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.

  • 1
    $\begingroup$ 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 $\endgroup$ – Ronald Apr 18 '12 at 21:58
  • $\begingroup$ @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. $\endgroup$ – Backslash Apr 18 '12 at 22:01
  • $\begingroup$ @Ronald if you could explain it in a little more depth I would be very grateful. $\endgroup$ – Backslash 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 $\frac{ABC}{4K} = R$.

| cite | improve this answer | |
  • $\begingroup$ I figured it out! I worked everything out, and now completely understand this equation, all thanks to you. Many many many many thanks. $\endgroup$ – Backslash Apr 19 '12 at 0:43
  • $\begingroup$ 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? $\endgroup$ – Backslash Apr 19 '12 at 3:18
  • 1
    $\begingroup$ 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 $\endgroup$ – Michael Biro Apr 19 '12 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.