Recursive Steiner Chains
Javascript ES6, 379 bytes

This solution was used to generate the example runs in the question.

f=(s,ls)=&gt;{with(V=document.createElement`canvas`)with(getContext`2d`)with(Math)return(width=height=s,translate(s/=2,s),(S=(o,d=0,n=ls[d],i=(o-o*sin(PI/n))/(sin(PI/n)+1),r=0)=&gt;{fillStyle=`rgba(0,0,0,${pow(d/ls.length,4)})`;beginPath(),arc(0,0,o,-PI,PI),fill();if(d++&lt;ls.length){S(i,d,n);for(;r&lt;n;++r){save();translate(0,(o+i)/2);S((o-i)/2,d);restore();rotate((2*PI)/n);}}})(s),V)}


Ungolfed:

f=(s,ls)=&gt;{                                        // define function that accepts image dimensions and a list of orders
 with(V=document.createElement`canvas`)            // create canvas to draw on, bring its functions into current scope chain
 with(getContext`2d`)                              // bring graphics functions into current scope chain
 with(Math)return(                                 // bring Math functions into current scope chain
  width=height=s,                                  // set width and height of image
  translate(s/=2,s),                               // center the transform on image
   (S=(o,d=0,                                      // define recursive function that accepts outer radius, depth, and optionally order
       n=ls[d],                                    // default chain order to corresponding order in input list
       i=(o-o*sin(PI/n))/(sin(PI/n)+1),            // calculate inner base circle radius
       r=0)=&gt;{                                     // initialize for loop var
    fillStyle=`rgba(0,0,0,${pow(d/ls.length,4)})`; // fill based on depth
    beginPath(),arc(0,0,o,-PI,PI),fill();          // draw circle
    if(d++&lt;ls.length){                             // if within recursion limit
     S(i,d,n);                                     //   recurse on inner circle
     for(;r&lt;n;++r){                                //   loop through all circles of the chain
      save();                                      //   save transform
      translate(0,(o+i)/2);                        //   translate origin to middle of the 2 base circles
      S((o-i)/2,d);                                //   recurse on chain circle
      restore();                                   //   restore transform
      rotate((2*PI)/n);                            //   rotate transform to next circle in chain
   }}})(s),                                        // begin the recursion
 V)}                                               // return the canvas


Note: f returns a canvas.

Example run (assumes there's a &lt;body&gt; to append to):

document.body.appendChild(f(600,[13,7,11,5,3]))


Should dump the following image to the page: