Question

A triangle has corners at `(4 ,7 )`, `(1 ,3 )`, and `(6 ,5 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`frac{725 pi}{98}`

Explanation:

The circumcised circle is just the circle that has those three points on it. We need the squared radius, so let's solve for the circle's equation.

`(x-a)^2 + (y-b)^2 = r^2`

Each point gives us an equation:

`(4-a)^2+(7-b)^2=r^2`
`(1-a)^2+(3-b)^2=r^2`
`(6-a)^2+(5-b)^2=r^2`

Expanding,

`16 - 8a + a^2 + 49 - 14b + b^2 = r^2`
`1 - 2a + a^2 + 9 - 6b + b^2 = r^2`
`36 - 12a + a^2 + 25 - 10b + b^2 =r ^2`

Subtracting pairs,

`65 - 10 - 8 a + 2a - 14b + 6b = 0`
`55= 6a + 8 b`
`65 - 61- 8 a + 12 a - 14 b + 10 b = 0`
`4 = -4 a + 4 b`
`1 = -a + b`
`8 = -8 a + 8 b`

Subtracting,

`55 - 8 = 6a - -8a`
`a = 47/14`
`b=a+1=61/14`

What we're really after is `r^2`. We substitute any one of the points:

`r^2 = (6 - 47/14)^2 + (5-61/14)^2`
`= frac{1}{14^2} ( ( 6(14)-47)^2 + (5(14) - 61)^2 )`

`r^2 = 725/98`

The area we seek is

`A = \pi r^2 = frac{725 pi}{98}`