A triangle has corners at #(5 ,6 )#, #(4 ,3 )#, and #(8 ,2 )#. What is the area of the triangle's circumscribed circle?

The standard Cartesian equation for a circle is:

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

where x and y correspond to any point, #(x, y)# on the circle, h and k correspond to the center point #(h, k)#, and r is the radius of the circle.

Before I use equation [1] and the 3 given points to write 3 equation, I will shift all 3 points so that one of them is the origin, #(0,0)#; this does not affect the area the circle and it makes the problem much easier.

#(4, 3) + (-4, -3) = (0,0)#
#(5,6) + (-4, -3) = (1, 3)#
#(8, 2) + (-4, -3) = (4, -1)#

Use equation [1] and the 3 shifted points to write 3 equations:

#(0 - h)^2 + (0 - k)^2 = r^2" [2]"#
#(1 - h)^2 + (3 - k)^2 = r^2" [3]"#
#(4 - h)^2 + (-1 - k)^2 = r^2" [4]"#

Equation [2] simplifies to:

#h^2 + k^2 = r^2" [5]"#

We can temporarily eliminate the variable, r, by substituting the left side of equation [5] into the right sides of equations [3] and [4]:

#(1 - h)^2 + (3 - k)^2 = h^2 + k^2" [6]"#
#(4 - h)^2 + (-1 - k)^2 = h^2 + k^2" [7]"#

Use the pattern, #(a - b)^2 = a^2 - 2ab + b^2#, to expand the squares:

#1 - 2h + h^2 + 9 - 6k + k^2 = h^2 + k^2" [8]"#
#16 - 8h + h^2 + 1 + 2k + k^2 = h^2 + k^2" [9]"#

#h^2 and k^2# are on both sides of both equations so they will sum to zero:

#1 - 2h + 9 - 6k = 0" [10]"#
#16 - 8h + 1 + 2k = 0" [11]"#

Collect the constant terms into a single term on the right:

#-2h - 6k = -10" [12]"#
#-8h + 2k = -17" [13]"#

Multiply equation [13] by 3 and add to equation [12] and then solve for h:

#-26h = -61#

#h = 61/26#

Substitute #61/26# for h into equation [13] and the solve for k:

#-8(61/26) + 2k = -17#

#-488/26 + 2k = -442/26#

#2k = 488/26 - 442/26#

#2k = 46/26#

#k = 23/26#

Substitute these values of h and k into equation [5] to obtain the value of #r^2#

#(61/26)^2 + (23/26)^2 = r^2#

#r^2 = 4250/676 = 2125/338#

The area of a circle is:

#Area = pir^2#

The area of the circumscribed circle is:

#Area = (2125pi)/338#