Question 1151686
<pre>
To draw this "Vienna sausage" arrangement:

{{{drawing(400,400,-4,4,-4,4,
circle(-2,0,1),circle(0,0,1),circle(2,0,1),circle(-1,sqrt(3),1),circle(1,sqrt(3),1),circle(-1,-sqrt(3),1),circle(1,-sqrt(3),1) )}}} 

You just need the coordinates of the centers and the radius. If you use 1 for
the radius and coordinate axes, you get this:

{{{drawing(400,400,-4,4,-4,4,
circle(-2,0,1),circle(0,0,1),circle(2,0,1),circle(-1,sqrt(3),1),circle(1,sqrt(3),1),circle(-1,-sqrt(3),1),circle(1,-sqrt(3),1),

line(-1,sqrt(3),1,sqrt(3)), line(1,sqrt(3),2,0), line(2,0,1,-sqrt(3)), line(1,-sqrt(3),-1,sqrt(3)), line(-1,sqrt(3),-2,0), line(-2,0,-1,sqrt(3)), line(-1,-sqrt(3),1,sqrt(3)), line(1,sqrt(3),-1,sqrt(3)), line(-2,0,-1,-sqrt(3)),
line(-1,-sqrt(3),1,-sqrt(3)),line(-1,-sqrt(3),-1,sqrt(3)), 
grid(1),line(1,-sqrt(3),1,sqrt(3))
 )}}} 

As you see, the centers form a regular hexagon, which is made up of 6
equilateral triangles, and each equilateral triangle is made up of 2
30°-60°-90° right triangles.  So there are 12 30°-60°-90° right triangles.

I used the radius of the circles as 1, but you can make it anything you like.

The shortest side of a 30°-60°-90° triangle is 1 times the length of the radius.
The medium side is sqrt(3) times the length of the radius,
The longest side (the hypotenuse) is 2 times the length of the radius.

That's all you need to write up the C++ coding.

Edwin</pre>