Question

A triangle has corners at `(4 ,7 )`, `(8 ,9 )`, and `(3 ,5 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`Area = (3665pi)/576`

Explanation:

Let's subtract 3 from every x coordinate and 5 from every y coordinate; this will make the 3rd point the origin.

`(1,2), (5,4) and (0,0)`

We do this because we are going to use the standard form of the equation of a circle:

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

and the 3 points to write 3 equations. This makes the equation for the point that is the origin very simple and useful:

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

Substitute the left side of equation [1] into equations [2] and [3]:

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

Expand the squares:

`1 - 2h+h^2 + 4 - 4k+k^2 = h^2 + k^2" [6]"`
`25 - 10h+h^2 + 16 - 8k+k^2 = h^2 + k^2" [7]"`

The square terms cancel:

`1 - 2h+cancel(h^2) + 4 - 4k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [6]"`
`25 - 10h+cancel(h^2) + 16 - 8k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [7]"`

Remove all of the cancelled terms and collect the constants on the right:

`-2h - 4k = -5" [8]"`
`-10h - 8k = -31" [9]"`

Multiply equation [8] by -2 and add to equation [9]:

`-6h = -21`

`h = 7/3`

Substitute into equation [9]:

`-10(7/3) - 8k = -31`

`k = 23/24`

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

`r^2 = (7/3)^ + (23/24)^2`

`r^2 = (7/3)^ + (23/24)^2`

`r^2 = 3665/576`

The area of the circle is:

`Area = pir^2`

`Area = (3665pi)/576`