Chords, Arcs & Segment Lengths

Two chord facts come up constantly.

A radius or diameter perpendicular to a chord bisects that chord and its arc. This creates a right triangle with the radius as hypotenuse, half the chord as one leg, and the distance from the centre as the other — which means the Pythagorean theorem finishes the problem.

Also: chords equally far from the centre are congruent, and longer chords sit closer to the centre. The diameter, at distance zero, is the longest.

Then there are the product rules for lengths:

• Two chords crossing inside: the products of their pieces are equal.
• Two secants from outside: (whole)(outside part) is the same for both.
• Tangent and secant: tangent² = (whole secant)(outside part).

For the outside rules, always use the *whole* secant, not just the far piece.

The perpendicular from the centre to a chord bisects both the chord and its arc, and conversely the perpendicular bisector of any chord passes through the centre — which is how a circle's centre is reconstructed from three points on it.

This produces a right triangle relating radius r, half-chord c/2 and centre distance d by r² = d² + (c/2)², the standard computational tool for chord problems.

Congruent chords are equidistant from the centre, and conversely; chord length decreases as centre distance increases.

The length relationships are all instances of the power of a point. For a point P and a circle, the product of the signed distances along any line through P to the two intersection points is constant, equal to d² − r² where d is the distance from P to the centre. Interior points give the two-chord rule ab = cd; exterior points give the secant-secant rule (whole₁)(outside₁) = (whole₂)(outside₂) and, in the limiting case where the two intersections coincide, the tangent-secant rule t² = (whole)(outside).

The recurring error is substituting the external segment where the whole secant belongs.