A triangle has corners at #(7 ,7 )#, #(1 ,3 )#, and #(6 ,5 )#. What is the area of the triangle's circumscribed circle?

Midpoint of corners #(7,7) and (1,3)# is given by
#((7+3)/2,(7+1)/2)#
#= (5,4)#
Slope of the line between the corners #(7,7) and (1,3)# is given by
#(3-7)/(1-7)#
#=(-4)/(-6)#
#=2/3#
Slope of the perpendicular bisector for the line between the corners #(7,7) and (1,3)# is given by
#=-1/(2/3)#
#=-3/2#
We have the equation of the perpendicular bisector for the line between the corners #(7,7) and (1,3)# given by
Point #(5,4)#
Slope #-3/2#
point slope form is
#(y-4)/(x-5)=-3/2#

Midpoint of corners #(1,3) and (6,5)# is given by
#((1+6)/2,(3+5)/2)#
#= (3.5,4)#
Slope of the line between the corners #(1,3) and (6,5)# is given by
#(5-3)/(6-1)#
#=(2)/(5)#
#=2/5#
Slope of the perpendicular bisector for the line between the corners #(1,3) and (6,5)# is given by
#=-1/(2/5)#
#=-5/2#
We have the equation of the perpendicular bisector for the line between the corners #(1,3) and (6,5)# given by
Point #(3.5,4)#
Slope #-5/2#
point slope form is
#(y-4)/(x-3.5)=-5/2#

Solving for the center of the circumcircle
#(y-4)/(x-5)=-3/2#
#2(y-4)=-3(x-5)#
#2y-8=-3x+15#
#2y+3x-15-8=0#
#3x+2y-23=0#

#(y-4)/(x-3.5)=-5/2#
#2(y-4)=-5(x-3.5)#
#2(y-4)+5(x-3.5)=0#
#2y-8+5x-17.5=0#
#5x+2y-24.5=0#

We have,
#3x+2y-23=0#
#5x+2y-24.5=0#
Eliminating y
#3x-5x-23+24.5=0#
#-2x+1.5=0#
#2x=1.5#
#x=0.75#
Substituting
#3(0.75)+2y-23=0#
#2.25+2y-23=0#
Simplifying
#2y=23-2.25#
#2y=20.75#
#y=10.375#
Center for the circum circle has the coordinates
#(0.75,10.375)#
Radius r=distance from center to vertex
#(0.75,10.375)----(7,7)#
Area of the circum circle is
#A = pir^2#
#pi=3.141#
#r^2=(7-0.75)^2+(7-10.375)^2#
#r^2=(6.25)^2+(-3.375)^2#
#r^2=50.453#
Hence, Area is
#A=3.141*50.453#
Area of the circum circle is #158.503#