Question

A triangle has corners at `(9 ,5 )`, `(2 ,1 )`, and `(3 ,6 )`. What is the area of the triangle's circumscribed circle?

Answer 1

Shift ALL the points so that one is the origin. Use the standard Cartesian form for the equation of a circle and the new points to write 3 equations. Use the 3 equations to solve for `r^2`.

Explanation:

Shift all 3 points so that one of them is the origin:

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

This is the standard Cartesian form for the equation of a circle:

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

Use the new points to write 3 equations:

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

Expand the squares:

`h^2 + k^2 = r^2" [5]"`
`49 - 14h + h^2 + 16 - 8k + k^2 = r^2" [6]"`
`1 - 2h + h^2 + 25 - 10k + k^2 = r^2" [7]"`

Subtract equation [6] from equation [5] and equation [7] from equation [5]:

`-49 + 14h - 16 + 8k = 0" [8]"`
`-1 + 2h - 25 + 10k = 0" [9]"`

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

`14h + 8k = 65" [10]"`
`2h + 10k = 26" [11]"`

Multiply equation [11] by -7 and add to equation [10]:

`-62k = -117`

`k = 117/62`

Substitute the value for k into equation [11] and solve for h:

`2h + 1170/62 = 26" [11]"`

`h = 221/62`

Use equation [5] to solve for `r^2`:

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

`r^2 = (221/62)^2 + (117/62)^2`

`r^2 = 62530/3844 = 31265/1922`

The area of the circle is:

`A = (31265pi)/1922`