Find an equation of the circle that has center \( (6,-6) \) and passes through \( (2,-1) \).

To find the equation of a circle, we use the standard form of the equation of a circle, which is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center of the circle and \(r\) is the radius. Given the center of the circle \((h, k) = (6, -6)\), we can substitute these values into the equation: \[ (x - 6)^2 + (y + 6)^2 = r^2 \] Next, we need to find the radius \(r\). The radius is the distance from the center of the circle to any point on the circle. We are given that the circle passes through the point \((2, -1)\). We can calculate the distance from the center \((6, -6)\) to the point \((2, -1)\) using the distance formula: \[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the coordinates of the center and the point: \[ r = \sqrt{(2 - 6)^2 + (-1 + 6)^2} \] Calculating the differences: \[ r = \sqrt{(-4)^2 + (5)^2} \] Calculating the squares: \[ r = \sqrt{16 + 25} = \sqrt{41} \] Now that we have the radius \(r = \sqrt{41}\), we can substitute \(r^2\) into the equation of the circle: \[ r^2 = (\sqrt{41})^2 = 41 \] Thus, the equation of the circle becomes: \[ (x - 6)^2 + (y + 6)^2 = 41 \] This is the equation of the circle with center \((6, -6)\) that passes through the point \((2, -1)\). Therefore, the final answer is: \[ \boxed{(x - 6)^2 + (y + 6)^2 = 41} \]