A triangle has corners at $(4 ,4 )$ , $(8 ,2 )$ , and $(3 ,1 )$ . What is the area of the triangle's circumscribed circle?

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Answer 1 · 2018-06-05T22:07:32

${325 pi}/49$

The radius of the circumcircle equals the product of the sides of the triangle divided by four times the area of the triangle.

This is more useful squared:

$r^2 = {a^2 b^2 c^2}/{16 \ text{area}^2}$

Archimedes' Theorem says

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

That all we need to do our problem.

$a^2 = (4-8)^2+(4-2)^2=20$

$b^2 = (8-3)^2+(2-1)^2=26$

$c^2=(4-3)^2+(4-1)^2=10$

$r^2 = {20(26)(10)}/{ 4(10)(20)-(26-20-10)^2} = 325/49$

I permuted $a^2, b^2$ and $c^2$ , necessarily permissible in an area formula.

Anyway the area of the circumcircle is of course $pi r^2 =$

${325 pi}/49$

A triangle has corners at (4 ,4 )(4 ,4 ), (8 ,2 )(8 ,2 ), and (3 ,1 )(3 ,1 ). What is the area of the triangle's circumscribed circle?