Question

A triangle has corners at `(7 ,3 )`, `(5 ,8 )`, and `(4 ,2 )`. What is the area of the triangle's circumscribed circle?

Answer 1

`color(blue)((5365pi)/578)` units squared.

Explanation:

The vertices of the triangle are all points on the circumference of the given circle. The equation for a circle is given by:

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

Where `bbh and bbk` are the `bbx` and `bby` coordinates of the centre respectively anf `bbr` is thr radius.

We can make 3 equations from this:

`(7-h)^2+(3-k)^2=r^2 \ \ \ \ \ [1]`

`(5-h)^2+(8-k)^2=r^2 \ \ \ \ \ [2]`

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

Subtracting `[3]` from `[2]`

`69-2h-12k=0 \ \ \ \ [4]`

Subtract `[2]` from `[1]`

`10k-4h-31=0 \ \ \ \ \ [5]`

From `[5]`:

`k=(4h+31)/10`

Substituting in `[4]`

`69-2h-12((4h+31)/10)=0`

`69-2h-(6(4h+31))/5=0`

`345-10h-24h-186=0`

`h=159/34`

Substituting this in `[5]`

`10k-4(159/34)-31=0`

`k=(31+4(159/34))/10`

`k=169/34`

So we have the centre, we now find the radius:

Using vertex `(4,2)`

`(4-159/34)^2+(2-169/34)^2=r^2`

`r^2=5365/578`

Area:

`(5365pi)/578`