Using the standard equation of a circle, (x - h)^2 + (y - k)^2 = r^2 we can use the 3 given points to write 3 equations:
[1] (2 - h)^2 + (8 - k)^2 = r^2
[2] (3 - h)^2 + (9 - k)^2 = r^2
[3] (4 - h)^2 + (7 - k)^2 = r^2
Because r^2 = r^2 we can set the left side of equation [1] equal to the left side of equation [2]:
(2 - h)^2 + (8 - k)^2 = (3 - h)^2 + (9 - k)^2
And the same for equations [1] and [3]:
(2 - h)^2 + (8 - k)^2 = (4 - h)^2 + (7 - k)^2
Expand the squares for both equations, using the pattern (a - b)^2 = a^2 - 2ab + b^2:
4 - 4h + h^2 + 64 - 16k + k^2 = 9 -6h + h^2 + 81 -18k+ k^2
4 - 4h + h^2 + 64 - 16k + k^2 = 16 -8h + h^2 + 49 -14k + k^2
The square terms cancel:
4 - 4h + 64 - 16k = 9 - 6h + 81 -18k
4 - 4h + 64 - 16k = 16 -8h + 49 -14k
Collect all of the constant terms on the left:
-4h -16k = -6h -18k + 22
-4h -16k = -8h-14k - 3
Collect all the h terms on the right:
-16k = -2h -18k + 22
-16k = -4h-14k - 3
Collect all of the k terms on the left:
[4] 2k = -2h + 22
[5] -2k = -4h - 3
Divide equation 4 by 2 and equation 5 by -2:
[6] k = -h + 11
[7] k = 2h + 3/2
Because k = k set the right side of equation [7] equal to the right side of equation [6]
2h + 3/2 = -h + 11
3h = 11 - 3/2
h = 19/6
Substitute 19/6 for h in equation [6]
k = -19/6 + 11
k = 47/6
To find the value of r^2, substitute 19/6 for h and 47/6 for k in equation [1]
(2 - 19/6)^2 + (8 - 47/6)^2 = r^2
(-7/6)^2 + (1/6)^2 = r^2
49/36 + 1/36 = r^2
r^2 = 50/36
The area, A, of the circle is pir^2
A = (50pi)/36
