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`