Question

A triangle has corners at `(2 ,4 )`, `(3 ,6 )`, and `(4 ,7 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`65/2 pi`

Explanation:

Just did one just like this.

The circumscribed circle is just the circle with the three vertices on it. The general equation is

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

Substituting three points,

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

Expanding,

`20 = 4a + 8b + r^2-a^2-b^2`
`45 = 6a + 12 b + r^2-a^2-b^2`
`65 = 8a + 14b + r^2-a^2-b^2`

Subtracting pairs,

`25 = 2a + 4b`
`20 = 2a + 2b`

Subtracting,

`5 = 2b`
`b=5/2`
`a = 10 - b = 15/2`

Substituting,

`r^2 = (2-15/2)^2 + (4-5/2)^2= 65 /2`

So the circle's area is

`A = \pi r^2 = 65/2 pi`