# What is the equation of the perpendicular bisector of a chord of a circle?

Apr 9, 2016

For a chord AB, with $A \left({x}_{A} , {y}_{A}\right) \mathmr{and} B \left({x}_{B} , {y}_{B}\right)$. the equation of the perpendicular bisector is $y = - \frac{{x}_{B} - {x}_{A}}{{y}_{B} - {y}_{A}} \cdot \left(x - \frac{{x}_{A} + {y}_{B}}{2}\right) + \frac{{y}_{A} + {y}_{B}}{2}$

#### Explanation:

Supposing a chord AB with
$A \left({x}_{A} , {y}_{A}\right)$
$B \left({x}_{B} , {y}_{B}\right)$

The midpoint is
$M \left(\frac{{x}_{A} + {x}_{B}}{2} , \frac{{y}_{A} + {y}_{B}}{2}\right)$

The slope of the segment defined by A and B (the chord) is
$k = \frac{\Delta y}{\Delta x} = \frac{{y}_{B} - {y}_{A}}{{x}_{B} - {x}_{A}}$
The slope of the line perpendicular to the segment AB is
$p = - \frac{1}{k}$ => $p = - \frac{{x}_{B} - {x}_{A}}{{y}_{B} - {y}_{A}}$

The equation of the line required is

$y - {y}_{M} = p \left(x - {x}_{M}\right)$
$y - \frac{{y}_{A} + {y}_{B}}{2} = - \frac{{x}_{B} - {x}_{A}}{{y}_{B} - {y}_{A}} \cdot \left(x - \frac{{x}_{A} + {x}_{B}}{2}\right)$
Or
$y = - \frac{{x}_{B} - {x}_{A}}{{y}_{B} - {y}_{A}} \cdot \left(x - \frac{{x}_{A} + {x}_{B}}{2}\right) + \frac{{y}_{A} + {y}_{B}}{2}$

Note: the center of the circle, assummedly point C$\left({x}_{C} , {y}_{C}\right)$, is in the line defined by the aforementioned equation, with radius $r = A C = B C$. Therefore the perpendicular bisector of the question is also the locus of the possible centers of circles with the chord AB

Jul 5, 2016

The equation of the perpendicular bisector of a chord of a circle is the equation of a diameter of the circle.

#### Explanation:

Let the equation of the circle be standard one having center at origin and radius r

$\textcolor{b l u e}{\text{The equation of the circle: } {x}^{2} + {y}^{2} = {r}^{2}}$

The coordinates of the end points of the chord AB

color(red)(A->(rcosa,rsina)" & " B->(rcosb,rsinb)

The coordinate of the middle point C (x',y') of AB

$x ' = \frac{r}{2} \left(\cos a + \cos b\right) = r \cos \left(\frac{a + b}{2}\right) \cos \left(\frac{a - b}{2}\right)$
$y ' = \frac{r}{2} \left(\sin a + \sin b\right) = r \sin \left(\frac{a + b}{2}\right) \cos \left(\frac{a - b}{2}\right)$

$\therefore \frac{y '}{x '} = \tan \left(\frac{a + b}{2}\right)$

The slope of AB

$m = \frac{r \sin a - r \sin b}{r \cos a - r \cos b}$

$= - \frac{2 \cos \left(\frac{a + b}{2}\right) \sin \left(\frac{a - b}{2}\right)}{2 \sin \left(\frac{a + b}{2}\right) \sin \left(\frac{a - b}{2}\right)}$

$= - \cot \left(\frac{a + b}{2}\right)$

Slope of the perpendicular bisector of AB is

$m ' = - \frac{1}{m} = \tan \left(\frac{a + b}{2}\right) = \frac{y '}{x '}$

$\implies m ' x ' = y '$

The equation of the perpendicular bisector of AB

$y - y ' = m ' \left(x - x '\right)$

$\implies y = m ' x - m ' x ' + y '$

$\implies y = m ' x - y ' + y '$

$\implies y = m ' x$

$y = \tan \left(\frac{a + b}{2}\right) x$

OR

$\textcolor{g r e e n}{y = \frac{y '}{x '} x}$

Obviously this is the equation of a straight line passing through the origin (0,0),the center of the circle. So the perpendicular bisector of the chord is a diameter of the circle.