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#
