Find the number of ways to place $n$ people into $n$ chairs (circular table) so that a particular pair has exactly $k$ people between them. 
Find the number of ways to place $n$ people into $n$ chairs (circular table) so that a particular pair has exactly $k$ people between them.

If they were seated in a row at a table, then the answer would be $2(n-k-1)(n-2)!$ since the pair can be placed in $n-k-1$ ways and they can interchange their seats in $2! = 2$ ways and the others can be seated in $(n-2)!$ ways. I can't figure out what changes with a circular table.
 A: See the comments following the question, and the comments following this answer, for a debate about whether the answer(s) below need to have a factor of $(2)$ applied [assuming that $n,k$ are not such that P-1 and P-2 are across from each other].

Denote the pair as P-1, P-2.  Without loss of generality, you can construe the position of P-1 as the head of the table.  Then, regardless of the value of $k$, the exact seating position of P-2 is fixed.
This implies that the overall computation is $(n-2)!.$

Note, that I am presuming that rotating everyone's position at the table does not alter the seating.  Alternatively, if you reject this presumption, then you have to multiply the above computation by a factor of $(n)$, since there are $n$ different people who could be at the head of the table.
Think of it like a Poker game at a round table, where once each person's seat is fixed, the dealer position rotates.
A: As mentioned by @Shubham, you should precise what definition you put on "between them". If an orientation is given (for example if you look at the number of chairs between them, counted in the anti clock sense), then I think that the answer is $2(k+1)(n-2)!$ plus the answer for a row table. Indeed, there are $k+1$ more options for chosing the two places, then you order the pair, then you order the others.
Try to bend your row table and join the extremities to see where the difference comes from.
A: Let's assume $k + 1 < n$ to avoid ambiguity and so that one person does not sit in the other's lap.
Suppose Amanda and Brian are the two people in question.  By convention, in a circular arrangement, only the relative order of the people matters.  Seat Amanda.  We will use her as our reference point.  If there are exactly $k$ people between them, then Brian can be seated in two ways unless $n = 2(k + 1)$, either $k + 1$ seats to Amanda's left or $k + 1$ seats to her right.  However, if $n = 2(k + 1)$, Brian is both $k + 1$ seats to her left and to her right, so he can be seated in only one way.  In either case, the remaining $n - 2$ people may be seated in the remaining $n - 2$ seats in $(n - 2)!$ ways as we proceed clockwise around the table from Amanda.  Hence, the number of ways to place $n$ people in $n$ chairs at a circular table if a particular pair has exactly $k$ people between them is
$$
\begin{cases}
2(n - 2)! & \text{if $n \neq 2(k + 1)$}\\
(n - 2)! & \text{if $n = 2(k + 1)$}
\end{cases}
$$
