Question

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

Answer 1

The area of the circle is `pir^2`:

`pir^2 = (221pi)/50`

Explanation:

We can use the points to write 3 equations for the circle with radius and center (h, k):
`r^2 = (2 - h)^2 + (4 - k)^2`
`r^2 = (6 - h)^2 + (5 - k)^2`
`r^2 = (4 - h)^2 + (2 - k)^2`

Set the right side of equation 1 equal to right side of equation 2 and equation 3:

`(2 - h)^2 + (4 - k)^2 = (6 - h)^2 + (5 - k)^2`
`(2 - h)^2 + (4 - k)^2 = (4 - h)^2 + (2 - k)^2`

I will use the pattern `(a - b)^2 = a² - 2ab + b²` expand the squares but I will not write the `b^2` terms, because they correspond to `h^2 or k^2` and will be common to both sides

`4 - 4h + 16 - 8k = 36 - 12h + 25 - 10k`
`4 - 4h + 16 - 8k = 16 - 8h + 4 - 4k`

Combine like terms with h and k terms on the left and constants on the right:

`8h + 2k = 41`
`4h - 4k = 0`

`h = k = 41/10`

Return to the first equation and substitute `40/10` for h and k:

`r^2 = (2 - 41/10)^2 + (4 - 41/10)^2`

`r^2 = (20/10 - 41/10)^2 + (40/10 - 41/10)^2`

`r^2 = (-21/10)^2 + (-1/10)^2`

`r^2 = 441/100 + 1/100`

`r^2 = 442/100 = 221/50`

The area of the circle is `pir^2`:

`pir^2 = (221pi)/50`