[ Coordinate method on the plane ] $\left[\begin{array}{l}\text { [Properties and characteristics of the tangent] }\end{array}\right.$

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

Let the radius of the desired circle be $R$. Then the distance from point $M$ to the given line is also $R$. Write the equation of this line in general form $(2 x-y+6=0)$ and use the formula for the distance between a point and 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_1b17058527caf5b5d058g-34.jpg?height=474&width=463&top_left_y=2161&top_left_x=800)

Answer

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

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 $N$.

## Solution

Notice that

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

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

Let $d$ be the distance from 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 point $Q$ onto the given line.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-35.jpg?height=640&width=815&top_left_y=989&top_left_x=633)

## Answer

5.