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.


circle-bisect-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.