Let's begin by moving the triangle to the left 4 and down 6:
#A = (3, -2), B = (0, 0), and C = (2, 1)#
this does not change the size of the triangle or the circumscribed circle.
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 centerpoint, and r is the radius.
Use the standard form and the 3 points to write 3 equations:
#(3 - h)^2 + (-2 - k)^2 = r^2" [1]"#
#(0 -h)^2 + (0 - k)^2 = r^2" [2]"#
#(2 - h)^2 + (1 - k)^2 = r^2" [3]"#
Please notice that equation [2] simplifies to #h^2 + k^2 = r^2#, therefore, we can temporarily eliminate the variable #r^2# by substitute this into equations [1] and [3]:
#(3 - h)^2 + (-2 - k)^2 = h^2 + k^2" [4]"#
#(2 - h)^2 + (1 - k)^2 = h^2 + k^2" [5]"#
Expand the squares, using the pattern #(a - b)^2 = a^2 - 2ab + b^2#:
#9 - 6h+ h^2 + 4 + 4k + k^2 = h^2 + k^2" [6]"#
#4 - 4h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [7]"#
The #h^2 and k^2# terms cancel:
#9 - 6h + 4 + 4k = 0" [6]"#
#4 - 4h + 1 - 2k = 0" [7]"#
Collect the constant terms on right:
#-6h + 4k = -13" [8]"#
#-4h -2k = -5" [9]"#
Multiply equation [9] by 2 and add to equation [8]:
#-14h + 0k = -23#
h = 23/14
Substitute into equation [9]
#-4(23/14) -2k = -5#
#-2k = 22/14#
#k = -11/14#
Use equation [2] to compute #r^2#
#r^2 = (23/14)^2 + (-11/14)^2#
#r^2 = 650/196#
#Area = pir^2#
#Area = (650pi)/196#
