# Congruent Chords in the Same Circle are Equidistant from the Center

Two congruent chords of a circle are equidistant from its center.
NOTE: Two chords are said to be congruent if they are equal in length.

Consider the equation of the circle
${x^2} + {y^2} = {r^2}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Suppose that $AB$ and $CD$ are congruent chords with $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$, $C\left( {{x_3},{y_3}} \right)$ and $D\left( {{x_4},{y_4}} \right)$ as shown in the given diagram. Since the circle passes through the points $A,\,B,\,C$ and $D$, the equation of the circle becomes
$\begin{gathered} {x_1}^2 + {y_1}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ {x_2}^2 + {y_2}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ {x_3}^2 + {y_3}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right) \\ {x_4}^2 + {y_4}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {\text{v}} \right) \\ \end{gathered}$

Since $M$ is the midpoint of the chord $AB$, so
$M\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)$

Since $N$ is the midpoint of the chord $CD$, so
$N\left( {\frac{{{x_3} + {x_4}}}{2},\frac{{{y_3} + {y_4}}}{2}} \right)$

Now we shall find the distance between $O$ and $M$, as follows:
$\begin{gathered} {\left| {OM} \right|^2} = {\left( {\frac{{{x_1} + {x_2}}}{2} – 0} \right)^2} + {\left( {\frac{{{y_1} + {y_2}}}{2} – 0} \right)^2} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{{{x_1} + {x_2}}}{2}} \right)^2} + {\left( {\frac{{{y_1} + {y_2}}}{2}} \right)^2} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {{x_1} + {x_2}} \right)}^2}}}{4} + \frac{{{{\left( {{y_1} + {y_2}} \right)}^2}}}{4} = \frac{{{{\left( {{x_1} + {x_2}} \right)}^2} + {{\left( {{y_1} + {y_2}} \right)}^2}}}{4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} + {y_1}^2 + {y_2}^2 + 2{y_1}{y_2}}}{4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{x_1}^2 + {y_1}^2 + {x_2}^2 + {y_2}^2 + 2{x_1}{x_2} + 2{y_1}{y_2}}}{4} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{r^2} + {r^2} + 2{x_1}{x_2} + 2{y_1}{y_2}}}{4} = \frac{{2\left( {{r^2} + {x_1}{x_2} + {y_1}{y_2}} \right)}}{4} \\ \Rightarrow {\left| {OM} \right|^2} = \frac{{{r^2} + {x_1}{x_2} + {y_1}{y_2}}}{2}\,\,\,\,{\text{ – – – }}\left( {{\text{vi}}} \right) \\ \end{gathered}$

Similarly, we can show that
$\Rightarrow {\left| {ON} \right|^2} = \frac{{{r^2} + {x_3}{x_4} + {y_3}{y_4}}}{2}\,\,\,\,{\text{ – – – }}\left( {{\text{vii}}} \right)$

Since $AB$ and $CD$ are congruent chords,  ${\left| {AB} \right|^2} = {\left| {CD} \right|^2}$
$\begin{gathered} \Rightarrow {\left( {{x_2} – {x_1}} \right)^2} + {\left( {{y_2} – {y_1}} \right)^2} = {\left( {{x_4} – {x_3}} \right)^2} + {\left( {{y_4} – {y_3}} \right)^2} \\ \Rightarrow {x_1}^2 + {x_2}^2 – 2{x_1}{x_2} + {y_1}^2 + {y_2}^2 – 2{y_1}{y_2} = \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_3}^2 + {x_4}^2 – 2{x_3}{x_4} + {y_3}^2 + {y_4}^2 – 2{y_3}{y_4} \\ \end{gathered}$

Using equations (2) and (3), we get the following result:
$\begin{gathered} \Rightarrow {r^2} + {r^2} – 2{x_1}{x_2} – 2{y_1}{y_2} = {r^2} + {r^2} – 2{x_3}{x_4} – 2{y_3}{y_4} \\ \Rightarrow – 2{x_1}{x_2} – 2{y_1}{y_2} = – 2{x_3}{x_4} – 2{y_3}{y_4} \\ \Rightarrow {x_1}{x_2} + {y_1}{y_2} = {x_3}{x_4} + {y_3}{y_4} \\ \end{gathered}$

Adding ${r^2}$ to both sides of the above equation, we have
$\begin{gathered} \Rightarrow {r^2} + {x_1}{x_2} + {y_1}{y_2} = {r^2} + {x_3}{x_4} + {y_3}{y_4} \\ \Rightarrow \frac{{{r^2} + {x_1}{x_2} + {y_1}{y_2}}}{2} = \frac{{{r^2} + {x_3}{x_4} + {y_3}{y_4}}}{2}\,\,\,{\text{ – – – }}\left( {{\text{viii}}} \right) \\ \end{gathered}$

Using equations (vi) and (vii) in equation (viii), we get
$\begin{gathered} {\left| {OM} \right|^2} = {\left| {ON} \right|^2} \\ \Rightarrow \left| {OM} \right| = \left| {ON} \right| \\ \end{gathered}$

This shows that the congruent chords of a circle are equidistant from its center.