# 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

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

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

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

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

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

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.