A triangle has corners at #(9 ,3 )#, #(4 ,1 )#, and #(2 ,8 )#. What is the area of the triangle's circumscribed circle?

The standard Cartesian form of the equation of a circle is:

#(x - h)^2 + (y - k)^2 = r^2" [1]"#

Because we only care about finding the square of the radius, I always shift all 3 points the same amount so that 1 of the points is the origin:

#(4-4,1-1)to(0,0)#

#(9-4,3-1) to (5, 2)#

#(2-4,8-1)to(-2,7)#

This makes 1 of the 3 equations that we will write using equation [1] become:

#h^2 + k^2 = r^2" [2]"#

The other 2 points gives us the following equations:

#(5 - h)^2 + (2 - k)^2 = r^2" [3]"#

#(-2 - h)^2 + (7 - k)^2 = r^2" [4]"#

Substitute the left side of equation [2] into the right side of equations [3] and [4]:

#(5 - h)^2 + (2 - k)^2 = h^2 + k^2" [5]"#

#(-2 - h)^2 + (7 - k)^2 = h^2 + k^2" [6]"#

Expand the squares:

#25 - 10h+h^2 + 4 - 4k+k^2 = h^2 + k^2" [7]"#

#4 + 4h+h^2 + 49 - 14k+k^2 = h^2 + k^2" [8]"#

The square terms cancel:

#25 - 10h+cancel(h^2) + 4 - 4k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [7]"#

#4 + 4h+cancel(h^2) + 49 - 14k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [8]"#

Leaving the following linear equations:

#10h + 4k = 29" [9]"#

#4h - 14k = -53" [10]"#

To obtain the value of h, multiply equation [9] by 7 and equation [10] by 2 and then add them:

#70h + 8h = 203 - 106#

#78h = 97#

#h = 97/78#

Substitute this value for h into equation [9]:

#10(97/78) + 4k = 29#

#4k = (2262 - 970)/78#

#k = 323/78#

Use equation [2] to obtain the value of #r^2#:

#r^2 = (97/78)^2+(323/78)^2#

#r^2 = 113738/6804 = 56869/3402#

The area of a circle is:

#A = pir^2#

The area of the circumscribed circle is:

#A = pi56869/3402#