# Calculating radius of circles which are a product of Tangent Intersections using a Regular Polygon

## Introduction

Lets have a regular polygon of $n$ sides inscribed in a circle of radius $H$, then construct tangents between the circle and each point of the polygon and draw new circle(s) trough the intersections of the tangents like this :

($n=3$ here)

($n=4$ here)

Lets call the radius of new blue circle $R_1$ which i have calculated to be: $$R_1= \frac{H}{cos(\frac{180}{n})}$$

Then if we set $n$ to $5$ and $6$ we get a second one:

($n=5$ here)

($n=6$ here)

Which I have calculated to be: $$R_2=H\times[ 1 + tan(\frac{180}{n })\times tan(\frac{360}{n}) ]$$

Then lets see the third one at $n=7$ :

Then I stopped before trying to do $R_3$, and thought If there could be a generalized formula for $R_m$?

## Visualization

See for example $n=32$ :

$n=100$

$n=100$ Zoomed Out:

## Question

How would one make a generalized formula to calculate $R_m$ if possible? Or else make a set of expressions to help calculate $R_1, R_2, R_3... R_m$ ?

This is also similar to my previous question in some sense:

Calculating radius of circles which are a product of Circle Intersections using Polygons

## 1 Answer

The answer is very simple: $$R_{kn}=H{1\over \cos(k\pi/n)},$$ where $k=1$ for the first circle, $k=2$ for the second one, and so on. This formula can be derived, for instance, by finding the intersection of a couple of tangent lines. As the line tangent at point $j$ for the polygon with $n$ sides centered at the origin has equation $x\cos{2j\pi\over n}+y\sin{2j\pi\over n}=1$ (I've set $H=1$ for simplicity) then one gets for the intersection points of line $i$ with line $j$: $$P_{ijn}= \left(\cos\frac{\pi(i+j)}{n} \sec\frac{\pi(i-j)}{n},\ \sin\frac{\pi (i+j)}{n} \sec\frac{\pi(i-j)}{n}\right).$$ Letting here $i-j=k$ one gets the above result for $R_{kn}=\sqrt{x_{ijn}^2+y_{ijn}^2}$.