Using the 3 points and the standard equation of a circle, r^2 = (x - h)^2 + (x - k)^2r2=(x−h)2+(x−k)2, we can write 3 equations:
r^2 = (3 - h)^2 + (5 - k)^2r2=(3−h)2+(5−k)2
r^2 = (7 - h)^2 + (9 - k)^2r2=(7−h)2+(9−k)2
r^2 = (4 - h)^2 + (2 - k)^2r2=(4−h)2+(2−k)2
Because r^2 = r^2r2=r2, we can set the right side of the first equation equal to the right side of the second equation and the third equation:
(3 - h)^2 + (5 - k)^2 = (7 - h)^2 + (9 - k)^2(3−h)2+(5−k)2=(7−h)2+(9−k)2
(3 - h)^2 + (5 - k)^2 = (4 - h)^2 + (2 - k)^2(3−h)2+(5−k)2=(4−h)2+(2−k)2
Expand the squares:
9 - 6h + h^2 + 25 - 10k + k^2 = 49 - 14h + h^2 + 81 - 18k + k^29−6h+h2+25−10k+k2=49−14h+h2+81−18k+k2
9 - 6h + h^2 + 25 - 10k + k^2 = 16 - 8h + h^2 + 4 - 4k + k^29−6h+h2+25−10k+k2=16−8h+h2+4−4k+k2
combine like terms and cancel the h^2h2 and k^2k2 terms:
8k = 96 - 8h8k=96−8h
-6k = -2h - 14−6k=−2h−14
Divide the first equation by 8:
k = 12 - hk=12−h
-6k = -2h - 14−6k=−2h−14
Substitution:
-6(12 - h) = -2h - 14−6(12−h)=−2h−14
-72 + 6h = -2h - 14−72+6h=−2h−14
8h = 588h=58
h = 29/4h=294
k = 48/4 - 29/4k=484−294
k = 19/4k=194
Substitute (29/4, 19/4)(294,194) for (h, k)(h,k):
r^2 = (3 - 29/4)^2 + (5 - 19/4)^2r2=(3−294)2+(5−194)2
r^2 = (-17/4)^2 + (1/4)^2r2=(−174)2+(14)2
r^2 = 290/16 = 145/8r2=29016=1458
The area of a circle is pir^2πr2
A = (145pi)/8A=145π8
