Question

A triangle has corners at `(4 ,6 )`, `(5 ,9 )`, and `(8 ,5 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`"Area" = pi2125/338`

Explanation:

Because we only care about the square of the radius of the circle, we can shift all 3 points so that one of them is the origin:

`(4-4,6-6)to (0,0)`
`(5-4,9-6)to (1,3)`
`(8-4,5-6) to(4,-1)`

We can use the general equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

And the three new points `(0.0)`, `(1,3)` and `(4,-1)` to write 3 equations. But we know that the equation for the first point is:

`h^2+k^2=r^2" [1]"`

The other 2 equation are:

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

Substitute the left side of equation [1] into the right sides of equations [2] and [3]

`(1-h)^2+(3-k)^2=h^2+k^2" [4]"`
`(4-h)^2+(-1-k)^2=h^2+k^2" [5]"`

Expand the squares:

`1-2h+h^2+9-6k+k^2=h^2+k^2" [6]"`
`16-8h+h^2+1+2k+k^2=h^2+k^2" [7]"`

Combine like terms and move the constant terms to the right:

`-2h-6k=-10" [8]"`
`-8h+2k=-17" [9]"`

Multiply equation [8] by -4 and add to equation [9]:

`26k = 23`

`k = 23/26`

Use equation [9] to find the value of h:

`-8h+2k=-17`

`h = 17/8+1/4(23/26)`

`h = 61/26`

Substitute the values of h and k into equation [1]:

`(61/26)^2+(23/26)^2 = r^2`

`r^2 = 4250/676 = 2125/338`

The area of a circle is:

`"Area" = pir^2`

Substitute the value of `r^2`:

`"Area" = pi2125/338`