Question

A triangle has corners at `(5 ,1 )`, `(2 ,9 )`, and `(4 ,3 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circle is `A = 1825/2pi`

Explanation:

When I do this type of problem, I always shift the 3 points so that one of them becomes the origin:

`(5,1) to (0,0)`
`(2,9) to (-3,8)`
`(4,3) to (-1, 2)`

The standard equation of a circle is:

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

where `(x, y)` is any point on the circle, `(h, k)` is the center, and r is the radius.

Use equation [1] and the 3 shifted points to write 3 equations:

`(0 - h)^2 + (0 - k)^2 = r^2" [2]"`
`(-3 - h)^2 + (8 - k)^2 = r^2" [3]"`
`(-1 - h)^2 + (2 - k)^2 = r^2" [4]"`

Equation [2] simplifies to a very useful equation:

`h^2 + k^2 = r^2" [5]"`

Substitute the left side of equation [5] into the right sides of equations [3] and [4]:

`(-3 - h)^2 + (8 - k)^2 = h^2 + k^2" [6]"`
`(-1 - h)^2 + (2 - k)^2 = h^2 + k^2" [7]"`

Expand the squares on the left side of equations [6] and [7]:

`9 + 6h + h^2 + 64 - 16k + k^2 = h^2 + k^2" [8]"`
`1 + 2h + h^2 + 4 - 4k + k^2 = h^2 + k^2" [9]"`

`h^2 + k^2` is on both sides of the equations and this sums to zero:

`9 + 6h + 64 - 16k = 0" [10]"`
`1 + 2h + 4 - 4k = 0" [11]"`

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

`6h - 16k = -73" [12]"`
`2h - 4k = -5" [13]"`

Multiply equation [13] by -4 and add to equation [12]:

`-2h = -53`

`h = 53/2`

Substitute `53/2` for h in equation [13};

`2(53/2) - 4k = -5`

`-4k = -58`

`k = 29/2`

Use equation [5] to compute `r^2`:

`r^2 = (53/2)^2 + (29/2)^2`

`r^2 = 2809/4 + 841/4`

`r^2 = 3650/4`

`r^2 = 1825/2`

The area of the circle is:

`A = pir^2`

`A = 1825/2pi`