4. The arithmetic sequence $\left\{a_{n}\right\}$ satisfies $a_{1}^{2}+a_{10}^{2} \leqslant 10$. Then the range of values for $S=a_{10}+a_{11}+\cdots+a_{19}$ is $\qquad$
4. $-50 \leqslant S \leqslant 50$.

From the problem, we have
$$
\begin{array}{l}
S=a_{10}+a_{11}+\cdots+a_{19} \\
=10 a_{10}+d+2 d+\cdots+9 d \\
=10 a_{10}+45 d=10 a_{1}+135 d .
\end{array}
$$

Thus, $9 d=\frac{1}{15}\left(S-10 a_{1}\right)$.
Since $a_{1}^{2}+a_{10}^{2} \leqslant 10$, we have
$$
\begin{array}{l}
a_{1}^{2}+\left(a_{1}+9 d\right)^{2} \leqslant 10 \\
\Rightarrow 250 a_{1}^{2}+10 S a_{1}+S^{2}-2250 \leqslant 0 \\
\Rightarrow a_{1}^{2}+\frac{1}{25} S a_{1}+\frac{1}{250} S^{2}-9 \leqslant 0 \text { has a solution } \\
\Rightarrow \Delta=\left(\frac{1}{25} S\right)^{2}-4\left(\frac{1}{250} S^{2}-9\right) \geqslant 0 \\
\Rightarrow-50 \leqslant S \leqslant 50 .
\end{array}
$$