Question

A triangle has corners at `(9 ,3 )`, `(4 ,1 )`, and `(2 ,8 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circumscribed circle is:

`A = pi56869/3402`

Explanation:

The standard Cartesian form of the equation of a circle is:

`(x - h)^2 + (y - k)^2 = r^2" [1]"`

Because we only care about finding the square of the radius, I always shift all 3 points the same amount so that 1 of the points is the origin:

`(4-4,1-1)to(0,0)`

`(9-4,3-1) to (5, 2)`

`(2-4,8-1)to(-2,7)`

This makes 1 of the 3 equations that we will write using equation [1] become:

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

The other 2 points gives us the following equations:

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

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

Substitute the left side of equation [2] into the right side of equations [3] and [4]:

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

`(-2 - h)^2 + (7 - k)^2 = h^2 + k^2" [6]"`

Expand the squares:

`25 - 10h+h^2 + 4 - 4k+k^2 = h^2 + k^2" [7]"`

`4 + 4h+h^2 + 49 - 14k+k^2 = h^2 + k^2" [8]"`

The square terms cancel:

`25 - 10h+cancel(h^2) + 4 - 4k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [7]"`

`4 + 4h+cancel(h^2) + 49 - 14k+cancel(k^2) = cancel(h^2) + cancel(k^2)" [8]"`

Leaving the following linear equations:

`10h + 4k = 29" [9]"`

`4h - 14k = -53" [10]"`

To obtain the value of h, multiply equation [9] by 7 and equation [10] by 2 and then add them:

`70h + 8h = 203 - 106`

`78h = 97`

`h = 97/78`

Substitute this value for h into equation [9]:

`10(97/78) + 4k = 29`

`4k = (2262 - 970)/78`

`k = 323/78`

Use equation [2] to obtain the value of `r^2`:

`r^2 = (97/78)^2+(323/78)^2`

`r^2 = 113738/6804 = 56869/3402`

The area of a circle is:

`A = pir^2`

The area of the circumscribed circle is:

`A = pi56869/3402`