Circles
Chords and tangents
In this note
- 4.8Use: the perpendicular from the centre bisects a chord (the perpendicular bisector of a chord passes through the centre).
- 4.8Use: equal chords are the same distance from the centre.
- 4.8Use: tangents from an external point are equal.
Before you start: Circle theorems.
Formula bank
Chords
The line from the centre at right angles to a chord cuts it in half. That makes a right-angled triangle with the radius as hypotenuse: \(r^{2}=d^{2}+(\text{half-chord})^{2}\). Equal chords are the same distance from the centre.
General methodHow to find a chord length or its distance from the centre
- Draw the perpendicular from the centre to the chord: it meets the chord at its midpoint.
- Right-angled triangle: hypotenuse = radius, one side = half the chord, other side = distance d.
- Pythagoras; double the half-chord if the whole chord is asked.
Example A chord is 5 cm from the centre of a circle of radius 13 cm. Find its length.
Half-chord by Pythagoras, then double.
- half-chord \(=\sqrt{13^{2}-5^{2}}=12\)
Answer 24 cm
Try it yourself
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Answer 8
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Answer 8
Two tangents from a point
Right angles at A and B (tangent ⟂ radius) make OAPB a kite: angle AOB + angle APB = 180°, and OP cuts both angles in half. Triangle PAB is isosceles.
General methodHow to find angles with two tangents
- Mark the right angles at the touching points.
- Angles in the kite OAPB add to 360°, so angle AOB + angle APB = 180°.
- PA = PB: triangle PAB is isosceles; OP halves angle APB and angle AOB.
Example PA and PB are tangents from P; angle AOB = 112°. Find angle APB and angle APO.
Kite: 360° with two right angles.
- angle APB = 360 − 90 − 90 − 112 = 68°
- angle APO = 68 ÷ 2
Answer 68° and 34°
Try it yourself
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Answer 55
Ways they test it
Type 1 Chord distance
"A chord of length 12 cm is in a circle of radius 10 cm. Find its distance from the centre." (3 marks)
Spot it a chord and a radius.
- Half-chord.
- Pythagoras with the radius as hypotenuse.
Type 2 Tangent kite
"PA and PB are tangents. Angle AOB = 112°. Find angle APB." (2 marks)
Spot it two tangents from one point.
- Right angles at A and B.
- 360 − 180 − 112.
Type 3 Combined
"PA, PB are tangents; angle APB = 42°; C on the major arc. Find angle ACB." (3 marks)
Spot it tangents and a point on the circle.
- Angle AOB = 138°.
- Centre theorem: halve it.
For example AOB = 138°
Answer 69°
Mistakes to avoid
- ✗ Using the whole chord in Pythagoras✓ half the chordThe right angle is at the midpoint.
- ✗ Radius as a shorter side✓ radius = hypotenuseIt is opposite the right angle.
- ✗ Kite angles summed to 180°✓ 360°It is a quadrilateral.
Summary
- Perpendicular from the centre bisects a chord.
- \(r^{2}=d^{2}+(\text{half-chord})^{2}\).
- Tangents from a point are equal; OAPB is a kite with two right angles.
If you see …
- a chord and a radiushalf-chord and Pythagoras.
- two tangents from one pointkite: angle AOB + angle APB = 180°.
Practice
Practice: easy, medium, hard
Easy
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Answer 24
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Answer 6
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Answer 65
Medium
Guide me
What is it asking? an angle
What do we know? two right angles in the kite
Plan 360 − 180 − 112
Show working
- 68°
Answer 68
Guide me
What is it asking? an angle
What do we know? kite
Plan 360 − 180 − 50
Show working
- 130°
Answer 130
Guide me
What is it asking? a length
What do we know? right angle at the touching point
Plan \(\sqrt{169-25}\)
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- 12
Answer 12
Hard
Guide me
What is it asking? an angle
What do we know? angle AOB = 138°
Plan centre theorem: halve
Show working
- 360 − 180 − 42 = 138
- 138 ÷ 2 = 69
Answer 69
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| AOB = 138 | M1A1 | ||
| 69 | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? a distance
What do we know? distances 5 and 12 from the centre
Plan add (opposite sides)
Show working
- \(\sqrt{169-144}=5\)
- \(\sqrt{169-25}=12\)
- 5 + 12 = 17
Answer 17
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| 5 and 12 | M1A1 | ||
| 17 | A1 | ||
| Total: 3 marks | |||
Guide me
What is it asking? an angle
What do we know? angle APB = 68°
Plan OP halves it
Show working
- 34°
Answer 34
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| 34 | B1 | ||
| Total: 1 mark | |||
Can I do this?
- I can find chord lengths and distances from the centre.
- I can use equal tangents and the tangent kite.
- I can combine these with other circle theorems.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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