Circles - Perpendicular from the Centre to a Chord

A chord is a line segment joining any two points on the circumference of a circle. The perpendicular distance from the centre to this chord is the shortest distance between them. Visually, imagine a circle with centre $O$ and a chord $AB$; the perpendicular $OM$ represents this shortest distance.

Theorem 1: The perpendicular from the centre of a circle to a chord bisects the chord. If $OM \perp AB$ where $M$ is a point on chord $AB$, then $AM = MB = \frac{1}{2}AB$. Visually, this creates two congruent right-angled triangles $\triangle OMA$ and $\triangle OMB$.

Theorem 2 (Converse): The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. If $M$ is the midpoint of chord $AB$ (such that $AM = MB$), then the line segment $OM$ must be perpendicular to $AB$ (i.e., $\angle OMA = \angle OMB = 90^{\circ}$).

Right Triangle Relationship: By connecting the centre $O$ to one end of the chord $A$, we form a right-angled triangle $\triangle OMA$. In this triangle, the radius $OA$ acts as the hypotenuse, the perpendicular distance $OM$ is the altitude, and half the chord $AM$ is the base.

Pythagoras Theorem Application: In the right-angled triangle formed by the radius ($r$), the distance of the chord from the centre ($d$), and half the length of the chord ($l/2$), the relationship is defined as $r^2 = d^2 + (\frac{l}{2})^2$.

Equidistant Chords: Chords of a circle that are equal in length are equidistant from the centre. Conversely, chords that are at an equal distance from the centre are equal in length. Visually, if two chords $AB$ and $CD$ are equal, their perpendicular distances from $O$ will be identical.

Circle through Three Points: There is one and only one circle passing through three given non-collinear points. This unique circle's centre is found at the intersection of the perpendicular bisectors of the lines joining these points.