Encircle a formula with arrow pointing from the circle

Question

Is there perhaps a package that already does something like this?

In THIS topic encircling a formula is given, but I don't understand the syntax to modify the 'circle' to have an arrow point outward from somewhere.

The end goal is to have something like this.

I tried to use the example given in the linked topic as a base

\documentclass{article}

\usepackage{tikz}

\makeatletter
\newcommand*{\encircled}[1]{\relax\ifmmode\mathpalette\@encircled@math{#1}\else\@encircled{#1}\fi}
\newcommand*{\@encircled@math}[2]{\@encircled{$\m@th#1#2$}}
\newcommand*{\@encircled}[1]{%
  \tikz[baseline,anchor=base]{\node[draw,circle,outer sep=0pt,inner sep=.2ex] {#1};}}
\makeatother

\begin{document} not important at the  moment
\end{document}

There are a lot of embellishments that have something to do with preserving proper formating or some such. The core of the problem is to modify the \node[...] bit which results in a circle with an arrow pointing out from it, but I feel that's a dead-end.

Answer 1

The best approach depends on whether your sketch is meant as a single image---in this case use a single tikzpicture environment---or whether the items appear mixed with text on a page---in this case use the \tikznode command from this post. Here are examples for both situations. Note that you have to run LaTeX twice on the second example.

Everything in a single picture: The position of the elements can be specified explicitly relative to each other.

\documentclass[border=1pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning}
\begin{document}
\begin{tikzpicture}
  \node[circle,draw](fx){$f(x)$};
  \node(a)[below right=of fx]{$a$};
  \draw[->] (fx) -- (a);
\end{tikzpicture}
\end{document}

Graphical elements as part of text: The position of the elements is defined implicitly by the typesetting engine that positions the elements. For examples of the use of \tikznode see e.g. Breaking numbers Simple addition, How to add arrow in equations and matrix? or how can you point the head of an arrow from tikzpicture to an equation in LaTex?.

\documentclass{article}
\usepackage{tikz}
\newcommand\tikznode[3][]%
   {\tikz[remember picture,baseline=(#2.base)]
      \node[minimum size=0pt,inner sep=0pt,#1](#2){#3};%
   }
\begin{document}
\[ \frac{g(x)}{\tikznode[circle,draw]{fx}{$f(x)$}}
  =\frac{g(x)}{\tikznode{a}{$a$}}
\]
\begin{tikzpicture}[remember picture,overlay]
  \draw[->] (fx) edge[bend right] (a);
\end{tikzpicture}
\end{document}

Answer 2

Maybe something like this?

\documentclass{article}
\usepackage{tikz,blindtext}
\newcommand{\mycircle}[2]{\begin{tikzpicture}[remember picture]
                            \node[draw,circle] (#2) {$#1$};
                            \draw[->] (#2) --+ (1,-1) node[below right] {$#2$};
                          \end{tikzpicture}
                         }
\begin{document}
    \blindtext\par
    \mycircle{f(x)}{a}\par
    \blindtext
\end{document}

Here is the output:

EDIT: The denominator isn't overlapped:

MWE:

\documentclass{article}
\usepackage{tikz,blindtext}
\newcommand{\mycircle}[2]{\begin{tikzpicture}[remember picture]
                            \node[draw,circle] (#2) {$#1$};
                            \draw[->] (#2.south east) --+ (1,-1) node[below right] {$#2$};
                          \end{tikzpicture}
                         }
\begin{document}
    \blindtext\par
    \mycircle{f(x)}{a}\par
    \[
        \frac{f(x)}{\mycircle{g(x)}{a}}
    \]
    \blindtext\par
\end{document}

Output:

Answer 3

\documentclass{article} \usepackage{tikz} \newcommand\tikznode[3][]% {\tikz[remember picture,baseline=(#2.base)] \nodeminimum size=0pt,inner sep=0pt,#1{#3};% } \begin{document} [ \frac{g(x)}{\tikznode[circle,draw]{fx}{$f(x)$}} =\frac{g(x)}{\tikznode{a}{$a$}} ] \begin{tikzpicture}[remember picture,overlay] \draw[->] (fx) edge[bend right] (a); \end{tikzpicture} \end{document}

what if I wanted it to be on top and not on the bottom as it is in the code? for example, if instead of doing in the denominator you want to do in the numerator? I'm trying to do, but it's getting low.