0580

Circle Theorems

Geometry

Three further facts

  • Equal chords are equidistant from the centre, and the statement works in reverse too, so chords equidistant from the centre have equal length
  • The perpendicular bisector of a chord passes through the centre, which means a radius drawn to the midpoint of a chord meets it at a right angle and cuts it exactly in half
A circle with centre O and a chord across it, with a line drawn from O to the chord meeting it at a right angle and the two halves of the chord marked equal by matching dashes
Source: Circles & chords by Save My Exams
  • Tangents from an external point are equal in length, which is the syllabus's own wording, and the two tangents with the two radii form a kite that has a line of symmetry
  • The chord fact creates a right-angled triangle from the radius, half the chord and the distance to the centre, which is where Pythagoras comes in
A chord PQ with its midpoint M joined to the centre O, the right angle at M marked, PM and MQ marked equal, and the radius OQ of 6 cm and the 40 degree angle at Q shown, so triangle OMQ can be solved
Source: Circles & chords by Save My Exams
  • The tangent fact creates two congruent right-angled triangles back to back
Worked example

Pythagoras on a bisected chord

A circle has centre O and radius 13 cm. A chord PQ is 24 cm long. Find the distance from O to the chord.

A circle with centre O. A horizontal chord PQ below the centre is labelled 24 cm, and the radius OP is labelled 13 cm. A dashed line runs from O down to the point M on the chord, meeting it at a marked right angle. The length of that dashed line is not labelled.

Solution:

  • Let M be the midpoint of PQ, and join OM
  • The perpendicular bisector of a chord passes through the centre, so OM meets PQ at 90° and PM = MQ
  • So PM = 24 ÷ 2 = 12 cm
  • Triangle OMP has a right angle at M, with OP = 13 cm as the hypotenuse
  • By Pythagoras, OM² = 13² − 12² = 169 − 144 = 25
  • OM = √25 = 5 cm
  • Check: 5² + 12² = 25 + 144 = 169 = 13²