The standard Cartesian form of the 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 point, and r is the radius.
When I do a problem of this type, I shift all 3 points so that one point is then origin, (0, 0)(0,0), because this simplifies the problem and it does not change the area of this circle:
(2,3) to (0,0)(2,3)→(0,0)
(1,4) to (-1,1)(1,4)→(−1,1)
(7,5) to (5, 2)(7,5)→(5,2)
Use equation [1] and the new points to write 3 equation:
(0 - h)^2 + (0 - k)^2 = r^2" [2]"(0−h)2+(0−k)2=r2 [2]
(-1 - h)^2 + (1 - k)^2 = r^2" [3]"(−1−h)2+(1−k)2=r2 [3]
(5 - h)^2 + (2 - k)^2 = r^2" [4]"(5−h)2+(2−k)2=r2 [4]
Equation [2] simplifies into:
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]:
(-1 - h)^2 + (1 - k)^2 = h^2 + k^2" [6]"(−1−h)2+(1−k)2=h2+k2 [6]
(5 - h)^2 + (2 - k)^2 = h^2 + k^2" [7]"(5−h)2+(2−k)2=h2+k2 [7]
Expand the squares:
1 + 2h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [8]"1+2h+h2+1−2k+k2=h2+k2 [8]
25 - 10h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"25−10h+h2+4−4k+k2=h2+k2 [9]
The h^2 and k^2h2andk2 terms sum to zero:
1 + 2h + 1 - 2k = 0" [10]"1+2h+1−2k=0 [10]
25 - 10h + 4 - 4k = 0" [11]"25−10h+4−4k=0 [11]
Collect the constant terms into a single term on the right:
2h - 2k = -2" [10]"2h−2k=−2 [10]
-10h -4k = -29" [11]"−10h−4k=−29 [11]
Multiply equation [10] by -2 and add to equation [11]:
-14h = -25−14h=−25
h = 25/14h=2514
Substitute #25/14 for h in equation [10] and the solve for k:
2(25/14) - 2k = -22(2514)−2k=−2
k = 1 + 25/14k=1+2514
k = 39/14k=3914
Substitute the values for h and k into equation [5]
r^2 = (25/14)^2 + (39/14)^2r2=(2514)2+(3914)2
r^2 = 2146/196r2=2146196
The area of a circle is:
Area = pir^2Area=πr2
The area of the circumscribed circle is:
Area = 2146/196piArea=2146196π
