Consider a polynomial $P\in{\mathbb F}[X]$ where ${\mathbb F}$ is a finite field and $P$ is of degree $f$.
Given the set of points $(1,y_1),\ldots,(n,y_n)$ where $n=3f+1$ and $f$ is the upper bound on the number of noisy points, that is, let $M\subset[n]$ such that $|M|\leq f$ we know that for all $i\in[n]\smallsetminus M$ it follows that $P(i)=y_i$ whereas for $i\in M$ we have $P(i)\neq P(i)$.
It is known that we can decode $P$ from $(1,y_1),\ldots,(n,y_n)$ in this case, so these points are correctable encoding of $P$.
However, when the degree of $P$ is $2f$ it is impossible to decode $P$ from $(1,y_1),\ldots,(n,y_n)$ with $f$ noisy points.
My question is, if I get another new point $(x^\star, y^\star)$ which is known to be correct (i.e. $P(x^\star)=y^\star$), can I do something better? Can I detect even a single faulty point? Can I even pick up 2 points such that one of them is faulty? - that would be helpful as well.
If I cannot do anything - is there other encoding that allows me to do so when I obtain a known correct point?
If you are given that $P(x^\star)=y^\star$, then this tells you that $P(x)=Q(x)(x-x^\star)+y^\star$ for some polynomial $Q$ of degree $f-1$. Therefore, this transforms the problem as follows: $$ \begin{array}{lcl} \text{Find }P(x) & \implies & \text{Find }Q(x) \\ \text{deg }P=f & \implies & \text{deg }Q=f-1 \\ P(i)=y_i \text{ for at least $n-m$ values of }i & \implies & Q(i)=\frac{y_i-y^\star}{i-x^\star} \text{ for at least $n-m$ values of }i \\ \end{array} $$ In other words, you have the exactly same problem, with different $y_i$ values, except that the degree has decreased. Therefore, your question becomes