Circles and tangents: tangent theorems, alternate segment, chords and secants
A tangent is perpendicular to the radius at the point of contact. Tangents from a point are equal, and the tangent-chord angle equals the angle in the alternate segment.
Tangent theorems
- A tangent is a line that touches a circle at exactly one point, the point of contact. A secant is a line that cuts the circle at two points.
- The tangent at a point is perpendicular to the radius through that point. Only one tangent can be drawn at a point on the circle.
- From a point outside the circle, exactly two tangents can be drawn. From a point inside, none can be drawn.
- The two tangents from an outside point are equal in length. They make equal angles with the line joining the point to the centre.
- Length of a tangent = √(d² − r²). Here d is the distance of the point from the centre and r is the radius.
- Two circles with radii R and r touch if they meet at exactly one point. That point lies on the line joining their centres. If they touch externally, the distance between the centres is R + r. If they touch internally, it is R − r.
Alternate segment theorem: figure and proof
Figure: draw a circle with centre O. Draw a tangent PQ touching it at A. Draw a chord AB and mark C on the major arc. Draw the diameter AD. Join BD, BC and AC. The **alternate segment** is the part of the circle on the other side of the chord from the angle.
- 1Statement: the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment. To prove: ∠BAQ = ∠ACB.
- 2∠DAQ = 90°, so ∠BAQ = 90° − ∠DAB.tangent ⊥ radius
- 3∠ABD = 90°, so in ΔABD, ∠ADB = 90° − ∠DAB.angle in a semicircle
- 4From steps 2 and 3, ∠BAQ = ∠ADB.
- 5∠ADB = ∠ACB.angles in the same segment
- 6Therefore ∠BAQ = ∠ACB. Hence proved.
Chords and secants
| Situation | Result | Example |
|---|---|---|
| Chords AB and CD cut at P inside the circle | PA × PB = PC × PD | PA = 4, PB = 6, PC = 3 (cm): PD = 24/3 = 8 cm |
| Secants PAB and PCD meet at P outside the circle | PA × PB = PC × PD | PA = 3, PB = 8, PC = 4 (cm): PD = 24/4 = 6 cm |
| Secant PAB and tangent PT from the same point P | PA × PB = PT² | PA = 4, PB = 9 (cm): PT² = 36, so PT = 6 cm |
Example 1: two tangents and the centre
PA and PB are tangents from P to a circle with centre O. ∠APB = 64°. Find ∠AOB.
- ∠OAP = ∠OBP = 90° (tangent ⊥ radius).
- OAPB is a quadrilateral, so its angles add up to 360°.
- ∠AOB = 360° − 90° − 90° − 64°
- ∠AOB = 116°
∠AOB = 116°
Example 2: length of a tangent (word problem)
A circular lotus pond has radius 5 m. A straight path from a gate P, 13 m from the centre O, just touches the pond at T. Find PT.
- OT ⊥ PT (radius ⊥ tangent), so ΔOTP is right-angled at T.
- PT² = OP² − OT²
- PT² = 13² − 5² = 169 − 25 = 144
- PT = √144 = 12
PT = 12 m
A bicycle wheel touches the road at just one point, so the road is a tangent to the wheel. The belt over two pulleys in a flour mill runs along tangents to both wheels.
2 marks: “Find the length of the tangent from a point 10 cm away from the centre of a circle of radius 6 cm.” Length = √(10² − 6²) = √(100 − 36) = √64 = 8 cm.
5 marks: “State and prove the alternate segment theorem.” Statement and figure: tangent PQ at A, chord AB, C in the alternate segment, diameter AD. ∠DAQ = 90° and ∠ABD = 90°, so ∠BAQ = ∠ADB. Then ∠ADB = ∠ACB (same segment).
2 marks: “Chords AB and CD of a circle intersect at P inside it. AP = 5 cm, PB = 6 cm and CP = 3 cm. Find PD.” AP × PB = CP × PD, so 5 × 6 = 3 × PD and PD = 10 cm.
Tangent ⊥ radius; two tangents equal; tangent-chord angle = angle across the chord; part × part = part × part.