# How does this equation to find the radius from 3 points actually work?

I had searched online and found an equation that solves the radius of a circle from 3 points that are located on the circumference of that specific circle. Where I had found this formula did not state its derivation or anything of the likes, however it is to find the radius. With the 3 points, you should form a triangle. Let us call this Triangle $ABC$.

Using the distance formula, we can calculate the distance between $AB$, $BC$, and $AC$. To simplify things, let us call these distances A, B, and C respectively. We also need the area of the triangle. After finding an altitude of the specified triangle, we can use the area of a triangle equation to solve for it. Let us call the area of this triangle $K$.

This is the formula:

$r = \dfrac{ABC}{4K}$

Which is essentially saying

$radius = \dfrac{\text{Product~of~the~triangle~side~lengths}}{\text{The~area~of~the~triangle~multiplied~by~4}}$

This above pseudo equation was just to clarify the formula, I understand everyone here are exceptional mathematicians, and as I mature, I also hope to become one as well.

This, to my astonishment, finds the correct radius, however I am unable to comprehend how this method works. I hope one of the kind people on Math StackExchange are willing to help me out on understanding this formula and its derivation. Many thanks.

-
the radius the same as a distance from a vertex to the 'circumcentre' of the triangle and simplifies to the Law of Sines. en.wikipedia.org/wiki/Law_of_sines#Relation_to_the_circumcircle –  Ronald Apr 18 '12 at 21:58
@Ronald yes, and normally, to find the circumcentre I would find the perpendicular bisector of 2 sides, and the POI of those 2 lines would be the circumcentre. However, in this method I can't understand how it works to find the radius. –  Qasim Apr 18 '12 at 22:01
@Ronald if you could explain it in a little more depth I would be very grateful. –  Qasim Apr 18 '12 at 22:06

Let $a$ be the angle opposite to side $A$. First show that if $R$ is the radius of the circle, then $\frac{A}{\sin a} = 2R$. This isn't hard (just drop a perpendicular from $O$ to $A$, and use the definition of $\sin$ on the similar triangles).
Then, the area of the triangle $ABC$, $K = \frac{1}{2}BC \sin a$, and so $\frac{A}{\sin a} = \frac{A}{\frac{2K}{BC}} = \frac{ABC}{2K} = 2R$, and so $\frac{ABC}{4K} = R$.
I have only one question, however. Why is there a $\text sina$ after the area of a triangle equation? Is that supposed to be there? –  Qasim Apr 19 '12 at 3:18
That is the general formula for the area of a triangle in terms of the side lengths and angles. If we have a right triangle, then $\sin a = 1$ and this reduces to base x height. en.wikipedia.org/wiki/Triangle#Computing_the_area_of_a_triangle –  Michael Biro Apr 19 '12 at 4:59