The common area is obtained by clipping against one of the circles and filling the other.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,through}
\makeatletter % from https://tex.stackexchange.com/a/127045/121799
\tikzset{use path/.code=\tikz@addmode{\pgfsyssoftpath@setcurrentpath#1}}
\makeatother
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (1.25,0.25);
\node (E) [name path=E,draw,circle through=(B),save path=\pathA] at (A) {};
\node (F) [name path=F,draw,circle through=(A),save path=\pathB] at (B) {};
\begin{scope}
\clip[use path=\pathA];
\fill[blue,use path=\pathB];
\end{scope}
\path [name intersections={of=E and F, by={[label=above:$C$]C,[label=below:$C'$]C'}}];
\end{tikzpicture}
\end{document}

In case you are concerned by the fact that the circle contour gets partly overpainted, use backgrounds.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,through,backgrounds}
\makeatletter % from https://tex.stackexchange.com/a/127045/121799
\tikzset{use path/.code=\tikz@addmode{\pgfsyssoftpath@setcurrentpath#1}}
\makeatother
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (1.25,0.25);
\node (E) [name path=E,draw,circle through=(B),save path=\pathA] at (A) {};
\node (F) [name path=F,draw,circle through=(A),save path=\pathB] at (B) {};
\begin{scope}[on background layer]
\clip[use path=\pathA];
\fill[blue,use path=\pathB];
\end{scope}
\path [name intersections={of=E and F, by={[label=above:$C$]C,[label=below:$C'$]C'}}];
\end{tikzpicture}
\end{document}
You can always fill intersection segments. (You can combine this with the background stuff above.)
\documentclass{article}
\usepackage{tikz}
\usepackage{pgfplots}
\usepgfplotslibrary{fillbetween}
\pgfplotsset{compat=1.16}
\usetikzlibrary{through}
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (1.25,0.25);
\node (E) [name path=E,draw,circle through=(B)] at (A) {};
\node (F) [name path=F,draw,circle through=(A)] at (B) {};
\path [name intersections={of=E and F, by={[label=above:$C$]C,[label=below:$C'$]C'}}];
\path[%draw,red,thick,
fill=blue,
intersection segments={of=E and F,sequence={L1--R2--L3}}];
\end{tikzpicture}
\end{document}
(Same output as above.)
The analytic determination of the arcs is another possibility.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections,through,calc}
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (1.25,0.25);
\node (E) [name path=E,draw,circle through=(B)] at (A) {};
\node (F) [name path=F,draw,circle through=(A)] at (B) {};
\path [name intersections={of=E and F, by={[label=above:$C$]C,[label=below:$C'$]C'}}];
\path[fill=blue] let \p1=($(A.center)-(B.center)$),\p2=($(C.center)-(A.center)$),
\p3=($(C'.center)-(A.center)$),\p4=($(C.center)-(B.center)$),
\p5=($(C'.center)-(B.center)$),
\n1={veclen(\x2,\y2)}, % radius A
\n2={veclen(\x4,\y4)}, % radius B
\n3={atan2(\y2,\x2)}, % angle A 1
\n4={atan2(\y3,\x3)}, % angle A 2
\n5={atan2(\y4,\x4)}, % angle B 1
\n6={atan2(\y5,\x5)} % angle B 2
in (C) arc(\n3:\n4:\n1) arc(\n6:\n5-360:\n2);
\end{tikzpicture}
\end{document}
pgfplotssolution as well – subham soni Mar 17 '19 at 03:39pgfplotshas a very careful version management, socompat=1.16indicates that version 1.16 is to be used. Almost certainly this has no effect here. However, I follow the practice to indicate the version because there are cases where that matters. – Mar 17 '19 at 04:02