Let $f=f(t),g=g(t) \in k[t]$, $f \neq g$, and $\lambda,\mu \in k$.
(1) Is there a relation between $\gcd(f,g)$ and $\gcd(f-\lambda,g-\mu)$?
If we only wish to know what is the relation between the number of common roots of $f$ and $g$ and the number of common roots of $f-\lambda$ and $g-\mu$ (without computing them), then a 'weaker' version of question $(1)$ is:
(2) Is there a relation between $\deg(rad(fg))$ and $\deg(rad(f-\lambda)(g-\mu))$? Notice that $\deg(rad(fg))=\deg(f)+\deg(g)-\deg(\gcd(f,g))$.
For example: $f=(x^2-1)(x-2)=x^3-2x^2-x+2$, $g=(x^2-1)(x-3)=x^3-3x^2-x+3$, $\lambda=2$, $\mu=3$.
We have, $\gcd(f,g)=x^2-1$ and $\gcd(f-\lambda,g-\mu)= \gcd(f-2,g-3)=\gcd(x^3-2x^2-x, x^3-3x^2-x)=\gcd(x(x^2-2x-1),x(x^2-3x-1))=x$
In this specific example, $\gcd(f,g)=x^2-1$, $\gcd(f-\lambda,g-\mu)=x$, $\deg(rad(fg))=4$ and $\deg(rad(f-\lambda)(g-\mu))=5$.
(3) Is there a relation between the resultant of $f$ and $g$ and the resultant of $f-\lambda$ and $g-\lambda$? (in the above example, both resultants are zero).
Perhaps this question is relevant, so we should consider the ideal generated by $\{f,g\}$, and the ideal generated by $\{f-\lambda,g-\lambda\}$.
Another example: $f=x^{10}$, $g=x^9$, $\lambda=1$, $\mu=0$. We have $\gcd(f,g)=\gcd(x^{10},x^9)=x^9$ and $\gcd(f-1,g)=\gcd(x^{10}-1,x^9)=1$, so probably there is nothing interesting that we can say about the degrees (so there is no answer to question $(1)$), unless we make further assumptions. For example, if we further assume that $k(f,g)=k(t)$ then by this answer there exist $\hat{\lambda},\hat{\mu} \in k$ such that $\deg(\gcd(f-\hat{\lambda},g-\hat{\mu}))=1$, but this does not tell much about the degree of arbitrary $\gcd(f-\epsilon,g-\delta)$, $\epsilon,\delta \in k$, as this last example shows (notice that $\gcd(f-1,g-1)=\gcd(x^{10}-1,x^9-1)=x-1$).
Edit: I have just found this question; I hope that it is possible to apply the very nice ideas in its answers here. In particular, I hope that Mason-Stothers theorem or Stothers theorem may help here.
Any comments are welcome!