A triangle has corners at #(9 ,5 )#, #(2 ,5 )#, and #(3 ,6 )#. What is the area of the triangle's circumscribed circle?

Shift the 3 points so that one of them is the origin:

#(2,5)-(2,5) to (0,0)#
#(9,5)-(2,5) to (7,0)#
#(3,6)-(2,5)to (1,1)#

Use the standard Cartesian form for the equation of a circle,

#(x - h)^2 + (y - k)^2 = r^2#

and the 3 new points to write 3 equations:

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

Equation [1] reduces to, #h^2 + k^2 = r^2" [4]"#

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

#(7 - h)^2 + (0 - k)^2 = h^2 + k^2" [5]"#
#(1 - h)^2 + (1 - k)^2 = h^2 + k^2" [6]"#

Expand the squares on the left side of equations [5] and [6], using the pattern #(a - b)^2 = a^2 -2ab + b^2#:

#49 - 14h+ h^2 + k^2 = h^2 + k^2" [7]"#
#1 - 2h+h^2 + 1 - 2k+ k^2 = h^2 + k^2" [8]"#

The #h^2 and k^2# terms cancel in both equations:

#49 - 14h = 0" [9]"#
#1 - 2h + 1 - 2k =" [10]"#

Equation [9] allows us to solve for h:

#h = 7/2#

Substitute #7/2# for h into equation [10] and solve for k:

#1 - 2(7/2) + 1 - 2k =" [10]"#

#k = 5/2#

Use equation [4] to solve for #r^2#:

#r^2 = (7/2)^2 + (5/2)^2#

#r^2 = 74/4 = 37/2#

The area of a circle is #A = pir^2#

The area of the circumscribed circle is #A = (37pi)/2#