This note covers the circle theorems: the angle at the centre, angles in a semicircle and in the same segment, cyclic quadrilaterals, tangents, the alternate segment theorem and chords. Each theorem lets you find an unknown angle or length from a circle diagram, and each needs to be recognised from the shape it makes.
Key ideas
A circle has a centre \(O\). A radius joins the centre to the circle. A chord joins two points on the circle. A diameter is a chord through the centre. A tangent is a straight line that touches the circle at exactly one point. An arc is part of the circumference, and a segment is the region between a chord and an arc.
The angle in a semicircle and the tangent-radius facts are Core. The other theorems below are Extended.
The theorems
- Angle at the centre. The angle at the centre is twice the angle at the circumference when both stand on the same arc.
- Angle in a semicircle. The angle at the circumference standing on a diameter is \(90^\circ\).
- Angles in the same segment. Angles at the circumference standing on the same arc (or chord) and on the same side of it are equal.
- Cyclic quadrilateral. If all four vertices of a quadrilateral lie on a circle, opposite angles add up to \(180^\circ\).
- Tangent and radius. A tangent meets the radius at the point of contact at \(90^\circ\).
- Two tangents from a point. Tangents drawn from the same outside point to a circle have equal lengths. The triangle they form with the chord joining the contact points is isosceles.
- Alternate segment theorem. The angle between a tangent and a chord equals the angle in the alternate segment, which is the segment on the opposite side of the chord.
- Chords. A line from the centre that is perpendicular to a chord bisects the chord. Equally, the line from the centre to the midpoint of a chord is perpendicular to the chord.
Two other facts are very useful. Any triangle made from two radii is isosceles, so its base angles are equal. Also, the radius and tangent facts often combine with Pythagoras' theorem, for example when finding the length of a tangent or a chord.
Some questions need two theorems in sequence. Mark each angle on the diagram as you find it. Write down the name of each theorem next to the step that uses it, so your reasoning is easy to follow.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Angle at centre \(= 2 \times\) angle at circumference | Both angles must be on the same arc. |
| Angle in a semicircle \(= 90^\circ\) | The chord is a diameter. |
| Angles in the same segment are equal | Same arc, same side of the chord. |
| Opposite angles of a cyclic quadrilateral | Add up to \(180^\circ\). |
| Tangent \(\perp\) radius | At the point of contact the angle is \(90^\circ\). |
| Equal tangents | Two tangents from one point have equal length. |
| Alternate segment theorem | Tangent-chord angle \(=\) angle in the alternate segment. |
| Perpendicular from centre to chord | Bisects the chord. Use with Pythagoras to find lengths. |
Worked examples
Example 1 (Extended) \(A\), \(B\) and \(C\) are points on a circle with centre \(O\). Angle \(AOB = 134^\circ\), and \(C\) is on the major arc. Find angle \(ACB\).
Step 1: angle \(AOB\) is at the centre and angle \(ACB\) is at the circumference. Both stand on arc \(AB\).
Step 2: the angle at the centre is twice the angle at the circumference, so \(ACB = 134 \div 2 = 67^\circ\).
Answer: \(67^\circ\) (angle at centre is twice angle at circumference).
Example 2 (Extended) \(ABCD\) is a cyclic quadrilateral. Angle \(ABC = 108^\circ\) and angle \(BAD = 85^\circ\). Find angles \(ADC\) and \(BCD\).
Step 1: angle \(ADC\) is opposite angle \(ABC\). Opposite angles of a cyclic quadrilateral add up to \(180^\circ\).
Step 2: \(ADC = 180 - 108 = 72^\circ\).
Step 3: angle \(BCD\) is opposite angle \(BAD\), so \(BCD = 180 - 85 = 95^\circ\).
Step 4: check. \(108 + 72 + 85 + 95 = 360\).
Answer: \(ADC = 72^\circ\) and \(BCD = 95^\circ\).
Example 3 (Extended) \(PA\) and \(PB\) are tangents from \(P\) to a circle with centre \(O\), touching at \(A\) and \(B\). Angle \(APB = 50^\circ\). Find angle \(OAB\).
Step 1: \(PA = PB\) (equal tangents), so triangle \(PAB\) is isosceles.
Step 2: angle \(PAB = (180 - 50) \div 2 = 65^\circ\).
Step 3: the radius \(OA\) is perpendicular to the tangent \(PA\), so angle \(OAP = 90^\circ\).
Step 4: angle \(OAB = 90 - 65 = 25^\circ\).
Step 5: check. Angle \(AOB = 360 - 90 - 90 - 50 = 130^\circ\) in quadrilateral \(OAPB\). Triangle \(OAB\) is isosceles, so \(OAB = (180 - 130) \div 2 = 25^\circ\).
Answer: \(25^\circ\)
Common errors
- Using the angle-at-the-centre theorem when the two angles are not on the same arc.
- Forgetting that the angle in a semicircle needs the chord to be a diameter. Check that the line passes through the centre.
- Adding up opposite angles of a quadrilateral that is not cyclic. All four corners must be on the circle.
- Matching the wrong angle in the alternate segment theorem. The equal angle is on the opposite side of the chord from the tangent-chord angle.
- Not using the isosceles triangle formed by two radii, which gives equal base angles.
Exam tip: when a diagram contains a tangent, mark the \(90^\circ\) angle with the radius straight away. When it has a centre and several radii, mark equal lengths and look for isosceles triangles. Give the full theorem name as your reason.
Check yourself
- \(AB\) is a diameter of a circle and \(C\) is on the circle. Angle \(CAB = 35^\circ\). Find angle \(ABC\).
Answer
Angle \(ACB = 90^\circ\) (angle in a semicircle). Angle \(ABC = 180 - 90 - 35 = 55^\circ\). - (Extended) An angle at the circumference is \(48^\circ\). What is the angle at the centre standing on the same arc?
Answer
\(2 \times 48 = 96^\circ\). - (Extended) A circle has radius 13 cm. A chord is 5 cm from the centre. Find the length of the chord.
Answer
The perpendicular from the centre bisects the chord. Half the chord is \(\sqrt{13^2 - 5^2} = \sqrt{144} = 12\) cm. The chord is \(2 \times 12 = 24\) cm. - (Extended) A tangent touches a circle at \(T\). A chord \(TA\) makes an angle of \(52^\circ\) with the tangent. What is the angle in the alternate segment?
Answer
By the alternate segment theorem, it is \(52^\circ\).