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Cambridge IGCSE Mathematics — 0580 Extended

Topic 4.8: Geometry — Circle Theorems II

Equal chords

Statement. Equal chords are equidistant from the centre. The converse is also true: chords the same distance from the centre are equal.

Distance means the length of the perpendicular from the centre to the chord. This is a fact about chords — do not quote it for tangents.

Two equal chords each have a perpendicular drawn from the centre. What is true of those perpendiculars?

Circle centre O with two equal chords and equal perpendiculars from O to each chord
The two perpendiculars from \(O\) have the same length. Equal chords \(\Leftrightarrow\) equal distances from the centre.

Reason you must write: equal chords are equidistant from the centre.

Perpendicular from the centre bisects a chord

Statement. The perpendicular from the centre to a chord bisects the chord. If \(OM \perp AB\) with \(M\) on \(AB\), then \(AM = MB\).

The converses are on the syllabus too: the line from the centre to the midpoint of a chord is perpendicular to the chord; and the perpendicular bisector of a chord passes through the centre.

Method

  1. Drop a perpendicular from \(O\) to chord \(AB\), meeting at \(M\).
  2. Then \(AM = MB\) and \(\angle OMA = \angle OMB = 90^\circ\).
  3. If you need the centre, construct the perpendicular bisector of a chord — it goes through \(O\).

Chord \(AB\) has midpoint \(M\), and \(OM\) is drawn. State the two facts shown.

Circle centre O with chord AB, midpoint M, and OM perpendicular to AB so AM equals MB
\(AM = MB\) and \(OM \perp AB\).

Reason you must write: perpendicular from the centre bisects the chord (or: perpendicular bisector of a chord passes through the centre).

Paper 2 (non-calculator)

If a chord has length \(10\,\text{cm}\) and the perpendicular from the centre meets it at \(M\), then each half is \(5\,\text{cm}\). You may then use Pythagoras in triangle \(OMA\) if a radius is given.

Two tangents from an external point

Statement. The two tangents from an external point to a circle are equal. If they touch at \(T_1\) and \(T_2\), then \(PT_1 = PT_2\).

Further facts that follow (and that you may use): the radii to the points of contact are perpendicular to the tangents (\(\angle OT_1P = \angle OT_2P = 90^\circ\)), and line \(OP\) bisects \(\angle T_1PT_2\).

Tangents from \(P\) touch the circle at \(T_1\) and \(T_2\). What lengths are equal, and what angles are \(90^\circ\)?

Circle centre O with two tangents from external point P touching at T1 and T2, so PT1 equals PT2 and radii perpendicular to the tangents
\(PT_1 = PT_2\). Radii \(OT_1\) and \(OT_2\) are perpendicular to the tangents at the points of contact.

Reason you must write: tangents from an external point are equal.

Try this

Chord \(AB = 10\,\text{cm}\). The perpendicular from the centre meets \(AB\) at \(M\). Find \(AM\). From an external point \(P\), two tangents touch at \(T_1\) and \(T_2\). If \(PT_1 = 8\,\text{cm}\), find \(PT_2\). Give a reason for each.

Show answer
Answer

  1. Perpendicular from the centre bisects the chord

    \[ AM = 5\,\text{cm} \]

  2. Tangents from an external point are equal

    \[ PT_2 = 8\,\text{cm} \]

Exam Traps

  • Do not mix “equal chords are equidistant from the centre” with “tangents from an external point are equal” — they are different theorems.
  • Do not write “circle theorem”. Name which one (equal chords, or tangents from an external point).
  • \(AM = MB\) only after the line from the centre is perpendicular to the chord (or \(M\) is already the midpoint).

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