I tried like that but did not get the required result.
Let the circle equation with center $C(-f,-g)$ is:
$x^2+y^2+2gx+2fy+c=0$.................(1)
\documentclass[10pt]{book}
\usepackage[paperheight=10in,paperwidth=9in, top=1in, bottom=0.8in, twocolumn, twoside]{geometry}
\setlength{\columnseprule}{0.4pt}
\usepackage{amssymb,amsfonts}
\usepackage{mathrsfs}
\usepackage[centertags]{amsmath}
\usepackage{amsthm}
\newtheorem{theorem}{Theorem}
\usepackage{epsfig}
\usepackage{graphicx}\graphicspath{{Graphics/}}
\usepackage{amsthm}
\usepackage{mathptmx}
\usepackage[square,sort&compress]{natbib}
\usepackage{pgf,tikz,pgfplots}
\pgfplotsset{compat=1.15}
\usetikzlibrary{arrows}
\usepackage[T1]{fontenc}
\usepackage{fancyhdr}\pagestyle{fancy}
\usepackage{xcolor}
\usepackage{setspace}
\usepackage{booktabs}
%\usepackage{hyperref}
\usetikzlibrary{arrows.meta}
\renewcommand{\baselinestretch}{1.5}
\newcommand\aug{\fboxsep=-\fboxrule\!\!\!\fbox{\strut}\!\!\!}
\theoremstyle{definition}
\newtheorem{Thm}{Theorem}[section]
\newtheorem{lem}[Thm]{Lemma}
\newtheorem{pro}[Thm]{Proposition}
\newtheorem{de}[Thm]{Definition}
\newtheorem{re}[Thm]{Remark}
\newtheorem{ex}[Thm]{Example}
\newtheorem{cor}[Thm]{Corollary}
\numberwithin{equation}{section}
\definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666}
\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\clip(5.,3.) rectangle (11.1,8.4);
\draw [->,line width=1.pt] (5.,4.) -- (11.,4.);
\draw [->,line width=1.pt] (6.,3.) -- (6.,8.);
\draw [line width=1.pt] (8.,6.) circle (1.7032909322837364cm);
\draw [line width=1.pt] (6.296709067716264,6.)-- (8.450890034689387,7.642527983511332);
\draw [line width=1.pt] (7.425867288241979,4.3963879430207005)-- (9.703175794491177,6.019804369703385);
\draw [line width=1.pt] (8.450890034689387,7.642527983511332)-- (9.703175794491177,6.019804369703385);
\draw [line width=1.pt] (6.296709067716264,6.)-- (7.425867288241979,4.3963879430207005);
\draw [line width=1pt,dotted] (6.856167263763389,5.2054666102157405)-- (9.07995919044889,6.8273742809484705);
\draw [line width=1.pt] (7.381953980879353,6.827481611880726)-- (8.60312548617443,5.2356156066094925);
\draw (7.6,7.0) node[anchor=north west] {$d_1$};
\draw (8.26,6.2) node[anchor=north west] {$d_2$};
\draw (7.24,5.7) node[anchor=north west] {$d_3$};
\draw (8.8,6.76) node[anchor=north west] {$d_4$};
\draw (5.96,6.78) node[anchor=north west] {$A$};
\draw (7.02,7.45) node[anchor=north west] {$P$};
\draw (8.24,8.25) node[anchor=north west] {$B$};
\draw (6.86,4.55) node[anchor=north west] {$D$};
\draw (9.86,6.4) node[anchor=north west] {$E$};
\draw (8.56,5.34) node[anchor=north west] {$Q$};
\draw (7.46,6.35) node[anchor=north west] {$C$};
\begin{scriptsize}
\draw [fill=black] (8.,6.) circle (2.0pt);
\draw [fill=black] (6.296709067716264,6.) circle (2.0pt);
\draw [fill=black] (8.450890034689387,7.642527983511332) circle (2.0pt);
\draw [fill=black] (7.425867288241979,4.3963879430207005) circle (2.0pt);
\draw [fill=black] (9.703175794491177,6.019804369703385) circle (2.0pt);
\draw [fill=black] (6.856167263763389,5.2054666102157405) circle (2.0pt);
\draw [fill=black] (9.07995919044889,6.8273742809484705) circle (2.0pt);
\draw [fill=black] (7.381953980879353,6.827481611880726) circle (2.0pt);
\draw [fill=black] (8.60312548617443,5.2356156066094925) circle (2.0pt);
\end{scriptsize}
\end{tikzpicture}
\end{document}
If $AB$ and $DE$ are the two chords of circle as shown in figure, then the coordinates of the end points of the chords $AB$ and $DE$ are respectively:
$A(x_1,y_1), B(x_3,y_3), D(x_2,y_2), E(x_4,y_4).$
The length of the chord $AB$ and $DE$ are equal, therefore $|AB|=|DE|.$
$\Rightarrow\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}=\sqrt{(x_4-x_2)^2+(y_4-y_2)^2}$
The equation of line $AB$ with two points formula
$y-y_1=\dfrac{y_3-y_1}{x_3-x_1}(x-x_1)$
$\Rightarrow \boxed{(y_3-y_1)x-(x_3-x_1)y-(x_1y_3-x_3y_1)=0}$.........(1)
Similarly, the equation of chord $DE$ with two points formula using the end points is:
$\boxed{(y_4-y_2)x-(x_4-x_2)y-(x_2y_4-x_4y_2)=0}$.............(2)
The perpendicular distance from center $C(-g,-f)$\quad to the chord\quad $AB$\quad is:
$d_1=\dfrac{|(y_3-y_1)(-g)-(x_3-x_1)(-f)-(x_1y_3-x_3y_1)|}{\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}}$
The perpendicular distance from center $C(-g,-f)$ to the chord $AB$ is:
$d_2=\dfrac{|(y_4-y_2)(-g)-(x_4-x_2)(-f)-(x_2y_4-x_4y_2)|}{\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}}.$
But the results $d_1\neq d_2$ at the end. Why?
If two chords are congruent, then their "semi-chords" are also congruent. Since the radii are also congruent, their distances to the center must be congruent by the Pythagorean Theorem.