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.