1. Given the triangle connecting $A(-3,0), B(0,-3)$, $C\left(\frac{15}{7}, \frac{24}{7}\right)$ and the circle $x^{2}+y^{2}=R^{2}(R>$ 0 ) always have common points. Then the range of the circle radius $R$ is
(A) $\left.\left(0, \frac{3 \sqrt{10}}{10}\right) \cup \frac{3 \sqrt{ }}{7},+\infty\right)$
(B) $\left(\frac{3 \sqrt{10}}{10}, \frac{3 \sqrt{89}}{7}\right)$
(C) $\left(0, \frac{3 \sqrt{2}}{2}\right) \cup(3,+\infty)$
(D) $\left(\frac{3 \sqrt{2}}{2}, 3\right)$
-1 (B).
As shown in Figure 11, from the given information, the equations of the lines on which the three sides lie are
$$
\begin{array}{l}
A B: y=-x-3, \\
A C: y=\frac{2}{3}(x+3), \\
B C: y=3(x-1) .
\end{array}
$$

Further, the distances from the center of the circle to the three vertices and the three sides can be calculated as
$$
\begin{array}{c}
|O A|=3,|O B|= \\
3,|O C|=\frac{3 \sqrt{89}}{7} ; \\
d_{A B}=\frac{3}{\sqrt{2}}, d_{A C}=\frac{6}{\sqrt{13}}, d_{B C}=\frac{3}{\sqrt{10}} .
\end{array}
$$

When $d_{B C} \leqslant R \leqslant|O C|$, the triangle and the circle must have common points, so the range of $R$ is $\left(\frac{3 \sqrt{10}}{10}, \frac{3 \sqrt{89}}{7}\right)$. $)$