We are given a circle with equation $x^2 + y^2 + 2x - 6y + 1 = 0$ and a line with equation $3x - y + 2 = 0$. We need to find the midpoint of the chord formed by the intersection of the line and the circle, and then find the length of the chord.

1. Problem Description

We are given a circle with equation $x^2 + y^2 + 2x - 6y + 1 = 0$ and a line with equation $3x - y + 2 = 0$. We need to find the midpoint of the chord formed by the intersection of the line and the circle, and then find the length of the chord.

2. Solution Steps

First, we find the center and radius of the circle. The equation of the circle can be rewritten as:

$(x^2 + 2x) + (y^2 - 6y) + 1 = 0$

Completing the square for the x terms: $(x^2 + 2x + 1) - 1$

Completing the square for the y terms: $(y^2 - 6y + 9) - 9$

So, the equation becomes:

$(x + 1)^2 - 1 + (y - 3)^2 - 9 + 1 = 0$

$(x + 1)^2 + (y - 3)^2 = 9$

This is in the form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

Therefore, the center of the circle is $(-1, 3)$ and the radius is $r = \sqrt{9} = 3$.

Next, we find the perpendicular distance from the center of the circle to the line. The formula for the distance $d$ from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is:

$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$

In our case, the point is $(-1, 3)$ and the line is $3x - y + 2 = 0$, so $a = 3$, $b = -1$, $c = 2$, $x_1 = -1$, and $y_1 = 3$.

$d = \frac{|3(-1) - (3) + 2|}{\sqrt{3^2 + (-1)^2}} = \frac{|-3 - 3 + 2|}{\sqrt{9 + 1}} = \frac{|-4|}{\sqrt{10}} = \frac{4}{\sqrt{10}}$

Now, let the midpoint of the chord be $M$. The line joining the center of the circle to the midpoint of the chord is perpendicular to the chord. The slope of the given line $3x - y + 2 = 0$ is $m = 3$.

The slope of the line joining the center $(-1, 3)$ to the midpoint $M(x, y)$ is $-\frac{1}{3}$ (since it's perpendicular).

So, $\frac{y - 3}{x + 1} = -\frac{1}{3}$

$3(y - 3) = -(x + 1)$

$3y - 9 = -x - 1$

$x + 3y = 8$

Also, the midpoint $M(x, y)$ lies on the line $3x - y + 2 = 0$. Thus,

$3x - y + 2 = 0$

Now we have a system of two linear equations:

$x + 3y = 8$

$3x - y = -2$

Multiply the second equation by 3:

$9x - 3y = -6$

Add the first equation to this:

$x + 3y + 9x - 3y = 8 - 6$

$10x = 2$

$x = \frac{1}{5}$

Substitute $x = \frac{1}{5}$ into $x + 3y = 8$:

$\frac{1}{5} + 3y = 8$

$3y = 8 - \frac{1}{5} = \frac{40 - 1}{5} = \frac{39}{5}$

$y = \frac{13}{5}$

So, the midpoint of the chord is $M(\frac{1}{5}, \frac{13}{5})$.

Let half the length of the chord be $l$. Then $l^2 + d^2 = r^2$ (Pythagorean theorem)

$l^2 = r^2 - d^2 = 3^2 - (\frac{4}{\sqrt{10}})^2 = 9 - \frac{16}{10} = 9 - \frac{8}{5} = \frac{45 - 8}{5} = \frac{37}{5}$

$l = \sqrt{\frac{37}{5}}$

The length of the chord is $2l = 2\sqrt{\frac{37}{5}} = 2\frac{\sqrt{37}}{\sqrt{5}} = 2\frac{\sqrt{37}\sqrt{5}}{5} = \frac{2\sqrt{185}}{5}$.

3. Final Answer

The midpoint of the chord is $(\frac{1}{5}, \frac{13}{5})$. The length of the chord is $\frac{2\sqrt{185}}{5}$.