The correct answer is $\boxed{\text{C}}$.
The eigenvalues of a matrix are the roots of its characteristic polynomial. The characteristic polynomial of a matrix $A$ is given by $$p(x) = \det(xI – A)$$ where $I$ is the identity matrix.
In this case, we have $$p(x) = \det \left( \begin{array}{cc} x – 2 & -3 \\ x & x – y \end{array} \right) = x^2 – (2+y)x + 2y – 6$$
We are given that the eigenvalues of $A$ are 4 and 8. Therefore, we must have $$4^2 – (2+y)4 + 2y – 6 = 0$$ and $$8^2 – (2+y)8 + 2y – 6 = 0$$
Solving these equations, we find that $y = -3$ and $x = 9$.
Therefore, the correct answer is $\boxed{\text{C}}$.
Here is a brief explanation of each option:
- Option A: $x = 4$, $y = 10$. This is not possible, because the eigenvalues of a matrix must be distinct.
- Option B: $x = 5$, $y = 8$. This is also not possible, because the eigenvalues of a matrix must be distinct.
- Option C: $x = -3$, $y = 9$. This is possible, because the eigenvalues of a matrix can be any two numbers that are not equal to each other.
- Option D: $x = -4$, $y = 10$. This is not possible, because the eigenvalues of a matrix must be distinct.