Question

A triangle has corners at `(5 ,6 )`, `(4 ,3 )`, and `(8 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

The area of the circumscribed circle is:

`Area = (2125pi)/338`

Explanation:

The standard Cartesian equation for a circle is:

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

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

Before I use equation [1] and the 3 given points to write 3 equation, I will shift all 3 points so that one of them is the origin, `(0,0)`; this does not affect the area the circle and it makes the problem much easier.

`(4, 3) + (-4, -3) = (0,0)`
`(5,6) + (-4, -3) = (1, 3)`
`(8, 2) + (-4, -3) = (4, -1)`

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

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

Equation [2] simplifies to:

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

We can temporarily eliminate the variable, r, by substituting the left side of equation [5] into the right sides of equations [3] and [4]:

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

Use the pattern, `(a - b)^2 = a^2 - 2ab + b^2`, to expand the squares:

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

`h^2 and k^2` are on both sides of both equations so they will sum to zero:

`1 - 2h + 9 - 6k = 0" [10]"`
`16 - 8h + 1 + 2k = 0" [11]"`

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

`-2h - 6k = -10" [12]"`
`-8h + 2k = -17" [13]"`

Multiply equation [13] by 3 and add to equation [12] and then solve for h:

`-26h = -61`

`h = 61/26`

Substitute `61/26` for h into equation [13] and the solve for k:

`-8(61/26) + 2k = -17`

`-488/26 + 2k = -442/26`

`2k = 488/26 - 442/26`

`2k = 46/26`

`k = 23/26`

Substitute these values of h and k into equation [5] to obtain the value of `r^2`

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

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

The area of a circle is:

`Area = pir^2`

The area of the circumscribed circle is:

`Area = (2125pi)/338`