Answer
$5y+x +13 = 0$
Work Step by Step
To find equation of the line containing the centers of the
two circles
$x^2 + y^2 - 4x + 6y + 4 = 0$
and
$x^2 + y^2 + 6x + 4y + 9 = 0$
Finding center of the two circles
In the form $(x-a)^2 + (y-b)^2 = r^2$ of equation $(a, b)$ is the center
$x^2 + y^2 - 4x + 6y + 4 = 0$ can be written as
$(x-2)^2 + (y-(-3))^2 = 3^2$
We get center as $(2, -3)$
$x^2 + y^2 + 6x + 4y + 9 = 0$ can be written as
$(x-(-3))^2 + (y-(-2))^2 = 2^2$
We get center as $(-3, -2)$
Equation of line through centers $(2, -3)$ and $(-3, -2)$ would be
$y-(-3) = \frac{-3-(-2)}{2-(-3)}(x-2)$
=>$ y +3 = \frac{-1}{5}(x-2) $
=> $5y+x +13 = 0$
