5. In a sports meet, ten athletes participate in a table tennis competition, where each pair of athletes plays exactly one match. During the competition, the first athlete wins $x_{1}$ games and loses $y_{1}$ games; the second athlete wins $x_{2}$ games and loses $y_{2}$ games; $\cdots \cdots$ the tenth athlete wins $x_{10}$ games and loses $y_{10}$ games. Let
$$
\begin{array}{l}
M=x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2} . \\
N=y_{1}^{2}+y_{2}^{2}+\cdots+y_{10}^{2} .
\end{array}
$$

Then ( ).
(A) $M < N$
(B) $M > N$
(C) $M = N$
(D) The relationship between $M$ and $N$ is uncertain
5. C.

Given that $x_{i}+y_{i}=9(i=1,2, \cdots, 10)$, and $x_{1}+x_{2}+\cdots+x_{10}=y_{1}+y_{2}+\cdots+y_{10}=45$.
Then $M-N$
$$
\begin{aligned}
= & \left(x_{1}^{2}-y_{1}^{2}\right)+\left(x_{2}^{2}-y_{2}^{2}\right)+\cdots+\left(x_{10}^{2}-y_{10}^{2}\right) \\
= & \left(x_{1}-y_{1}\right)\left(x_{1}+y_{1}\right)+\left(x_{2}-y_{2}\right)\left(x_{2}+y_{2}\right)+ \\
& \cdots+\left(x_{10}-y_{10}\right)\left(x_{10}+y_{10}\right) \\
= & 9\left[\left(x_{1}-y_{1}\right)+\left(x_{2}-y_{2}\right)+\cdots+\left(x_{10}-y_{10}\right)\right] \\
= & 0 .
\end{aligned}
$$