Question

A triangle has corners at `(9 ,1 )`, `(4 ,6 )`, and `(7 ,4 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`"Area" = (169pi)/2`

Explanation:

One way to solve this problem is as follows.

The vertices of the triangle are, also, points on the circumscribed circle; this allow us to use the Cartesian equation for a circle, `(x-h)^2+(y-k)^2=r^2`, and the 3 given `(x,y)` points, `(9,1), (4,6), and (7,4)` to write 3 equations:

`(9-h)^2+(1-k)^2=r^2" [1]"`
`(4-h)^2+(6-k)^2=r^2" [2]"`
`(7-h)^2+(4-k)^2=r^2" [3]"`

where `(h,k)` is the center of the circumscribed circle and `r` is its radius.

After a lot of non-linear algebra, you can verify that `r = 13/sqrt2`

The formula for the area of the circumscribed circle is:

`"Area" = pir^2`

Substitute `r = 13/sqrt2`:

`"Area" = (169pi)/2`