When the polynomial $x^4 - 6x^3 + 16x^ 2 - 25x + 10$ is divided by $x^2 - 2x + k,$ the remainder is $x + a.$  Enter the ordered pair $(k,a).$
The long division is shown below.

\[
\begin{array}{c|ccccc}
\multicolumn{2}{r}{x^2} & -4x & +(8 - k) & \\
\cline{2-6}
x^2 - 2x + k & x^4 & -6x^3 & +16x^2 & -25x & +10 \\
\multicolumn{2}{r}{x^2} & -2x^3 & + kx^2 \\ 
\cline{2-4}
\multicolumn{2}{r}{0} & -4x^3 & +(16 - k)x^2   \\
\multicolumn{2}{r}{} &- 4x^3 & +8x^2 & - 4kx \\ 
\cline{3-5}
\multicolumn{2}{r}{} & 0 & +(8 - k)x^2 & +(4k - 25)x   \\
\multicolumn{2}{r}{} & & +(8 - k)x^2 & +(2k - 16)x & +k(8 - k) \\
\cline{4-6}
\multicolumn{2}{r}{} & & 0 & +(2k - 9)x  & +(k^2 - 8k + 10) \\
\end{array}
\]Thus, the remainder is $(2k - 9)x + (k^2 - 8k + 10).$  We want this to be $x + a,$ so $2k - 9 = 1$ and $k^2 - 8k + 10 = a.$  Solving, we find $(k,a) = \boxed{(5,-5)}.$