Perpendicular from the Centre of Circle Bisects the Chord

The perpendicular dropped from the centre 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 ends of the chord as shown in the given diagram. Since the circle passes through the point A\left( {{x_1},{y_1}} \right) so equation of circle becomes

{x_1}^2  + {y_1}^2 = {r^2}\,\,\,\,{\text{ -  -  -  }}\left( {{\text{ii}}} \right)


Also equation of circle passes through the second point B\left( {{x_2},{y_2}} \right), so circle becomes

{x_2}^2  + {y_2}^2 = {r^2}\,\,\,\,{\text{ -   -  - }}\left( {{\text{iii}}}  \right)


Suppose that M be the midpoint of the chord AB, then by using midpoint formula we have

M\left(  {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)


Slope of the chord AB is given by m = \frac{{{y_2} - {y_1}}}{{{x_2}  - {x_1}}}
Slope of perpendicular   =  - \frac{{{x_2} - {x_1}}}{{{y_2} -  {y_1}}}
Equation of line passing through the centre 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 whether 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 centre of the circle bisects the chord.
Conversely, this can be proves that the perpendicular bisector of any chord of a circle passes through the centre of the circle.

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