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
