How to draw a gif with expanding circles that reveal lines connecting a non-centered point to the expanding circle using Tikz

Question

I’ve racked my brain for weeks trying to create a .gif which is a moving version of what the below diagram represents.

The goal is to start with point (A) and (B) at the same position (origin) with the black lines diverging away from that point.

Then, shift (B) to the right to an arbitrary point. As (B) shifts right, it creates a pulse (the ring) that radiates outward.

As the ring radiates outward, it reveals the red lines, which diverge away from (B) as it moves.

My current progress is:

\documentclass[tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}
  \foreach\radius in{1,...,10}{
    \begin{tikzpicture}
    \def\innerring{\radius/10-.08}
    \def\dottoinner{\radius/10+.08}
    \def\outerring{\radius/10}

    \draw [color=white] (-1,0) -- (1,0); %Used to keep frame the same size
    \draw [color=white] (0,-1) -- (0,1); %Used to keep frame the same size


    %When the dot is moving
    \ifnum\numexpr\radius<4
      %\draw [color=red] (\innerring, 0) -- (-\innerring,.2); %tried, failed, try again
      \draw [color=red] (\innerring, 0) -- (-\dottoinner,0);
      \node (A) at (\innerring, 0) [circle,fill,inner sep=.7pt]{};

    %Whent the dot is no longer moving
    \else
      %\draw [color=red] (.4-.08, 0) -- (-\innerring*.707,\innerring*.5); %tried, failed, try again
      \draw [color=red] (.4-.08, 0) -- (-\dottoinner,0);
      \node (B) at (.3, 0) [circle,fill,inner sep=.7pt]{};
    \fi

  %The pulse of radiation  
  \draw [thin] (0,0) circle[radius=\outerring] node (C) {};
  \draw [thin] (0,0) circle[radius=\innerring] node (D) {};

  \end{tikzpicture}
  }
\end{document}

Which produces the gif:

The crux of the issue is that I can't seem to come up with an expression to relate the non-horizontal lines from the point (B) to the inner circle without having them change angles as the circles expands.

All packages are available to me for this project.

Answer 1

It is a very simple computation. The x coordinate of the circle center B is given by

  B_x = r * a ,

where r is the radius of the circle and 0<a<1 is a velocity. The rays emitted from B under an angle alpha have the parametrization

 gamma(t) = (B_x + t*cos(alpha), t*sin(alpha)) .

Their intersection with the circle around A of radius r can be obtained by demanding that

|gamma(t_crit)|^2 = r^2 . 

This is a quadratic equation, which has a unique positive solution, a simple function tcrit that I punched in the following code.

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
\foreach \R in {0.2,0.3,...,4.5}
{\begin{tikzpicture}[declare function={a=0.6; % velocity
 tcrit(\r,\ang)=(-2*a*\r*cos(\ang)+sqrt(2)*sqrt(\r*\r*(2-a*a+a*a*cos(2*\ang))))/2;}]
 \clip (-5,-5) rectangle (5,5);
 \foreach \ang in {0,22.5,...,337.5}
 {\draw[very thick] (a*\R,0) -- ++(\ang:{tcrit(\R,\ang)}) -- 
 (\ang:1.2*\R) -- (\ang:10);}
 \draw[thick,even odd rule,fill=white,fill opacity=0.5] (0,0) circle[radius=\R] 
 circle[radius=1.2*\R];
\end{tikzpicture}}
\end{document}

Answer 2

I am not particularly happy with this answer as it takes a long time to render and frame 1 and 3 is a bit dodgy (still trying to fix this) but here it is:

\documentclass[tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}
  \foreach\radius in{1,...,10}{
    \begin{tikzpicture}[scale=10]
    \def\innerring{\radius/10-.08}
    \def\dottoinner{\radius/10+.08}
    \def\outerring{\radius/10}
    \def\dot{\radius/10-0.1}

    \node at (-1,-1) {};
    \node at (1,1) {};

    \foreach \x in {0,30,...,330} {
        \draw (\x:\outerring) -- (\x:1.5);
    }
    \draw [thin] (0,0) circle[radius=\outerring];
    \draw [thin] (0,0) circle[radius=\innerring];

    %When the dot is moving
    \ifnum\numexpr\radius<4
%      \draw [color=red] (\innerring, 0) -- (-\innerring,.2); %tried, failed, try again
%      \draw [color=red] (\innerring, 0) -- (-\dottoinner,0);
      \node (A) at (\dot, 0) [circle,fill,inner sep=2pt]{};
      \foreach \x in {0,30,...,330} {
            \begin{scope}
            \clip (0,0) circle[radius=\innerring];
            \draw[name path=P1,red] (A) -- +(\x:5);
            \end{scope}
            \draw[name path=P2,draw opacity=0] (0,0) circle (\innerring);
            \path [name intersections={of=P1 and P2,by=E}];
            \draw[black!50] (E) -- (\x:\outerring);
        }


    %Whent the dot is no longer moving
    \else
      %\draw [color=red] (.4-.08, 0) -- (-\innerring*.707,\innerring*.5); %tried, failed, try again
%      \draw [color=red] (.4-.08, 0) -- (-\dottoinner,0);
      \node at (A) [circle,fill,inner sep=2pt]{};
      \foreach \x in {0,30,...,330} {
        \begin{scope}
        \clip (0,0) circle[radius=\innerring];
        \draw[name path=P1,red] (A) -- +(\x:5);
        \end{scope}
        \draw[name path=P2,draw opacity=0] (0,0) circle (\innerring);
        \path [name intersections={of=P1 and P2,by=E}];
        \draw[black!50] (E) -- (\x:\outerring);
      }
    \fi

  \end{tikzpicture}
  }
\end{document}

I based it off the image you had at the beginning