8,9 Find the distance between the parallel lines $y=-3 x+5$ and $y=-3 x-4$.

#
## First method.

Since the coordinates of point $A(0 ; 5)$ satisfy the equation $y=-3 x+5$, this point lies on the first line. Since the lines are parallel, the distance between them is equal to the distance from point $A$ to the second line. Let's write the equation of this line in general form $(y+3 x+4=0)$ and use the formula for the distance from a point to a line:

$$
d=\frac{|5+3 \cdot 0+4|}{\sqrt{1^{2}+3^{2}}}=\frac{9}{\sqrt{10}}
$$

## Second method.

Let the first line intersect the $O Y$ axis at point $A(0 ; 5)$, and the second line at point $B(0 ;-4)$. Then $A B=5-(-4)=9$. If $\alpha$ is the acute angle between each of these lines and the $O X$ axis, then $\operatorname{tg} \alpha=3$. Then

$$
\cos \alpha=\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{1}{\sqrt{10}}
$$

Let $C$ be the projection of point $B$ onto the line $y=-3 x+5$. Then $\angle A B C=\alpha$, and the distance between the lines is equal to the length of segment $B C$.

From the right triangle $A B C$, we find that

$$
B C=A B \cos \alpha=9 \cdot \frac{1}{\sqrt{10}}=\frac{9}{\sqrt{10}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_ef1f6e5b136aa760f4dfg-30.jpg?height=598&width=418&top_left_y=2268&top_left_x=820)

![](https://cdn.mathpix.com/cropped/2024_05_06_ef1f6e5b136aa760f4dfg-31.jpg?height=106&width=1129&top_left_y=284&top_left_x=13)

Form the equation of the circle with center at point $M(3 ; 2)$, touching the line $y=2 x+6$.

## Solution

First method.

Let the radius of the desired circle be $R$. Then the distance from point $M$ to the given line is also $R$. Let's write the equation of this line in general form $(2 x-y+6=0)$ and use the formula for the distance from a point to a line:

$$
R=\frac{|2 \cdot 3-2+6|}{\sqrt{2^{2}+1^{2}}}=\frac{10}{\sqrt{5}}=2 \sqrt{5}
$$

Therefore, the desired equation is

$$
(x-3)^{2}+(y-2)^{2}=20
$$

Second method.

The desired equation is

$$
(x-3)^{2}+(y-2)^{2}=R^{2}
$$

Substitute $y=2 x+6$ into the left side of this equation. After obvious simplifications, we get the quadratic equation $5 x^{2}+10 x+25-R^{2}=0$. Since the line and the circle have only one common point, the obtained equation has exactly one solution, meaning its discriminant is 0, i.e.

$$
D=100-20\left(25-R^{2}\right)=20\left(5-25+R^{2}\right)=20\left(R^{2}-20\right)-0
$$

From this, we find that $R^{2}=20$. Therefore, the desired equation is

![](https://cdn.mathpix.com/cropped/2024_05_06_ef1f6e5b136aa760f4dfg-31.jpg?height=483&width=481&top_left_y=1559&top_left_x=797)

## Answer

$(x-3)^{2}+(y-2)^{2}=20$