4. Given real numbers $x, y$ satisfy $x^{2}+y^{2}=20$. Then the maximum value of $x y+8 x+y$ is $\qquad$ .
4. 42 .

By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x y+8 x+y)^{2} \\
\leqslant\left(x^{2}+8^{2}+y^{2}\right)\left(y^{2}+x^{2}+1^{2}\right) \\
=84 \times 21=42^{2} .
\end{array}
$$

Therefore, the maximum value sought is 42.