What is the relationship between a chord and its distance from the centre?

The longer the chord, the closer it is to the centre. The shortest chord passes through the farthest point from the centre, and the longest chord (diameter) passes through the centre itself (distance = 0). Formula: if a chord of length l is at distance d from the centre, and radius is r, then r² = d² + (l/2)².

How do you prove that the perpendicular from the centre bisects the chord?

Draw radii OA and OB to the ends of the chord AB. Drop perpendicular OM to AB (M on AB). In △OMA and △OMB: OA = OB (radii), OM = OM (common side), ∠OMA = ∠OMB = 90°. By RHS congruence rule, △OMA ≅ △OMB. Therefore AM = MB by CPCT — the chord is bisected.

Are equal chords always equidistant from the centre?

Yes! Theorem: Equal chords of a circle are equidistant from the centre — and the converse is also true. If AB = CD (equal chords in the same circle), then the perpendicular distances from centre O to each chord are equal (OM = ON). This works for chords in the same circle or in congruent circles.

Chord Properties and Theorems

Circles • Class 9 Mathematics • NCERT • CBSE

Key chord theorems: (1) Equal chords are equidistant from the centre. (2) The perpendicular from the centre to a chord bisects the chord. (3) The line drawn through the centre to bisect a chord is perpendicular to the chord. These are Chapter 10 theorems for Class 9.

Key Formulas

Perpendicular from centre bisects chord: AM = MB (if OM ⊥ AB)
Radius-chord-distance: r² = d² + (l/2)² where d = distance, l = chord length
Equal chords ↔ Equal distances from centre
Angle at centre = 2 × angle at circumference (same arc)
Angle in semicircle = 90°

Frequently Asked Questions

What is the relationship between a chord and its distance from the centre?: The longer the chord, the closer it is to the centre. The shortest chord passes through the farthest point from the centre, and the longest chord (diameter) passes through the centre itself (distance = 0). Formula: if a chord of length l is at distance d from the centre, and radius is r, then r² = d² + (l/2)².
How do you prove that the perpendicular from the centre bisects the chord?: Draw radii OA and OB to the ends of the chord AB. Drop perpendicular OM to AB (M on AB). In △OMA and △OMB: OA = OB (radii), OM = OM (common side), ∠OMA = ∠OMB = 90°. By RHS congruence rule, △OMA ≅ △OMB. Therefore AM = MB by CPCT — the chord is bisected.
Are equal chords always equidistant from the centre?: Yes! Theorem: Equal chords of a circle are equidistant from the centre — and the converse is also true. If AB = CD (equal chords in the same circle), then the perpendicular distances from centre O to each chord are equal (OM = ON). This works for chords in the same circle or in congruent circles.

Study With GyanAI

GyanAI's AI tutor can answer any question about this topic instantly. Try GyanAI free for step-by-step NCERT solutions aligned with the CBSE curriculum.