A `\coordinate` command ignored

Question

$\triangle[ABC]$ is a 30-60 right triangle, and its right angle is at C. A is at the origin. A circle is inscribed in it; its center is at

O = (2*sqrt(3)*(sqrt(3) - 1), 2*(sqrt(3) - 1))

and its radius is 12(sqrt(3) - 1). Leg AC is the shorter leg. The equation of the line through it is y = sqrt(3)*x. The line perpendicular to AC has slope -sqrt(3)/3, and the line through O with slope -sqrt(3)/3 is

y = (-sqrt(3)/3)*(x - 2*(sqrt(3))*(sqrt(3)-1)) + 2*(sqrt(3)-1) .

The two lines intersect on leg AC at

Q = (8*sqrt(3)*(sqrt(3)-1), 24*(sqrt(3)-1)) .

So, the command \draw (O) -- (Q); should draw a radius of the circle to leg AC. On my computer, the command renders a line segment through the other leg and ridiculously long. It seems to me that the command locating point Q has been ignored.

\documentclass{amsart}
\usepackage{amsmath}



\usepackage{tikz}
\usetikzlibrary{calc,intersections}


\begin{document}

\noindent \hspace*{\fill}
\begin{tikzpicture}

\path (0,0) coordinate (A) (8,0) coordinate (B) (2,{2*sqrt(3)}) coordinate (C);
\node[anchor=north, inner sep=0, font=\footnotesize] at (0,-0.15){\textit{A}};
\node[anchor=north, inner sep=0, font=\footnotesize] at ($(B) +(0,-0.15)$){\textit{B}};
\node[anchor=south, inner sep=0, font=\footnotesize] at ($(C) +(0,0.15)$){\textit{C}};
\draw (A) -- (B) -- (C) -- cycle;
\path let \n1={2*(sqrt(3))*(sqrt(3)-1)}, \n2={2*(sqrt(3)-1)} in coordinate (O) at (\n1,\n2);
\draw[fill] (O) circle (1.5pt);
\draw[blue] let \n1={2*(sqrt(3)-1)} in (O) circle (\n1);


\path let \n1={2*(sqrt(3))*(sqrt(3)-1)} in coordinate (P) at (\n1,0);
\node[anchor=north, inner sep=0, font=\footnotesize] at ($(P) +(0,-0.15)$){\textit{P}};
\draw (O) -- (P);
\path let \n1={8*sqrt(3)*(sqrt(3)-1)}, \n2={24*(sqrt(3)-1)} in coordinate (Q) at (\n1,\n2);
\draw[fill=green] (Q) circle (1.5pt);
\draw[green] (O) -- (Q);


\end{tikzpicture}

\end{document}

Answer 1

I am sorry, I cannot follow your equations at all. you ask TikZ to do

 \path let \n1={8*sqrt(3)*(sqrt(3)-1)}, \n2={8*3*(sqrt(3)-1)} in coordinate (Q) at (\n1,\n2); 

which is equivalent to

 \path ({8*sqrt(3)*(sqrt(3)-1)},{8*3*(sqrt(3)-1)}) coordinate (Q); 

(meaning you do not need calc for that), and this is where TikZ places the point. I cannot tell you everything that went wrong in your computation of Q, but here is one point: how is it possible that you do not need the coordinates of O in your way of doing things? You should be solving

 alpha * 1 = O_x + beta
 alpha * sqrt(3) = O_y - beta * sqrt(3)/3   

if you want to find the point where AC intersects with the line that is perpendicular and runs through O, but I cannot see you doing this. (BTW, there is intersection cs: specifically for that, you do not need to do such things by hand.)

Luckily these projections can be done with calc out of the box.

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\noindent \hspace*{\fill}
\begin{tikzpicture}
\draw (0,0) coordinate[label=below:$\scriptstyle A$] (A) --
({8*1},0) coordinate[label=below:$\scriptstyle B$] (B) --
({8*(1/4)},{8*sqrt(3)/4}) coordinate[label=above:$\scriptstyle A$] (C) -- cycle;

\draw[fill] ({8*(sqrt(3)/4)*(sqrt(3)-1)},{8*(1/4)*(sqrt(3)-1)}) 
 coordinate (O) circle (1.5pt);
\draw[blue]  (O) circle({8*(sqrt(3)-1)/4});

\path ($(A)!(O)!(C)$) coordinate[label=left:$\scriptstyle Q$] (Q)
 ($(A)!(O)!(B)$) coordinate[label=below:$\scriptstyle P$] (P);
\draw (O) -- (P);
\draw[fill=green] (Q) circle (1.5pt);
\draw[green] (O) -- (Q);
\end{tikzpicture}
\end{document}

Answer 2

To draw the circumscribed circle: Draw the perpendicular bisectors of AB and AC; their intersection is the center O of the circle.

\documentclass{amsart}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,through,intersections}

\begin{document}
\begin{tikzpicture}
\def\fangle{60}
\def\sangle{30}
\coordinate (A) at (0,0);
\coordinate (B) at (8,0);
\coordinate (C) at (2,{2*tan(\fangle)});
\path [draw,name path=AB](A)node[left]{$A$}--(B);
\path [draw,name path=BC](B)node[right]{$B$}--(C);
\path [draw,name path=CA](C)node[above]{$C$}--(A);
\path [name path=A-bisector] (A)--++(\fangle/2:8);
\path [name path=B-bisector] (B)--++(180-\sangle/2:8);
\path [name intersections={of=A-bisector and B-bisector, by={O}}];
\path [name path=radius] (O)--++(-90:8);
\path [name intersections={of=AB and radius, by={P}}];
\node [draw,name path=circle,blue] at (O) [circle through={(P)}] {};
\path [name intersections={of=circle and CA, by={Q}}];
\filldraw (O) circle (1.5pt);
\draw (O)--(P)node[below]{$P$};
\draw[green] (O)--(Q)node[left,color=black]{$Q$};
\filldraw (Q) circle (1.5pt);
\end{tikzpicture}

\end{document}