The vertices of triangle are (9, 3), \ (3, 9) & (5, 8) hence its sides say a, b & c are given by using distance formula as follows
a=\sqrt{(9-3)^2+(3-9)^2}=6\sqrt2
b=\sqrt{(9-5)^2+(3-8)^2}=\sqrt{41}
c=\sqrt{(3-5)^2+(9-8)^2}=\sqrt{5}
Now, the area (\Delta) of triangle with vertices (x_1, y_1)\equiv(9,3), (x_2, y_2)\equiv(3,9) & (x_3, y_3)\equiv(5,8) is given as follows
\Delta=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|
=\frac{1}{2}|9(9-8)+3(8-3)+5(3-9)|
=\frac{1}{2}|-6|
=3
hence, the radius (R) of circumscribed circle is given by following formula
R=\frac{abc}{4\Delta}
=\frac{6\sqrt2\cdot \sqrt{41}\cdot \sqrt5}{4\cdot 3}
=\sqrt{\frac{205}{2}}
hence, the area of circumscribed circle with radius R=\sqrt{205/2} is
=\pi R^2
=\pi(\sqrt{205/2})^2
=\frac{205\pi}{2}\ \text{sq. units}
