A triangle has corners at `(7 ,5 )`, `(2 ,3 )`, and `(1 ,4 )`. What is the area of the triangle's circumscribed circle?

The standard Cartesian form of the equation of a circle is:

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

where `(x, y)` is any point on the circle, `(h, k)` is the center point, and r is the radius.

When I do a problem of this type, I shift all 3 points so that one point is then origin, `(0, 0)`, because this simplifies the problem and it does not change the area of this circle:

`(2,3) to (0,0)`

`(1,4) to (-1,1)`

`(7,5) to (5, 2)`

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

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

Equation [2] simplifies into:

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

Substitute the left side of equation [5] into the right sides of equations [3] and [4]:

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

Expand the squares:

`1 + 2h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [8]"`
`25 - 10h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"`

The `h^2 and k^2` terms sum to zero:

`1 + 2h + 1 - 2k = 0" [10]"`
`25 - 10h + 4 - 4k = 0" [11]"`

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

`2h - 2k = -2" [10]"`
`-10h -4k = -29" [11]"`

Multiply equation [10] by -2 and add to equation [11]:

`-14h = -25`

`h = 25/14`

Substitute #25/14 for h in equation [10] and the solve for k:

`2(25/14) - 2k = -2`

`k = 1 + 25/14`

`k = 39/14`

Substitute the values for h and k into equation [5]

`r^2 = (25/14)^2 + (39/14)^2`

`r^2 = 2146/196`

The area of a circle is:

`Area = pir^2`

The area of the circumscribed circle is:

`Area = 2146/196pi`