Perpendicular from the Centre of Circle Bisects the Chord

The perpendicular dropped from the centre 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 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

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

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

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

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

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.