Question

A triangle has corners at `(4 ,7 )`, `(1 ,4 )`, and `(6 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`{841 pi}/98`

Explanation:

I worked out the general form [here.](https://socratic.org/questions/a-triangle-has-corners-at-2-5-3-1-and-8-2-what-is-the-area-of-the-triangle-s-cir)

There's a formula we can just apply, but let's just work from the conclusion:

The squared radius of the circumcircle is the product of the squared sides of the triangle divided by `16(text{area})^2`:

`r^2 = {a^2b^2 c^2}/{ 16(text{area})^2 }`

It's easier to leave everything squared. Since we need to generate the squared sides anyway, we note Archimedes' Theorem:

`16(text{area})^2 = 4 a^2 b^2 - (c^2 -a ^2 - b^2)^2`

`r^2 = {a^2b^2 c^2}/{ 4 a^2 b^2 - (c^2 -a ^2 - b^2)^2 }`

`(4,7), (1,4), and (6,2)`

We're free to assign `a,b,c` however we like. I prefer `c` to be the longest side.

`a^2 = (4-1)^2 + (7-4)^2=18`

`b^2 = (4-6)^2+(7-2)^2=29`

`c^2 = (6-1)^2 + (2-4)^2 = 29`

Oh, isosceles.

`r^2 = {(18)(29)(29)}/{ 4 (18)(29) - (18^2) } = 841/98`

The area of the circle is `pi r^2` or `{841 pi}/98`