When I do this type of problem, I always shift the 3 points so that one of them becomes the origin:
(5,1) to (0,0)(5,1)→(0,0)
(2,9) to (-3,8)(2,9)→(−3,8)
(4,3) to (-1, 2)(4,3)→(−1,2)
The standard equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2" [1]"(x−h)2+(y−k)2=r2 [1]
where (x, y)(x,y) is any point on the circle, (h, k)(h,k) is the center, and r is the radius.
Use equation [1] and the 3 shifted points to write 3 equations:
(0 - h)^2 + (0 - k)^2 = r^2" [2]"(0−h)2+(0−k)2=r2 [2]
(-3 - h)^2 + (8 - k)^2 = r^2" [3]"(−3−h)2+(8−k)2=r2 [3]
(-1 - h)^2 + (2 - k)^2 = r^2" [4]"(−1−h)2+(2−k)2=r2 [4]
Equation [2] simplifies to a very useful equation:
h^2 + k^2 = r^2" [5]"h2+k2=r2 [5]
Substitute the left side of equation [5] into the right sides of equations [3] and [4]:
(-3 - h)^2 + (8 - k)^2 = h^2 + k^2" [6]"(−3−h)2+(8−k)2=h2+k2 [6]
(-1 - h)^2 + (2 - k)^2 = h^2 + k^2" [7]"(−1−h)2+(2−k)2=h2+k2 [7]
Expand the squares on the left side of equations [6] and [7]:
9 + 6h + h^2 + 64 - 16k + k^2 = h^2 + k^2" [8]"9+6h+h2+64−16k+k2=h2+k2 [8]
1 + 2h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"1+2h+h2+4−4k+k2=h2+k2 [9]
h^2 + k^2h2+k2 is on both sides of the equations and this sums to zero:
9 + 6h + 64 - 16k = 0" [10]"9+6h+64−16k=0 [10]
1 + 2h + 4 - 4k = 0" [11]"1+2h+4−4k=0 [11]
Collect the constant terms into a single term on the right:
6h - 16k = -73" [12]"6h−16k=−73 [12]
2h - 4k = -5" [13]"2h−4k=−5 [13]
Multiply equation [13] by -4 and add to equation [12]:
-2h = -53−2h=−53
h = 53/2h=532
Substitute 53/2532 for h in equation [13};
2(53/2) - 4k = -52(532)−4k=−5
-4k = -58−4k=−58
k = 29/2k=292
Use equation [5] to compute r^2r2:
r^2 = (53/2)^2 + (29/2)^2r2=(532)2+(292)2
r^2 = 2809/4 + 841/4r2=28094+8414
r^2 = 3650/4r2=36504
r^2 = 1825/2r2=18252
The area of the circle is:
A = pir^2A=πr2
A = 1825/2piA=18252π
