3 square in a circle (geometric question) There is three square in a circle like the picture attached.
  How can find radius of circle  ?
I know that's not a hard problem,but I was away from geometry for years.  If possible give me hint or idea to start to solving.   Thanks in advance.

the smallest side is $6$
and medium square side is $6+6$
and big side is $6+6+6$
But I got stuck here and not to go over there ...
 A: It is only the Pythagorean theorem:

A: Coordinate Geometry works.
Draw the horizontal diameter of the circle.
Let the center of the line segment between the squares be the origin $(0,0)$.
Let the center of the circle be $ (x, 0 )$.   


 What is the coordinate of the top right blue point?    

$ $ 

 What is the coordinate of the top left blue point?    

$ $

 Hence, what is an equation that we can write?   

$ $

 Hence, what is the radius of the circle?    

A: 
Apply the Pythagorean theorem to the triangles ADO and CBO to match the overall length AB = 36,
$$\sqrt{r^2-9^2} + \sqrt{r^2-3^2} =36$$
Solve to obtain the radius $r=\sqrt{370}$.
A: 
Equations:
$$ a + b = 36 $$
$$ a^2 + 9^2 = r^2 $$
$$ b^2 + 3^2 = r^2 $$
Using 2 and 3:
$$ b^2 - a^2 = 72 $$
Factoring:
$$ (b-a)(b+a) = 72 $$
$$ (b-a)36 = 72 $$
$$ b-a = 2 $$
Adding the first:
$$ 2b = 38 $$
$$ b = 19 $$
$$ a = 17 $$
Adding 2 eq y 3 eq:
$$ 19^2 + 17^2 +9^2 + 3^2 = 2r^2 $$
$$ 370 = r^2 $$
$$ r = \sqrt{370} = 19.23.... $$
A: The circumradius of a triangle is
$$
R=\frac{abc}{4A}
$$
where $a,b,c$ are the sides of the triangle and $A$ is its area.

The sides are $18$, $\sqrt{36^2+6^2}=6\sqrt{37}$, and $\sqrt{36^2+12^2}=12\sqrt{10}$ and the area is $\frac12\,18\cdot36=324$. Therefore, the circumradius is
$$
R=\frac{18\cdot6\sqrt{37}\cdot12\sqrt{10}}{4\cdot324}=\sqrt{370}
$$
