The standard form of the equation of circle is:
#(x - h)^2 + (y - k)^2 = r^2#
where #(x,y)# any point on the circle, #(h, k)# is the center, and r is the radius.
Move all three points to the left 3 and down 8:
#A = (0,0), B = (2,1), and C = (1, -2)#
This does not change the size of the triangle or the circumscribed circle but it greatly simplifies the first equation that we write, using the points:
#h^2 + k^2 = r^2" [1]"#
#(2 - h)^2 + (1 - k)^2 = r^2" [2]"#
#(1 - h)^2 + (-2 - k)^2 = r^2" [3]"#
Substitute #h^2 + k^2# for #r^2# equations [2] and [3]:
#(2 - h)^2 + (1 - k)^2 = h^2 + k^2" [4]"#
#(1 - h)^2 + (-2 - k)^2 = h^2 + k^2" [5]"#
Expand the squares:
#4 - 4h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [6]"#
#1 - 2h + h^2 + 4 + 4k + k^2 = h^2 + k^2" [7]"#
The square terms cancel:
#4 - 4h + cancel(h^2) + 1 - 2k + cancel(k^2) = cancel(h^2) + cancel(k^2)" [8]"#
#1 - 2h + cancel(h^2) + 4 + 4k + cancel(k^2) = cancel(h^2) + cancel(k^2)" [9]"#
Collect the constant terms on the right:
#-4h - 2k = -5" [10]"#
#-2h + 4k = -5" [11]"#
Multiply equation [10] by 2 and add to equation [11]:
#-10h = -15#
Solve for h:
#h = 3/2#
Substitute #3/2# for h in equation [11]:
#-2(3/2) + 4k = -5#
#4k = -2#
#k = -1/2#
Use equation [1] to find the value of #r^2#
#r^2 = (3/2)^2 + (-1/2)^2#
#r^2 = 10/4#
The area of a circle is:
#Area = pir^2#
Substitute #10/4# for #r^2#:
#Area = (10pi)/4#
