Bsrgvty

I would like to draw this figure, where two circles are tangents to a semicircle, as in the following figure:

I can partially draw:

documentclass[12pt]article

usepackagetikz 

begindocument

begincenter

begintikzpicture[scale=0.3]

draw [ultra thick] (0,0) arc (0:180:6);

draw [ultra thick] (-12,0)--(0,0);

node at (-6,0) $bullet$;

node at (-12,0) $bullet$;

node at (0,0) $bullet$;

endtikzpicture

endcenter

enddocument

documentclass[12pt]article
usepackagetikz 
begindocument
begincenter
begintikzpicture[scale=0.3,declare
function=R=6;a=1;,bullet/.style=circle,fill,inner sep=1.5pt]
draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
pgfmathsetmacromyalphaasin(a/(R-a))
draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
draw[dashed] (L) -- (R);
path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
 (180-myalpha:R) node[bullet] (myalpha:R) node[bullet]; 
endtikzpicture
endcenter
enddocument

Here is a possible derivation.

documentclass[12pt,fleqn]article
usepackagetikz 
begindocument
Call the large radius $R$ and the small radius $a$. The condition that the
circles ``touch'' means that the slopes of the circles at the touching point
coincide. This implies that the circles touch at points with coinciding polar
angle $alpha$ (see figure~reffig:Computation). Therefore,
[ a+a,sinalpha~=~R,sinalpha]
and thus
[ alpha~=~arcsinfracaR-a;.]
This means that the centers of the circles are at
$(pm(R-a)cosalpha,a)=(pmsqrtR,(R-2 a),a)$.

beginfigure[!h]
centering
begintikzpicture[scale=0.8,declare function=R=6;a=1;]

draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
node at (-R,0) $bullet$;
node at (R,0) $bullet$;
node at (0,0) $bullet$;
pgfmathsetmacromyalphaasin(a/(R-a))
draw[red] (0,0) -- (180-myalpha:R)
 (-2,0) arc(180:180-myalpha:2) node[midway,left]$alpha$;
draw[red] ((a-R)*cos(myalpha)-2,a) -- ((a-R)*cos(myalpha),a) 
-- ++ (180-myalpha:2) 
((a-R)*cos(myalpha)-2,a) arc(180:180-myalpha:2) node[midway,left]$alpha$;
draw ((a-R)*cos(myalpha),a) -- ((a-R)*cos(myalpha),0)
node[midway,fill=white]$a$;
draw (0,0) -- (40:R) node[midway,fill=white]$R$;
draw[blue] ((a-R)*cos(myalpha),a) circle[radius=a]
((R-a)*cos(myalpha),a) circle[radius=a];
endtikzpicture
captionComputation of the center of the circle.
labelfig:Computation
endfigure
enddocument

Of course, you can vary a.

documentclass[tikz,border=3mm]standalone
begindocument
foreach X in 1,1.1,...,3,2.9,2.8,...,1.1
begintikzpicture[declare function=R=6;a=X;,bullet/.style=circle,fill,inner sep=1.5pt]
 path[use as bounding box] (-R-0.2,-0.2) rectangle (R+0.2,R+0.2);
 draw [ultra thick] (R,0) arc (0:180:R) -- cycle;
 pgfmathsetmacromyalphaasin(a/(R-a))
 draw[blue] ((a-R)*cos(myalpha),a) node[bullet] (L) circle[radius=a]
 ((R-a)*cos(myalpha),a) node[bullet] (R) circle[radius=a];
 draw[dashed] (L) -- (R);
 path (-R,0) node[bullet] (-R,0) node[bullet] (0,0) node[bullet]
 (180-myalpha:R) node[bullet] (myalpha:R) node[bullet]; 
endtikzpicture
enddocument