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4c · E4.7–E4.8

Circle theorems

Topic 4 Geometry · Core and Extended · Papers 1–4

🎯What you need to be able to do

  • Use the angle in a semicircle (\( 90^{\circ} \)) and the angle between a tangent and a radius (\( 90^{\circ} \)), giving reasons.
  • Use the angle at the centre, angles in the same segment, opposite angles of a cyclic quadrilateral and the alternate segment theorem EXTENDED.
  • Use the symmetry properties: equal chords are equidistant from the centre, the perpendicular bisector of a chord passes through the centre, and tangents from an external point are equal EXTENDED.
  • Give each reason using the wording of the property.

📚The mathematics

The theorems

Six small circle diagrams. 1: a triangle on a diameter with a right angle at the circumference. 2: a tangent meeting a radius at a right angle. 3: angle at the centre 2x and angle at the circumference x on the same arc. 4: two angles in the same segment, both x. 5: a cyclic quadrilateral with opposite angles x and 180 minus x. 6: the alternate segment theorem, where the angle between a tangent and a chord equals the angle in the opposite segment, both x.
The six angle properties. Core uses the first two; Extended uses all six.
  • Angle in a semicircle \( = 90^{\circ} \): the angle at the circumference standing on a diameter.
  • Tangent and radius meet at \( 90^{\circ} \).
  • EXTENDED Angle at the centre is twice the angle at the circumference standing on the same arc.
  • EXTENDED Angles in the same segment are equal.
  • EXTENDED Opposite angles of a cyclic quadrilateral add up to \( 180^{\circ} \).
  • EXTENDED Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.

EXTENDED Three symmetry facts complete the list: equal chords are equidistant from the centre; the perpendicular bisector of a chord passes through the centre (so the perpendicular from the centre bisects the chord, which gives a right-angled triangle for Pythagoras); and tangents from an external point are equal in length (so they make an isosceles triangle).

Almost every circle question also uses ordinary angle facts: radii are equal, so a triangle with two radii as sides is isosceles. Spotting those isosceles triangles is usually the key step. Name angles with three letters (angle \(ABC\)).

✏️Worked example

\(A\), \(B\), \(C\) and \(D\) lie on a circle with centre \(O\). \(AC\) is a diameter and \(TA\) is the tangent at \(A\), with \(T\) on the same side of \(AC\) as \(B\). Angle \( BAC = 34^{\circ} \). Find, giving reasons: (a) angle \(ABC\) [1] (b) angle \(ACB\) [1] (c) angle \(TAB\) [2] (d) EXTENDED angle \(ADB\) [1] (e) EXTENDED angle \(AOB\) [1]
A circle with centre O and diameter AC drawn horizontally. B is on the upper arc and D on the lower arc. The tangent at A is vertical, with T above A. Angle BAC is 34 degrees, angle ABC is a right angle, angle ACB is 56 degrees, angle TAB is 56 degrees, angle ADB is 56 degrees and angle AOB is 112 degrees.
Every answer can be read off one diagram — and three of them are equal, which is itself a check.

(a) \( 90^{\circ} \) — angle in a semicircle.

(b) \( 180 - 90 - 34 = 56^{\circ} \) — angles in a triangle add up to \( 180^{\circ} \).

(c) Angle \( TAC = 90^{\circ} \) (tangent and radius), so angle \( TAB = 90 - 34 = 56^{\circ} \).

(d) \( 56^{\circ} \) — angles in the same segment (angle \(ADB\) and angle \(ACB\) stand on the same chord \(AB\)).

(e) \( 2 \times 56 = 112^{\circ} \) — angle at the centre is twice the angle at the circumference.

Check it. Angle \(TAB\) (tangent–chord) equals angle \(ACB\) (alternate segment): both \( 56^{\circ} \) ✓. And triangle \(AOB\) is isosceles (\( OA = OB \)): \( (180 - 112) \div 2 = 34^{\circ} \), which matches angle \(BAC\) ✓.
Reasons that are not reasons. “Because it is a circle” or “semicircle rule” earns nothing. Write the property: “angle in a semicircle is \( 90^{\circ} \)”.

📝Practise

All non-calculator style. Give a reason for every step. EXTENDED marks Extended-only content.

1. \(PQ\) is a diameter of a circle and \(R\) is a point on the circle. Angle \( QPR = 27^{\circ} \). Find angle \(PQR\). [2]
Angle \( PRQ = 90^{\circ} \) (angle in a semicircle), so angle \( PQR = 180 - 90 - 27 = 63^{\circ} \) (angles in a triangle).
2. \(PT\) is a tangent to a circle with centre \(O\), touching it at \(T\). Angle \( TOP = 58^{\circ} \). Find angle \(OPT\). [2]
Angle \( OTP = 90^{\circ} \) (tangent and radius), so angle \( OPT = 180 - 90 - 58 = 32^{\circ} \).
3. EXTENDED An angle at the circumference is \( 47^{\circ} \). Find the angle at the centre standing on the same arc. [1]
\( 94^{\circ} \) (angle at the centre is twice the angle at the circumference).
4. EXTENDED In a cyclic quadrilateral, two opposite angles are \( (2x + 10)^{\circ} \) and \( (3x - 5)^{\circ} \). Find \(x\) and both angles. [3]
Opposite angles add to \( 180^{\circ} \): \( 5x + 5 = 180 \), \( x = 35 \). The angles are \( 80^{\circ} \) and \( 100^{\circ} \).
5. EXTENDED Two tangents from \(P\) touch a circle, centre \(O\), at \(A\) and \(B\). Angle \( APB = 40^{\circ} \). Find angle \(PAB\) and angle \(AOB\). [3]
\( PA = PB \) (tangents from an external point), so triangle \(PAB\) is isosceles: angle \( PAB = (180 - 40) \div 2 = 70^{\circ} \). In quadrilateral \(OAPB\), angles \(OAP\) and \(OBP\) are \( 90^{\circ} \), so angle \( AOB = 360 - 90 - 90 - 40 = 140^{\circ} \).
6. EXTENDED A tangent at \(A\) makes an angle of \( 64^{\circ} \) with the chord \(AB\). \(C\) is a point on the major arc \(AB\). Find angle \(ACB\). [1]
\( 64^{\circ} \) (alternate segment theorem).
7. (Calculator.) EXTENDED A chord of length 16 cm is in a circle of radius 10 cm. Find the distance of the chord from the centre. [2]
The perpendicular from the centre bisects the chord (perpendicular bisector of a chord passes through the centre), giving a right-angled triangle with hypotenuse 10 and one side 8: \( \sqrt{100 - 64} = 6 \) cm.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • GeoGebra — drag points round a circle and watch the angle at the centre stay twice the angle at the circumference
  • Corbettmaths — circle theorems, with the reasons written out