Area of circumcircle of a triangle

Let equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Passes through $(0,0) \implies c = 0$. Passes through $(4,0) \implies 16 + 8g = 0 \implies g = -2$. Passes through $(3,9) \implies 9 + 81 + 2(-2)(3) + 2f(9) = 0 \implies 90 - 12 + 18f = 0 \implies 18f = -78 \implies f = -13/3$.

Radius squared $R^2 = g^2 + f^2 - c = (-2)^2 + (-13/3)^2 = 4 + \frac{169}{9} = \frac{205}{9}$. Area of the circle = $\pi R^2 = \frac{205\pi}{9}$.