Finding circumradius of regular polygon such a circle is inscribed

Question

Say we want to inscribe a cicle with a regular polygon:

The circumradius is is the distance to the vertices of the regular n-ngon.

• We can see on the image, this circumradius of a regular n-gon is greater at lower n.
• And as n goes to infinity, the circumradius should approach the radius of the inscribed circle.

But what is the equation of the radius?

I've tried calculating it and google it, and the condition must be, the apothem of the regular n-ngon should equal the radius of the inscribed circle.

I would love your help on this.

Answer 1

Okay I had another go at it.

let the side length of a n-polygon be $l$, let the radius of the inner circle be $r$ and let the circumradius be $R$

The side length is given by

$l = 2r \tan(\pi / n) $

$R$, $r$ and $l/2$ forms a triangle with hypotenuse $R$. From Pythagorean we get:

$ R = \sqrt{r^2 + (\frac{l}{2})^2} $

insert value of $l$

$ R = \sqrt{r^2 + (\frac{2r \tan(\pi / n)}{2})^2} $

$ = \sqrt{r^2 + r^2 \tan^2(\pi / n))} $

$ = r \sqrt{\tan^2(\pi / n)) +1} $

When $n \to \infty$, $R \to r$

Answer 2

Perimeter of a regular polygon circumscribed around a circle of constant given radius $r$;

$$ p= 2 ~n ~r ~ \tan \theta = 2 ~n ~r ~ \tan \frac{2 \pi}{2n} \to ~2 \pi r, \text{as } n\to \infty. $$

You should draw the circle diameter contained in an equilateral triangle with same size as the other three.