking for common zers of $f_a=0, g_b=0$, for some $a,b \in \mathbb{C}$.
Examples:
- $a=b=0$, then solving $xy+y=0, xy+y=0$ yields two common zeros $(0,0), (-1,-1)$.
- $a=b=1$, then solving $xy+y=1, xy+y=1$ yields two common zeros $(\frac{-1+\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}), (\frac{-1-\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2})$.
- $a=b=-\frac{1}{4}$, then solving $xy+y=-\frac{1}{4}, xy+y=-\frac{1}{4}$ yields one common zeros $(-\frac{1}{2},-\frac{1}{2})$.
- $a=-4,b=-9$, then solving $xy+y=-4, xy+y=-9$ yields one common zeros $(2,-3)$.
Generally, if we take arbitrary $a,b \in \mathbb{C}$ and solve $xy+x=a, xy+y=b$, we get $x-y=a-b$, so $y^2+(a-b+1)y-b=0$. Over $\mathbb{C}$ this equation has two zeros or one zero (with multiplicity two). If we count multiplicity, then $f_a=0,g_b=0$ always has two common zeros, for every $a,b \in \mathbb{C}$.
Next, if we take $u=x^2+y^2, v=x^2+y^2$, then there are more options:
- $a=b=0$, then solving $x^2+y^2=0, x^2+y^2=0$ yields several common zeros $(0,0), (1,i), (1,-i), (-1,i), (-1,-i), (i,1), (i,-1), (-i,1), (-i,-1)$, and actually there are infinitely many common zeors, $\{(c,ci)|c \in \mathbb{C}\}$ etc.
- $a=1,b=2$, then $x^2+y^2=1, x^2+y^2=2$ has no common zeros.
Question: I wonder if there exists a known characterization for two arbitrary plane curves $U(x,y), V(x,y) \in \mathbb{C}[x,y]$, each with arbitrary total degree, for which there exist some $a,b \in \mathbb{C}$ such that $U=a,V=b$ has exactly one common zero (counting multiplicity).
I apologize if my question is too easy or too general.
Thank you very much!