A triangle has corners at `(9 ,4 )`, `(2 ,5 )`, and `(3 ,6 )`. What is the area of the triangle's circumscribed circle?

I always begin this type of problem by shifting all 3 points so that one of them is the origin. Let's shift `(2, 5)` to the origin:

`(2,5) to (0,0)`
`(9,4) to (7, -1)`
`(3,6) to (1, 1)`

I do this, because I am going use the standard equation of a circle:

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

to write 3 equations using the points. Because the first point is the origin, equation is simplified to become:

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

We can use equation [1] to substitute `h^2 + k^2` for `r^2` in equations [2] and [3]:

`(7 - h)^2 + (-1 - k)^2 = h^2 + k^2" [4]"`
`(1 - h)^2 + (1 - k)^2 = h^2 + k^2" [5]"`

Expand the left sides of equations [4] and [5]:

`49 - 14h + h^2 + 1 + 2k + k^2 = h^2 + k^2" [6]"`
`1 - 2h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [7]"`

cancel the square terms:

`49 - 14h + cancel(h^2) + 1 + 2k + cancel(k^2) = cancel(h^2) + cancel(k^2)" [6]"`
`1 - 2h + cancel(h^2) + 1 - 2k + cancel(k^2) = cancel(h^2) + cancel(k^2)" [7]"`

Collect the constant terms on the right:

`-14h + 2k = -50" [8]"`
`-2h - 2k = -2" [9]"`

Add equation [8] to equation [9]:

`-16h = -52`

`h = 13/4`

A simplified version of equation [9] will help us find the value k:

`13/4 + k = 1`
`k = -9/4`

Use equation [1] to find the value of `r^2`

`r^2 = (13/4)^2 + (9/4)^2`

`r^2 = 250/16 = 125/8`

The area of the circle is:

`Area = pir^2`

`Area = (125pi)/8`