Question

A triangle has corners at `(3 ,5 )`, `(4 ,9 )`, and `(7 ,6 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circumscribed circle is:

`A = 5.78pi`

Explanation:

When I do this type problem, I always shift all of the corners so that one of them is the origin. This does not change the area of the circumscribed circle and it makes one of the 3 equations (that we must write) become very simple and useful.

Subtract 3 from every x coordinate and subtract 5 from every y coordinate:

`(3,5) to (0,0)`
`(4,9) to (1, 4)`
`(7,6) to (4, 1)`

Use the standard form of the equation of a circle,

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

, and the 3 shifted points to write three equations:

`(0 - h)^2 + (0 - k)^2 = r^2" [1]"`
`(1 - h)^2 + (4 - k)^2 = r^2" [2]"`
`(4 - h)^2 + (1 - k)^2 = r^2" [3]"`

Please observe that equation [1] simplifies to:

`h^2 + k^2 = r^2" [4]"`

This allows us to substitute `h^2 + k^2` for `r^2` in equations [2] and [3]:

`(1 - h)^2 + (4 - k)^2 = h^2 + k^2" [5]"`
`(4 - h)^2 + (1 - k)^2 = h^2 + k^2" [6]"`

Expand the squares, using the pattern #(a - b)^2 = a^2 - 2ab + b^2:

`1 - 2h + h^2 + 16 - 8k + k^2 = h^2 + k^2" [5]"`
`16 - 8h + h^2 + 1 - 2k + k^2 = h^2 + k^2" [6]"`

`h^2 and k^2` are on both side of the equation, therefore, they sum to zero:

`1 - 2h + 16 - 8k = 0" [7]"`
`16 - 8h + 1 - 2k = 0" [8]"`

collect the constant terms into a single term on the right:

`-2h - 8k = -17" [7]"`
`-8h - 2k = -17" [8]"`

Multiply equation [8] by -4 and add to equation [7]:

`30h = 51`

`h = 1.7`

Substitute 1.7 for h in equation [8]:

`-8(1.7) - 2k = -17`

`-2k = -3.4`

`k = 1.7`

Use equation [4] to find the value of `r^2`:

`r^2 = 1.7^2 + 1.7^2`

`r^2 = 5.78`

The area of a circle is:

`A = pir^2`

The area of the circumscribed circle is:

`A = 5.78pi`