Question

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

Answer 1

`{425 pi}/98`

Explanation:

We need to find the radius `R` of the circumscribed circle `C`. Let the coordinates of the center of `C` be `(x,y)`. This point is at a distance of `R` from each of the three corners of the triangle.

Using the distance formula we get three equations

`(x-3)^2 + (y-8)^2 = R^2`
`(x-5)^2 + (y-9)^2 = R^2`
`(x-4)^2 + (y-5)^2 = R^2`

The first two of these give us

`(x-3)^2 + (y-8)^2 = (x-5)^2 + (y-9)^2`

and the last two

`(x-5)^2 + (y-9)^2 = (x-4)^2 + (y-5)^2`

(You could try to form another of these by taking the first and the third, but that one will turn out to be equivalent to these two)

Changing sides and using `a^2 -b^2 = (a+b)(a-b)` we can simplify the two equations above to

`4x + 2y = 33`
`2x + 8y=65`

This pair of linear simultaneous equations can be solved easily to yield

`x = {67}/{14}, y = {97}/{14}`

Substituting these in any of the first three equations will give us

`R^2 = {425}/{98}`

and the area of the circumscribed circle is simply `pi R^2`.