A triangle has corners at #(3 ,8 )#, #(5 ,9 )#, and #(8 ,2 )#. What is the area of the triangle's circumscribed circle?

ABC is the triangl and D,E,F are the midpoints of AB,BC & CA respectively.

Point D has coordinates as #((3+5)/2,(8+9)/2)=(4,17/2)#
Similarly, E coordinates #((5+8)/2,(9+2)/2)=(13/2,11/2)#
F coordinates #((8+3)/2,(2+8)/2)=(11/2,5)#

Slope of AB #=(9-8)/(5-3)=1/2#
Slope of #C’D=-2# where C’ is the circum center and C’D is perpendicular bisector of AB.
Circum-center C’ is the meeting point of perpendicular bisectors of the Sides.
Equation of C’D is #y-(17/2)=-2*(x-4)#
#2y-17=-4x+16#
#color(red)(2y+4x=33) #. Eqn (1)

Slope of BC #=9-2/5-8=-(7/3)#
Slope ofC’E perpendicular bisector of BC is #=3/7#
Equation of C’E is #y-(11/2)=(3/7)(x-(13/2))#
#14y-77=6x-39#
#color(red)(14y-6x=38)#. Eqn (2)

Solving equations (1) & (2) will give point C’.
#6y+12x=99#
#28y-12x=76#
Adding both equations,
#34y=175#
#y=175/34#
#4x=33-2(175/34)==33-(175/17)=(561-175)/68=386/68=193/34#
Coordinates of C’ is #color (red)(193/34,175/34)#

Radius of circum-circle is C’A=C’B=C’C
C’A #=sqrt((3-(193/34))^2+(8-(175/34)^2)#
#=sqrt(7.16+8.14)=color(blue)3.91#

Verification:
C’B #=sqrt(5-(193/34)^2+(9-(175/34))^2)#
#=sqrt(0.46+14.85)=color(blue)(3.91)#

Area of circle #=pir^2=(22/7)*(3.91)^2= color(green)(48.05)#