Finding radius of the central radius?

Question

Is there any easy way to get the radius of the central radious?

Answer 1

This is a more general approach : We consider several layers of circles with $n$ circles per layer. The layers are numbered $ 1 , 2, 3, ... k, ...$. The small inner circle is not included ( there are $n$ circles in the layer number 1 ).

The next figure shows the notations and how the recurrence formula is derived, leading to the relationship between the radius of the circles of the k-ieme layer and the radius of the circles of the first layer.

Comming back to the case considered in the question raised by "Complex Guy", there are 2 layers. The radius of the circles in layer 2 is given. Applying the above formula leads to the formula for the radius of the small inner circle.

Answer 2

Let $(0,0)$ be the centre of the figure and $(a,0)$ the center of one of the radius $r$ circles and let the tiny circle be a unit circle. Let $(b,c)$ be the center of one of the meium sized circles touching the $x$-axis. Then by Pythagoras $$\tag1b^2+c^2=(c+1)^2$$ and $$\tag2(a-b)^2+c^2=(r+c)^2.$$ Moreover, by the symmetries of the figure $$\tag3r:a = c:(c+1) = \sin 36^\circ=\frac{\sqrt{2(5-\sqrt 5)}}{4}.$$ From $(3)$ we find $$c=\frac{\sqrt{2(5-\sqrt 5)}}{4-\sqrt{2(5-\sqrt 5)}},$$ then with $(1)$ $$ b=\sqrt{\frac{4+\sqrt{2(5-\sqrt 5)}}{4-\sqrt{2(5-\sqrt 5)}}}.$$ (I was too lazy to simplify these expressions). Using these values and $(3)$ to eliminate $a$, $(2)$ becomes a quadratic equation in $r$ (which looks a bit involved, so one should really spend a few thoughts on simplifying all the surd expressins first). Ultimately, we have computed $r$ as the ratio of the radii of the big circles and the small circle. Invert to compute the radius of the small circle if the big radius is given.