3. The circle $(x+3)^{2}+(y-4)^{2}=50$ and the line $y=2 x-5$ intersect at two points. Determine the distance between these two points.
Answer: $2 \sqrt{5}$
Solution: From the given, we know $(x+3)^{2}+[(2 x-5)-4]^{2}=50$.
$$
\begin{aligned}
(x+3)^{2}+(2 x-9)^{2} & =50 \\
\left(x^{2}+6 x+9\right)+\left(4 x^{2}-36 x+81\right) & =50 \\
5 x^{2}-30 x+40 & =0 \\
5(x-2)(x-4) & =0
\end{aligned}
$$

Thus, $x=2$ or $x=4$. If $x=2$, then $y=-1$. If $x=4$, then $y=3$. The distance between $(2,-1)$ and $(4,3)$ is $\sqrt{2^{2}+4^{2}}=2 \sqrt{5}$.