Find the equation of the line containing centers of the 2 circles...

x^2 + y^2 - 4x + 6y + 4 = 0 . . . . AND . . . . x^2 + y^2 + 6x + 4y + 9 = 0
I got Center: (2, -3) . . . . . . . . . . . . . . . . . Center: (-3, -2)
Radius: 3 . . . . . . . . . . . . . . . . . . . . . . . . Radius: 2

what do I do after that?

Use the two-point form of the equation of a line: y - y₁ = ((y₂ - y₁)/(x₂ - x₁)) *(x - x₁).
Here, x₁ = 2; y₁ = -3; x₂ = -3; y₂ = -2. Substitute these into the formula, giving
y - (-3) = ((-2 - (-3))/(-3 - 2)) * (x - 2), or
y + 3 = ((-2 + 3)/(-5)) * (x - 2), or
y + 3 = (1/(-5)) * (x - 2), or
y + 3 = -(1/5)(x - 2), or
y + 3 = -(1/5)x + 2/5, or
y = -(1/5)x - 13/5.

Find the slop:

( difference between the y's coordinate / difference between the x's coordinate)

= (-2- -3)/ (-3-2) = -0.2

then:
take any pt. let me take (2, -3)
y- -3=slope*(x-2)

=> y+3= -0.2*x + 0.4
finally we get:

y = -0.2x - 2.6