Chords and the angles they make
A chord is a straight segment joining two points of a circle. The longest chord passes through the centre: it is a diameter (2 × radius).
Join the ends A and B of a chord to the centre O. The angle ∠AOB is the angle the chord subtends (makes) at the centre.
Theorem: equal chords subtend equal angles at the centre. Why: if AB = CD, then ΔAOB ≅ ΔCOD by SSS (OA = OC = OB = OD = radius). So ∠AOB = ∠COD. The converse is true too: chords that subtend equal angles at the centre are equal (SAS).
The perpendicular from the centre bisects the chord
Theorem: if OM ⟂ AB, then AM = MB.
Why: in ΔOMA and ΔOMB, ∠OMA = ∠OMB = 90°, OA = OB (radii), and OM is shared. By RHS they are congruent, so AM = MB.
Converse: the line from the centre to the midpoint of a chord is perpendicular to the chord (SSS).
A useful result: through three points not on one line there is exactly one circle. Its centre is where the perpendicular bisectors of two chords meet.
Equal chords are equidistant from the centre
The distance of a chord from the centre means the length of the perpendicular from O to it.
Theorem: equal chords are at equal distances from the centre. With OM ⟂ AB and ON ⟂ CD: AM = ½ AB = ½ CD = CN, and OA = OC. By RHS, ΔOMA ≅ ΔONC, so OM = ON.
Converse: chords at equal distances from the centre are equal.
By Pythagoras, OM² + AM² = r². So the closer a chord is to the centre, the longer it is.
The angle at the centre is double
Let arc AB subtend ∠AOB at the centre and ∠APB at a point P on the rest of the circle.
Theorem: ∠AOB = 2 ∠APB.
Why (one case): join PO and extend it to X. OA = OP, so ΔOAP is isosceles and ∠OAP = ∠OPA. The exterior angle ∠AOX = ∠OAP + ∠OPA = 2∠OPA. In the same way ∠BOX = 2∠OPB. Add them: ∠AOB = 2(∠OPA + ∠OPB) = 2∠APB.
Consequences:
- Angles in the same segment are equal: every P on the same arc sees AB at the same angle.
- Angle in a semicircle is 90°: if AB is a diameter, ∠AOB = 180°, so ∠APB = 90°.
Concyclic points and cyclic 4-gons
Points are concyclic if one circle passes through all of them. A 4-gon with all four corners on a circle is a cyclic quadrilateral.
Theorem: opposite angles of a cyclic 4-gon add to 180°. Why: ∠A is half the centre angle of arc BCD and ∠C is half the centre angle of arc BAD. These two centre angles together make 360°, so ∠A + ∠C = 180°. The same is true for ∠B + ∠D.
Converse: if a pair of opposite angles of a 4-gon adds to 180°, the 4-gon is cyclic. Also, if a segment AB subtends equal angles at two points C and D on the same side of it, then A, B, C, D are concyclic.
Example: every rectangle is cyclic (90° + 90° = 180°), but a rhombus that is not a square is not.
Try it at home
Trace round a bangle or a steel plate to draw a circle. Draw any chord and fold the paper so the two ends of the chord meet. The fold line passes through the centre and cuts the chord in half. Do it for a second chord: the two folds cross at the centre.
Now draw a diameter and pick any point on the circle. Join it to both ends and measure the angle with a protractor or the corner of a notebook: it is always 90°. In the 3D free play, set the chord to 180° and move P to see the same.
Key formulas and definitions
- Equal chords ⇔ equal angles at the centre
- OM ⟂ AB ⇒ AM = MB
- Equal chords ⇔ equal distances from the centre
- OM² + (½ AB)² = r²
- ∠AOB = 2 ∠APB; angle in a semicircle = 90°
- Cyclic 4-gon: ∠A + ∠C = 180°, ∠B + ∠D = 180°
Worked examples
1. A chord is 8 cm long in a circle of radius 5 cm. How far is it from the centre?
Step 1: the perpendicular from O cuts the chord in half, so AM = 4 cm. Step 2: OM² = OA² − AM² = 25 − 16 = 9. Step 3: OM = 3 cm.
2. In a circle of radius 13 cm, a chord is 5 cm from the centre. Find its length.
Step 1: AM² = 13² − 5² = 169 − 25 = 144. Step 2: AM = 12 cm. Step 3: chord AB = 2 × 12 = 24 cm.
3. ∠AOB = 110° at the centre. Find ∠APB for a point P on the major arc.
Step 1: the angle at the centre is double the angle at the circle. Step 2: ∠APB = 110° ÷ 2 = 55°.
4. AB is a diameter and C is a point on the circle with ∠CAB = 35°. Find ∠CBA.
Step 1: angle in a semicircle: ∠ACB = 90°. Step 2: angle sum in ΔABC: ∠CBA = 180 − 90 − 35 = 55°.
5. ABCD is a cyclic 4-gon with ∠A = 72° and ∠B = 104°. Find ∠C and ∠D.
Step 1: opposite angles add to 180°. Step 2: ∠C = 180 − 72 = 108°. Step 3: ∠D = 180 − 104 = 76°.
6. Two equal chords AB and CD of a circle are 4 cm from the centre, and the radius is 5 cm. Find the length of each chord.
Step 1: half-chord² = 5² − 4² = 9, so half-chord = 3 cm. Step 2: each chord = 6 cm. Step 3: both are 4 cm from the centre, which agrees with 'equal chords are equidistant'.
7. Points P and Q lie on the same arc, and chord AB subtends ∠APB = 40°. Find ∠AQB and ∠AOB.
Step 1: angles in the same segment are equal, so ∠AQB = 40°. Step 2: the angle at the centre is double: ∠AOB = 80°.
Common mistakes
- Halving when you should double. The angle at the centre is the bigger one: ∠AOB = 2 ∠APB.
- Forgetting to halve the chord before using Pythagoras. Use ½ AB, not AB.
- Adding next angles of a cyclic 4-gon to 180°. It is the opposite angles (A and C, B and D) that add to 180°.
- Using the minor-arc angle when P is on the minor arc. Then ∠APB is half of the reflex angle AOB, which is 360° − ∠AOB.