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Let's say that the radius of the bigger circle is R. Every circle inside touches the perimeter of the bigger circle and two other circles. How to find the radius of the smaller circles (all are identical).

To illustrate:

enter image description here

Any tips and ideas would be awesome.

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The middle stanza of Soddy's Kiss Precise gives the formula:

Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.

Applied here it says $$\frac 3{r^2}+\frac 1{R^2}=\frac 12 \left(\frac 3r-\frac 1R\right)^2\\\frac 3{r^2}+\frac 1{R^2}=\frac {9}{2r^2}+\frac 1{2R^2}-\frac 3{rR}$$ As all the we can get is the ratio, let $r=1$ and we have $$3+\frac 1{R^2}=\frac 92+\frac 1{2R^2}-\frac 3R\\0=3R^2-6R-1\\R=\frac 16(6\pm\sqrt{48})=\frac 13(3\pm2\sqrt{3})$$ and we want the plus sign.

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Call the radius of the smaller circles $r$. Their centers form an equilateral triangle of side $2 r$. The centers of the small circles are at a distance of $R - r$ from the large circle's center, and $r$ from the large circumference. The tangency points of the smaller and large circle are also an equilateral triangle. I believe that drawing all the triangles mentioned gives you enough in terms of angles to find relations among $r$, $R$ and $R - r$ to get $r$ by trigonometry.

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  • $\begingroup$ Why must the centers form a triangle of size $2r$? In other words, why must the triangle intersect where the circles touch? $\endgroup$ – SomeStrangeUser Feb 6 '14 at 20:06
  • $\begingroup$ Each radius is perpendicular to the tangent line. $\endgroup$ – André Nicolas Feb 6 '14 at 20:08
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Perhaps you can use Descartes' Theorem here: http://en.wikipedia.org/wiki/Descartes%27_theorem

Note: There is also a circle internally tangent to the three tangent circles.

(I get (-3 + 2 √3)R = r for the relationship between the outer circle of radius R to the inner circle(s) of radius r.)

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  • $\begingroup$ The Kiss Precise by Frederick Soddy seems unfortunately to be 425 characters too long! $\endgroup$ – Alan Feb 7 '14 at 1:12
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Two more instances of same question:

The inner/outer circle is $3 \pm 2 \sqrt{3}$ times the radius of the three circles.

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For a ring of n circles inscribed within a larger circle this will calculate the radius or diameter for either the small circles or the larger circle.

$n$ = the number of small circles
$r$ = the radius of the small circles
$R$ = the Radius of the large perimeter circle formed by the outer ring of small circles
$d$ = the diameter of the small circles
$D$ = the Diameter of the large perimeter circle formed by the outer ring of small circles

$r = R ⋅ \sin(π ÷ n) ÷ [\sin(π ÷ n) + 1]$
$R = r ⋅[\sin(π ÷ n) + 1] ÷ \sin(π ÷ n)$

$d = D ⋅ \sin(π ÷ n) ÷ [\sin(π ÷ n) + 1]$
$D = d ⋅[\sin(π ÷ n) + 1] ÷ \sin(π ÷ n)$

Note: the Radians function within Excel is not used for these formulas.

Source: http://www.had2know.com/academics/inner-circular-ring-radii-formulas-calculator.html

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I strongly recommend drawing out the diagram I am going to explain, with the initial diagram given in mind, as even I wouldn't be able to imagine this from my explanation. With that being said,

If we were to label the centers of the smaller circles $a , b , c $ with each one of the circles having a radius of $r$, we could pass a line joining the points $a \Rightarrow c , a \Rightarrow b, b \Rightarrow c$ and take the perpendicular bisectors of these lines, we know they would intersect at the center of the larger circle(proof of which is found here https://proofwiki.org/wiki/Perpendicular_Bisector_of_Chord_Passes_Through_Center), call this point $d$. WLOG , consider $\triangle abd$. It is then simple to see that $|ad| + r =R $ as $R$ is simply the radius of the bigger circle. It is then a matter of finding $|ad|$. Draw a perpendicular bisector through the $120°$ angle from $d$ to the line joining $a$ and $b$. You will now be left with a $30,60,90$ triangle and simple trigonometry solves the problem $$ \cos(30) = \frac{r}{|ad|} \implies |ad| = \frac{2r}{\sqrt3}$$ Returning to our original idea, $$ \frac{2r}{\sqrt3} + r = R \implies R = \frac{r(\sqrt3+2)}{\sqrt3} \implies r = \frac{R\sqrt3}{\sqrt3 +2}$$

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Let the smaller circles radii be $1$ unit. Let $x$ be central part of diagram bound by the three circles. Let $A$, $B$, and $C$ be the centres of inner circles.

As the lengths $AB = AC =BC$, so all angles of triangle $ABC$ are of $60^{\circ}$. Let two circles meet at $D$ then $ABD$ is a right triangle. Then $\sin(60^{\circ})=\frac{AD}{AB}$ i.e., $\sin(60^{\circ}) =\frac{x+1}{2}$, so $x\approx .732$. If $R$ is the radius of bigger circle then

$$R= 2r +\frac{x}{2}$$ i.e., $R \approx 2 +\frac{0.732}{2}$, so $R \approx 2.366$.

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Let $R$ be radius of enclosing circle and $r$ be the radii of the three internal circles. Since three internal circles kiss each other their centers form a equilateral triangle with sides of $2r$. The centroid of this equilateral triangle is also center of enclosing Circle of radius $R$.There is a smaller space filling Circle that kisses the three inner circles. This Circle has radius $z$ where

$$R=2r+z$$

Centroid of triangle is $2/3$ length from vertex to midpoint of opposite side. For equilateral triangle this is $\sqrt 3 r/2$ This gives $z$ as function of $r$. Also, $R = 2r + z$.

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