8. Given $\odot M:(x-1)^{2}+(y-3)^{2}=4$, there exists a secant line $P B A$ of $\odot M$ through the point $P(a, 0)$ on the $x$-axis, such that $P B=P A$. Then the range of the $x$-coordinate $a$ of point $P$ is $\qquad$
8. $1-3 \sqrt{3} \leqslant a \leqslant 1+3 \sqrt{3}$.

The center of the circle $M(1,3)$, diameter $d=4$.
As shown in Figure 5, draw a secant through point $P$ and
$M$. By the secant theorem, we have
$$
\begin{array}{l}
\left(P M+\frac{d}{2}\right) . \\
\left(P M-\frac{d}{2}\right) \\
= P B \cdot P A . \\
\text { Hence } P B=B A \\
\Leftrightarrow P M^{2}-\frac{d^{2}}{4}=2 A B^{2} \leqslant 2 d^{2} \\
\Leftrightarrow|P M| \leqslant \frac{3}{2} d \Leftrightarrow(a-1)^{2}+3^{2} \leqslant 6^{2} \\
\Leftrightarrow 1-3 \sqrt{3} \leqslant a \leqslant 1+3 \sqrt{3} .
\end{array}
$$