A triangle has corners at #(4 ,6 )#, #(5 ,9 )#, and #(8 ,5 )#. What is the area of the triangle's circumscribed circle?

Because we only care about the square of the radius of the circle, we can shift all 3 points so that one of them is the origin:

#(4-4,6-6)to (0,0)#
#(5-4,9-6)to (1,3)#
#(8-4,5-6) to(4,-1)#

We can use the general equation of a circle:

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

And the three new points #(0.0)#, #(1,3)# and #(4,-1)# to write 3 equations. But we know that the equation for the first point is:

#h^2+k^2=r^2" [1]"#

The other 2 equation are:

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

Substitute the left side of equation [1] into the right sides of equations [2] and [3]

#(1-h)^2+(3-k)^2=h^2+k^2" [4]"#
#(4-h)^2+(-1-k)^2=h^2+k^2" [5]"#

Expand the squares:

#1-2h+h^2+9-6k+k^2=h^2+k^2" [6]"#
#16-8h+h^2+1+2k+k^2=h^2+k^2" [7]"#

Combine like terms and move the constant terms to the right:

#-2h-6k=-10" [8]"#
#-8h+2k=-17" [9]"#

Multiply equation [8] by -4 and add to equation [9]:

#26k = 23#

#k = 23/26#

Use equation [9] to find the value of h:

#-8h+2k=-17#

#h = 17/8+1/4(23/26)#

#h = 61/26#

Substitute the values of h and k into equation [1]:

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

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

The area of a circle is:

#"Area" = pir^2#

Substitute the value of #r^2#:

#"Area" = pi2125/338#