Determine the equation of the circle with center \( (-5,-3) \) containing the point \( (3,-18) \).

To determine the equation of the circle with a center at \( (-5,-3) \) and passing through the point \( (3,-18) \), we first need to find the radius. The radius \( r \) can be calculated using the distance formula between the center and the point on the circle: \[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \( (x_1, y_1) = (-5, -3) \) and \( (x_2, y_2) = (3, -18) \). Calculating: \[ r = \sqrt{(3 - (-5))^2 + (-18 - (-3))^2} \] \[ r = \sqrt{(3 + 5)^2 + (-18 + 3)^2} \] \[ r = \sqrt{(8)^2 + (-15)^2} \] \[ r = \sqrt{64 + 225} \] \[ r = \sqrt{289} \] \[ r = 17 \] Now that we have the radius, we can write the equation of the circle using the standard form: \[ (x - h)^2 + (y - k)^2 = r^2 \] Substituting \( h = -5 \), \( k = -3 \), and \( r = 17 \): \[ (x + 5)^2 + (y + 3)^2 = 17^2 \] \[ (x + 5)^2 + (y + 3)^2 = 289 \] Thus, the equation of the circle is: \[ \boxed{(x + 5)^2 + (y + 3)^2 = 289} \]