Equation of family of concentric circles to the circle $x^2 + y^2 + 6x + 8y - 5 = 0$ is $x^2 + y^2 + 6x + 8y + \lambda = 0$ which is similar to $x^2 + y^2 + 2gx + 2fy + c = 0$ Thus, the point $(-3, 2)$ lies on the cirlce $x^2 + y^2 + 6x + 8y + c = 0$ $\therefore \, (-3)^2 + (2)^2 + 6(-3) + 8 (2) + c = 0$ $\Rightarrow \, 9 + 4 - 18 + 16 + c = 0 $ $ \Rightarrow c = -11$