Question

1. what is the equation of a line through the center of $x^{2}+y^{2}+4x + 12y+36 = 0$ parallel to $9x - 5y=4$?
2. find the equation of a line through $(0,-4)$ and the center of $x^{2}+y^{2}-3x - 4y = 0$.

Explanation:

Step1: Rewrite the circle equation to find the center

For the circle $x^{2}+y^{2}+4x + 12y+36 = 0$, we complete the square.
The general form of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center.
$x^{2}+4x+y^{2}+12y+36 = 0$ can be rewritten as $(x + 2)^{2}-4+(y + 6)^{2}-36+36 = 0$, so the center is $(-2,-6)$.
The line $9x-5y = 4$ can be rewritten as $y=\frac{9}{5}x-\frac{4}{5}$, and its slope $m=\frac{9}{5}$.
Using the point - slope form $y - y_{1}=m(x - x_{1})$, with $(x_{1},y_{1})=(-2,-6)$ and $m=\frac{9}{5}$, we have $y+6=\frac{9}{5}(x + 2)$.
Expanding gives $y+6=\frac{9}{5}x+\frac{18}{5}$, or $9x-5y=-12$.

Step2: Rewrite the second circle equation to find the center

For the circle $x^{2}+y^{2}-3x-4y = 0$, we complete the square.
$x^{2}-3x+y^{2}-4y = 0$ becomes $(x-\frac{3}{2})^{2}-\frac{9}{4}+(y - 2)^{2}-4=0$. The center is $(\frac{3}{2},2)$.
The slope between $(0,-4)$ and $(\frac{3}{2},2)$ is $m=\frac{2+4}{\frac{3}{2}-0}=\frac{6}{\frac{3}{2}} = 4$.
Using the point - slope form with $(x_{1},y_{1})=(0,-4)$ and $m = 4$, we get $y+4=4(x - 0)$, or $y = 4x-4$.

Answer:

1. $9x-5y=-12$
2. $y = 4x-4$