How to find the radius of this middle circle arranged as shown. There is this maths competition geometry problem and my approach.
And this is my initial approach.

From the picture, the shaded circle looks slightly bigger.
What we are looking for is the $x$ which is half of the length of the side of the red square. But it seems that we have two unknowns, the $x$ and the hypotenuse of the triangle.
Are there any other way besides this? Or any useful theorem? But I don't think it is that complicated.
Many many thanks! Helps are greatly appreciated.
 A: Hint without words (except these):

A: The radius of the shaped circle is $( \sqrt{3} + \sqrt{2} - 2 )\times 110 \approx 126.089080693617$ units.
For simplicity of derivation, let us rescale all the unshaded circles to have radius $1$.
Let $r$ be the radius of the shaded circle. If we choose the coordinate system
such the the shaded circle is centered at $A = (0, 0)$ and the $x$-axis passing through the nearest unshaded circles on its right, the centers of 3 of the unshaded circles on its right are given by
$$B = (r+1,0),\; C = (r+3,0)\quad\text{ and }\quad D = (x_D,y_D) = (r+2,\sqrt{3})$$
Construct a line passing through $A$ tangent to the circle centered at $D$. Let the line
touch the circle at point $E = (x_E,y_E)$. By symmetry, it is clear $\angle BAE = 45^\circ$. It is easy to see
$$
\begin{cases}
x_E &= x_D - \cos45^\circ = r + 2 - \frac{1}{\sqrt{2}}\\
y_E &= y_D + \sin45^\circ = \sqrt{3} + \frac{1}{\sqrt{2}}
\end{cases}
$$
As a result,
$$\angle BAE = 45^\circ
\quad\implies\quad
x_E = y_E
\quad\implies\quad 
r = \sqrt{3} + \sqrt{2} - 2$$
Following is a figure illustrating the arrangement of the circles.
The unshaded/shaded circles in question have been colored in red/blue in this figure.
$\hspace1in$ 
A: Hint.  Sorry, not good at diagrams so will answer in words, but I'm sure you can draw your own diagram.
Look at the five circles which form a "ring" in the top left of the diagram.  Specifically, start with the shaded circle then go left, up/left, up/right, down/right and down to the shaded circle again.  Call the centres of the circles $A,B,C,D,E$ respectively.  Then you should be able to work out the lengths $BC,\,CD,\,DE$, and you can see that $AB=EA$.  You should also be able to find the five angles in the pentagon, and this will be enough to solve the problem.
