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}#
