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
