The standard form for the equation of a circle is:
#(x - h)^2 + (y - k)^2 = r^2#
where #(x,y)# is any point on the circle, #(h,k)# is the center, and r is the radius.
We still have the same size circumscribe circle, if we move the triangle to the left 2 and down 1:
#A = (3, 1), B = (0, 0), and C = (1,3)#
Using standard form and the 3 new points, write 3 equations:
#(3 - h)^2 + (1 - k)^2 = r^2" [1]"#
#(0 - h)^2 + (0 - k)^2 = r^2" [2]"#
#(1 - h)^2 + (3 - k)^2 = r^2" [3]"#
Please notice that equation [2] reduces to, #h^2 + k^2 = r^2#, therefore, we can make this substitution for #r^2# into equation [1] and [3]
#(3 - h)^2 + (1 - k)^2 = h^2 + k^2" [4]"#
#(1 - h)^2 + (3 - k)^2 = h^2 + k^2" [5]"#
Using the pattern #(a - b)^2 = a^2 + 2ab + b^2#, expand the squares:
#9 - 6h +h^2 + 1 - 2k + k^2 = h^2 + k^2" [6]"#
#1 - 2h + h^2 + 9 - 6k + k^2 = h^2 + k^2" [7]"#
The #h^2 and k^2# terms cancel:
#9 - 6h + 1 - 2k = 0" [8]"#
#1 - 2h + 9 - 6k = 0" [9]"#
Multiply both equations by -1 and move the 10 to right side:
#6h + 2k = 10" [10]"#
#2h + 6k = 10" [11]"#
Multiply equation [11] by -3 and add to equation 10:
#0h - 16k = -20" [12]"#
#2h + 6k = 10" [13]"#
#k = 5/4#
Substitute #5/4# for k in equation [13] and the solve for h:
#2h + 6(5/4) = 10#
#2h + 15/2 = 10#
#2h = 10 - 15/2#
#h = 5 - 15/4#
#h = 5/4#
Substitute #5/4# for h and k into the reduced form of equation [2]:
#2(5/4)^2 = r^2#
#r^2 = 25/8#
The area of the circle is:
#Area = pir^2#
#Area = (25pi)/8#
