[ Coordinate method on the plane ] [Triangle inequality (other).]

Point $M$ lies on the line $3 x-4 y+34=0$, and point $N-$ lies on the circle $x^{2}+y^{2}-8 x+2 y-8=0$. Find the minimum distance between points $M$ and $N$.
Notice that

$$
x^{2}+y^{2}-8 x+2 y-8=0 \Leftrightarrow x^{2}-8 x+16+y^{2}+2 y+1=25 \Leftrightarrow(x-4)^{2}+(y+1)^{2}=5^{2}
$$

This means the center of the circle is the point $Q(4, -1)$, and the radius is 5.

Let $d$ be the distance from the point $Q$ to the line $3 x-4 y+34=0$. Then

$$
d=\frac{|3 \cdot 4-4 \cdot(-1)+34|}{\sqrt{3^{2}+4^{2}}}=\frac{50}{5}=10>5
$$

This means that all points on the given line lie outside the given circle. Therefore, for each point $M$ on the given line and each point $N$ on the given circle,

$$
M N>M Q-Q N=M Q-5>d-5=10-5=5
$$

and this distance equals 5 if $M$ is the projection of the point $Q$ onto the given line.

![](https://cdn.mathpix.com/cropped/2024_05_06_ef1f6e5b136aa760f4dfg-32.jpg?height=629&width=809&top_left_y=423&top_left_x=630)

## Answer

5.