Find an equation of the circle that has center \( (5,-2) \) and passes through \( (1,-6) \).

To find the equation of the circle, we'll start with the standard form of a circle's equation: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \( (h, k) \) is the center of the circle and \( r \) is the radius. In your case, the center is \( (5, -2) \), so \( h = 5 \) and \( k = -2 \). Next, we need to find the radius \( r \). We can calculate the radius by finding the distance between the center of the circle and the point \( (1, -6) \). The distance \( r \) can be found using the distance formula: \[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the coordinates: \[ r = \sqrt{(1 - 5)^2 + (-6 + 2)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \] Now we can substitute \( h \), \( k \), and \( r \) back into the circle equation: \[ (x - 5)^2 + (y + 2)^2 = (4\sqrt{2})^2 \] Calculating \( (4\sqrt{2})^2 \): \[ (4\sqrt{2})^2 = 16 \cdot 2 = 32 \] Finally, the equation of the circle is: \[ (x - 5)^2 + (y + 2)^2 = 32 \]