两个半圆之间的交角

Question

我正在尝试重新创建以下图像，但到目前为止，我只能复制两个相交的半圆。

    \documentclass[a4paper,12pt]{article} 
    \usepackage{tikz}
    \usetikzlibrary{angles, calc, intersections, positioning, quotes, backgrounds}
    \usepackage{pgfplots,caption}
    \pgfplotsset{compat=1.9}
    \begin{document}

    \begin{center}
            \begin{tikzpicture}
                \begin{scope}
                     \clip (-3.5,0) rectangle (3.5,2);
                     \draw (0,0) circle(1.1);
                     \draw(1.8,0) circle(1.6); 
                \end{scope}
                        \draw (-2,0) -- (4,0);

         \node at (-1.1, 0)[below] (1){ -6};
         \node at (1.1, 0)[below] (2){  6};
         \node at (0.2, 0)[below] (3){  $4\sqrt{2}$};
         \node at (3.4, 0)[below] (4){  $6\sqrt{2}$};
            \end{tikzpicture}
        \end{center}
\end{document}

到目前为止我只有这个

Answer 1

一个可能的解决方案可能是这样的。

\documentclass[a4paper,12pt]{article} 
\usepackage{tikz}
\usetikzlibrary{angles, calc, intersections, positioning, quotes, backgrounds}
\begin{document}

\begin{tikzpicture}[xscale=1.5, yscale=1.5]
    \begin{scope}
        \clip (-3.5,0) rectangle (3.5,2);
        \draw[name path=C1] (0,0) coordinate(C-1-0) circle(1.1);
        \draw[name path=C2] (1.8,0) coordinate(C-2-0) circle(1.6); 

        \draw [red, name intersections={of=C1 and C2}] (intersection-1) -- ($(intersection-1)!1cm!-90:(C-2-0)$) coordinate(A);
        \draw [red] (intersection-1) -- ($(intersection-1)!1cm!90:(C-1-0)$) coordinate(C);

        \begin{scope}
            \draw[clip] (0,0) circle(1.1);
            \draw[clip] (1.8,0) circle(1.6); 
            \draw pic["$\psi$", angle radius=0.5cm, draw] {angle = A--intersection-1--C};
        \end{scope}
    \end{scope}

    \draw (-2,0) -- (4,0);

    \node at (-1.1, 0)[below] (1){ -6};
    \node at (1.1, 0)[below] (2){  6};
    \node at (0.2, 0)[below] (3){  $4\sqrt{2}$};
    \node at (3.4, 0)[below] (4){  $6\sqrt{2}$};
\end{tikzpicture}
\end{document}