Find an equation of the line containing the conters of the two circles whose equations are given below. \[ \begin{array}{l} (x-4)^{2}+(y+5)^{2}=25 \\ (x+3)^{2}+(y-6)^{2}=16 \end{array} \] The general form of the equation of the line \( \square \) (Simplify your answer.)

To find the equation of the line that contains the centers of the two circles given by the equations, we need to identify the centers of each circle. 1. For the circle \((x-4)^{2}+(y+5)^{2}=25\), the center is at \( (4, -5) \). 2. For the circle \((x+3)^{2}+(y-6)^{2}=16\), the center is at \( (-3, 6) \). Next, we find the slope \( m \) of the line that passes through these two points \((4, -5)\) and \((-3, 6)\): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-5)}{-3 - 4} = \frac{11}{-7} = -\frac{11}{7}. \] Now, we can use point-slope form to write the equation of the line. Using the point \((4, -5)\): \[ y + 5 = -\frac{11}{7}(x - 4). \] Simplifying this equation: \[ y + 5 = -\frac{11}{7}x + \frac{44}{7}. \] Subtract 5 (which is \(\frac{35}{7}\)) from both sides: \[ y = -\frac{11}{7}x + \frac{44}{7} - \frac{35}{7} = -\frac{11}{7}x + \frac{9}{7}. \] To express this in general form, multiply every term by 7: \[ 7y = -11x + 9. \] Rearranging gives: \[ 11x + 7y - 9 = 0. \] Thus, the general form of the equation of the line is: \[ \boxed{11x + 7y - 9 = 0}. \]