# Perpendicular from the Center of a Circle Bisects the Chord

The perpendicular dropped from the center of a circle on a chord bisects the chord. Consider the equation of the circle
${x^2} + {y^2} = {r^2}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Let the points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ be the ends of the chord as shown in the given diagram. Since the circle passes through the point $A\left( {{x_1},{y_1}} \right)$, the equation of the circle becomes
${x_1}^2 + {y_1}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Also the equation of the circle passes through the second point $B\left( {{x_2},{y_2}} \right)$, so the circle becomes
${x_2}^2 + {y_2}^2 = {r^2}\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

Suppose that $M$ is the midpoint of the chord $AB$, then by using the midpoint formula we have
$M\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)$

The slope of the chord $AB$ is given by $m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}$

The slope of the perpendicular $= – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}$

The equation of the line passing through the center $O$ and perpendicular to the chord is

$\begin{gathered} y – 0 = – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}\left( {x – 0} \right) \\ \Rightarrow \left( {{y_2} – {y_1}} \right)y = – \left( {{x_2} – {x_1}} \right)x \\ \Rightarrow \left( {{x_2} – {x_1}} \right)x + \left( {{y_2} – {y_1}} \right)y = 0\,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right) \\ \end{gathered}$

It is observed that if the perpendicular (iv) bisects the chord, we check whether $M$ satisfies (iv). Putting the values $x = \frac{{{x_1} + {x_2}}}{2},\,\,y = \frac{{{y_1} + {y_2}}}{2}$ in equation (iv), we get

$\begin{gathered} \left( {{x_2} – {x_1}} \right)\frac{{{x_1} + {x_2}}}{2} + \left( {{y_2} – {y_1}} \right)\frac{{{y_1} + {y_2}}}{2} = 0 \\ \Rightarrow \left( {{x_2} – {x_1}} \right)\left( {{x_1} + {x_2}} \right) + \left( {{y_2} – {y_1}} \right)\left( {{y_1} + {y_2}} \right) = 0 \\ \Rightarrow {x_2}^2 – {x_1}^2 + {y_2}^2 – {y_1}^2 = 0 \\ \Rightarrow {x_2}^2 + {y_2}^2 – \left( {{x_1}^2 + {y_1}^2} \right) = 0 \\ \Rightarrow {r^2} – {r^2} = 0\,\,\,\,\,\,\,\,\,\,\because {x_2}^2 + {y_2}^2 = {r^2},\,\,\,{x_1}^2 + {y_1}^2 = {r^2} \\ \Rightarrow 0 = 0 \\ \end{gathered}$

Thus, $M$ satisfies equation (iv), so the perpendicular dropped from the center of the circle bisects the chord.
Conversely, this can prove that the perpendicular bisector of any chord of a circle passes through the center of the circle.