Question

A triangle has corners at `(6 ,3 )`, `(5 ,8 )`, and `(4 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Shift all 3 points so that one is the origin. Use the standard Cartesian equation of a circle and the new points to write 3 equations. Solve the 3 equations for `r^2`. Use `r^2` to compute the area.

Explanation:

Shift all three points so that one is the origin:

`(6,3) - (4,2) = (2,1)`
`(5,8) - (4,2) = (1,6)`
`(4,2) - (4,2) = (0,0)`

Use the standard Cartesian equation of a circle:

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

and the three new points to write 3 equations:

`(2 - h)^2 + (1 - k)^2 = r^2" [1]"`
`(1 - h)^2 + (6 - k)^2 = r^2" [2]"`
`(0 - h)^2 + (0 - k)^2 = r^2" [3]"`

Expand the squares using the pattern `(a - b)^2 = a^2 - 2ab + b^2`:

`4 - 4h + h^2 + 1 - 2k + k^2 = r^2" [4]"`
`1 - 2h + h^2 + 36 - 12k + k^2 = r^2" [5]"`
`h^2 + k^2 = r^2" [6]"`

Subtract equation [6] from equation [4] and [5]:

`4 - 4h + 1 - 2k = 0" [7]"`
`1 - 2h + 36 - 12k = 0" [8]"`

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

`-4h - 2k = -5" [9]"`
`-2h- 12k = -37" [10]"`

Multiply equation [9] by -6 and add to equation [10]:

`22h = -7`

`h = -7/22`

Substitute `-7/22` for h into equation [10] and solve for k:

`-2(-7/22)- 12k = -37`

`7/11- 12k = -37`

`-12k = -37 -7/11`

`k = 69/22`

Use equation [6], to compute the value of `r^2`

`r^2 = (-7/22)^2 + (69/22)^2`

`r^2 = 4810/484 = 2405/242`

The formula for the area of the circle is

`A = pir^2`

Substitute the value for `r^2`:

`A = (2405pi)/242`