Let $A,B,C,D \in \mathbb{C}[x,y]$, with $\deg(A),\deg(B),\deg(C),\deg(D) \geq 1$.
Assume that the ideal generated by $A$ and $B$ equals the ideal generated by $C$ and $D$, namely, $\langle A,B \rangle =\langle C,D \rangle$.
Let $\lambda,\mu \in \mathbb{C}$.
$\langle A-\lambda,B-\mu \rangle = $ ? in terms of $C$ and $D$.
What we know is that $A=F_1C+G_1D$ and $B=F_2C+G_2D$, for some $F_1,G_1,F_2,G_2 \in \mathbb{C}[x,y]$.
Therefore, $A-\lambda=F_1C+G_1D-\lambda$ and $B-\mu=F_2C+G_2D-\mu$, but I guess that this does not tell much..
I do not mind to concentrate on the special case where $C \in \mathbb{C}[x]$ is separable (= has no multiple roots) and $D=y$, namely:
If $\langle A,B \rangle =\langle f(x),y \rangle$ with $f$ separable, then $\langle A-\lambda,B-\mu \rangle = $ ? in terms of $f(x)$ and $y$.
Remarks for the special case:
(1) It seems that $\langle f(x),y \rangle$ with $f$ separable is a radical ideal.
(2) $A=F_1f(x)+G_1y$ and $B=F_2f(x)+G_2y$, for some $F_1,G_1,F_2,G_2 \in \mathbb{C}[x,y]$. Take $y=0$ and get: $A(x,0)=F_1(x,0)f(x)$ and $B(x,0)=F_2(x,0)f(x)$. (This shows that $f(x)$ divides $\gcd(A(x,0),B(x,0))$).
Then we can write: $A=\epsilon y+A(x,0)= \epsilon y+F_1(x,0)f(x)=\epsilon y +uf(x)$ and $B=\delta y+B(x,0)= \delta y+F_2(x,0)f(x)=\delta y +vf(x)$, where $\epsilon,\delta \in \mathbb{C}[x,y]$ and $u=F_1(x,0),v=F_2(x,0) \in \mathbb{C}[x]$.
Perhaps similar arguments to the ones presented in the answer to this question (but one has to be careful, since there we had $(y,x-c)$ and here we have $(y,f(x))$) show that $\epsilon v- \delta u \in \mathbb{C}$. Therefore, $\deg_y(\epsilon)=\deg_y(\delta)$, which is quite a restrictive condition.
Thank you very much!
