A triangle has corners at #(3 ,7 )#, #(2 ,5 )#, and #(8 ,4 )#. What is the area of the triangle's circumscribed circle?

Perhaps there is a simpler way, but here is how I would do it:

Part 1: Find the Lengths of the Three Sides
Let #a# be the side #(3,7):(2,5)#
#color(white)("XXX")abs(a)=sqrt((2-3)^2+(5-7)^2)=sqrt(1+4)=sqrt(5)#

Let #b# be the side #(2,5):(8,4)#
#color(white)("XXX")abs(b)=sqrt((8-2)^2+(4-5)^2)=sqrt(36+1)=sqrt(37)#

Let #c# be the side #(8,4):(3,7)#
#color(white)("XXX")abs(c)=sqrt((3-8)^2+(7-4)^2)=sqrt(25+9)=sqrt(34)#

Part 2: Find the Perimeter, Semi-Perimeter, and Area of the Triangle
(I will need the help of a calculator here)
Perimeter: #p=sqrt(5)+sqrt(37)+sqrt(34)~~14.1497824026#

Semi-perimeter: #s=p/2~~7.0748912013#

Area of triangle (using Heron's formula):
#color(white)("XXX")"Area_triangle=sqrt(s(s-a)(s-b)(s-c))~~6.5#

Part 3: Find the Radius and Area of the Circumscribed Circle
(more calculator usage)
Radius of Circumscribed Circle: #r=("Area"_triangle)/s=0.9187420435#
(ask as a separate question if you are unfamiliar with this relation).

Area of Circle: #"Area"_circ = pir^2 =2.6517773377#

The standard form for the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2#

where #(h, k)# is the center and #r# is the radius

Because the triangles vertices are points on the circumscribed circle, we can use the 3 points and the standard form to write 3 equations:

#(3 - h)^2 + (7 - k)^2 = r^2##" [1]"#
#(2 - h)^2 + (5 - k)^2 = r^2##" [2]"#
#(8 - h)^2 + (4 - k)^2 = r^2##" [3]"#

We have 3 equations and we can use them to find the values of, h, k, and #r^2#.

Temporarily eliminate #r^2# by setting the left side of equation [1] equal to the left side of equation [2] and the same for left side of equation [3]:

#(3 - h)^2 + (7 - k)^2 = (2 - h)^2 + (5 - k)^2##" [1 = 2]"#
#(3 - h)^2 + (7 - k)^2 = (8 - h)^2 + (4 - k)^2##" [1 = 3]"#

Expand the squares, using the pattern, #(a - b)^2 = a^2 - 2ab + b^2#:

#9 -6h + h^2 + 49 -14k + k^2 = 4 -4h + h^2 + 25 -10k + k^2#
#9 -6h + h^2 + 49 -14k + k^2 = 64 -16h + h^2 + 16 -8k + k^2#

Please notice that for every #h^2 and k^2# on the left there is a corresponding one on the right so they cancel:

#9 -6h + 49 -14k = 4 -4h + 25 -10k#
#9 -6h + 49 -14k = 64 -16h + 16 -8k#

Combine all of the constant terms into a single term on the right:

#- 6h - 14k = - 4h- 10k - 29#
#- 6h- 14k =- 16h- 8k + 22#

Combine all of the h terms into a single term on the right:

#- 14k = 2h- 10k - 29#
#- 14k =- 10h- 8k + 22#

Combine all of the k terms into a single term on the left:

#-4k = 2h - 29##" [4]"#
#-6k = -10h + 22##" [5]"#

Multiply equation [4] by -3 and equation [5] by 2 and add them:

#12k - 12k = -6h - 20h + 87 + 44#

#0 = -26h + 131#

#h = 131/26#

Substitute #131/26# for h in equation [4]:

#-4k = 2(131/26) - 29#

#-4k = -492/26#

#k = 123/26#

Substitute the values for h and k into equation [3]

#(8 - 131/26)^2 + (4 - 123/26)^2 = r^2#

#(77/26)^2 + (-19/26)^2 = r^2#

#r^2 = 6290/676#

The area of the circle is #A = pir^2#

#A = (6290pi)/676#