The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is

Equation of circle $$x^2+y^2+2gx+2fy+c=0$$

It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $$c=0$$

=> On substituting the value (4,0) in the above equation, we obtain:  $$16+0+8g+0 = 0$$ ; $$g=-2$$

=> On substituting the value (3,9) in the above equation, we obtain: $$9+81-12+18f = 0$$ ; $$f= -13/3$$

Radius of the circle r = $$\sqrt{\ g^2+f^2-c}$$ => $$r^2=\frac{205}{9}$$

Therefore, Area = $$\pi\ r^2=\frac{205\pi\ }{9}$$