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Circle geometry · Lesson 4 of 5

Tangents

A tangent meets the radius at 90°, two tangents from a point are equal, and the angle between a tangent and a chord equals the angle in the alternate segment.

15 minYou should already know: Angles, triangles & polygons
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5

A tangent is a straight line that touches a circle at exactly one point, called the point of contact, without cutting into it. Tangents bring three facts. The first two are about lengths and right angles; the third, the alternate segment theorem, is the one exam questions test most.

Tangent and radius

Tangents from a pointDrag P
OT₁T₂P
9.9 cmPT₁=9.9 cmPT₂90°∠OT₁P
A tangent touches the circle at one point only, and it is always at 90° to the radius drawn to that point. From any point outside, you can draw exactly two tangents, and they are always the same length: here 9.9 cm each.

Drag P around, near the circle and far from it.

  • Each tangent meets its radius at the point of contact at exactly 90°, wherever P is.
  • The two tangents from P are always the same length.
OTtangent
Tangent and radius∠OTP = 90°
OPT₁T₂equal
Two tangentsPT₁ = PT₂

Why the two tangents are equal. Press Show why they are equal. Triangles OT1POT_1P and OT2POT_2P each have a right angle, they share the side OP (the hypotenuse), and OT1=OT2OT_1 = OT_2 because both are radii. Two right-angled triangles with the same hypotenuse and another equal side are congruent (the RHS rule from triangles). So every matching part is equal, including PT1=PT2PT_1 = PT_2.

Because of the two right angles, the angles of the kite OT1PT2OT_1PT_2 at P and at O always add up to 360∘−90∘−90∘=180∘360^\circ - 90^\circ - 90^\circ = 180^\circ.

Look at that last answer: 60∘60^\circ is exactly half of 120∘120^\circ. That’s no accident, and it leads to the next theorem.

Chords: the perpendicular from the centre

The perpendicular from the centre to a chord cuts the chord in half. That makes a right-angled triangle with the radius as its hypotenuse, so Pythagoras links the radius, the distance from the centre and half the chord.

OABMr
Perpendicular from the centreOM ⊥ AB, so AM = MB and OB² = OM² + MB²

More: chord lengths

The alternate segment theorem

Draw a tangent at T and a chord TA from the point of contact. The chord cuts the circle into two segments. The alternate segment is the one on the other side of the chord from the angle you’re looking at.

The alternate segmentDrag the gold points
55°55°TSAX
55°tangent and chord, ∠ATS=55°in the alternate segment, ∠TXA
The angle between the tangent and the chord TA is 55°. The shaded part is the alternate segment: the one on the other side of the chord. Any angle standing on TA in it is also 55°. Drag X around the shaded arc, then move A.
  1. Drag X around the shaded arc. The angle at X never changes, and it always equals the angle between the tangent and the chord.
  2. Move A. Both angles change together.
  3. Press Use the angle on the other side. The shading jumps to the other segment, and the rule still holds.
TAXyy
Alternate segment∠ATS = ∠TXA

Why it’s true. Make the tangent angle acute and press Show why it works. Draw the diameter TD.

  1. ∠TAD=90∘\angle TAD = 90^\circ (angle in a semicircle, lesson 2).
  2. ∠DTS=90∘\angle DTS = 90^\circ (tangent meets radius at 90∘90^\circ).
  3. So ∠ATS=90∘−∠ATD\angle ATS = 90^\circ - \angle ATD. In triangle TAD, ∠TDA=180∘−90∘−∠ATD=90∘−∠ATD\angle TDA = 180^\circ - 90^\circ - \angle ATD = 90^\circ - \angle ATD as well. So ∠ATS=∠TDA\angle ATS = \angle TDA.
  4. X and D are in the same segment, so ∠TXA=∠TDA\angle TXA = \angle TDA (lesson 2). That gives ∠ATS=∠TXA\angle ATS = \angle TXA.

Lengths: tangents, secants and crossing chords

Two more length rules appear in JAMB questions.

  • Tangent and secant from the same point. If a line from PP cuts the circle at AA and BB, and the tangent from PP touches at TT, then PT2=PA×PBPT^2 = PA \times PB.
  • Two chords that cross. If chords ABAB and CDCD meet at XX inside the circle, then AX×XB=CX×XDAX \times XB = CX \times XD.
PABTPT² = PA × PB
Tangent and secantPT² = PA × PB, both measured from P
ABCDXAX × XB = CX × XD
Crossing chordsThe two parts of one chord multiply to the same as the other's

More: tangent, secant and chord lengths

A past question, step by step

Worked example · JAMB 1994

JAMB 1994 · UME · Q31

In the diagram, PTSPTS is a tangent to the circle TQRTQR at TT, ∠QRT=60∘\angle QRT = 60^\circ and ∠QTR=50∘\angle QTR = 50^\circ. Calculate ∠RTS\angle RTS.

60°50°?TRQSP

  1. Which theorem?

    A tangent (PTS) and a chord from the point of contact (TR): that’s the alternate segment theorem. ∠RTS\angle RTS is the angle between the tangent and chord TR.

    Think first. ∠RTS\angle RTS is between the tangent TS and the chord TR. Where is the alternate segment, and which angle of the triangle is in it?

  2. Find the angle in the alternate segment

    Across chord TR from ∠RTS\angle RTS is the point Q on the circle. So we need ∠RQT\angle RQT, the angle at Q standing on TR.

    Think first. What is ∠RQT\angle RQT?

  3. Angles of triangle QRT

    ∠RQT=180∘−60∘−50∘=70∘\angle RQT = 180^\circ - 60^\circ - 50^\circ = 70^\circ

  4. Use the theorem

    ∠RTS=∠RQT=70∘(angle in the alternate segment)\angle RTS = \angle RQT = 70^\circ \quad \text{(angle in the alternate segment)}

    The answer is B.

Your turn

WAEC 2024 · Paper 1 · Q45

In the diagram, TUTU is a tangent to the circle at PP. If ∠PTS=44∘\angle PTS = 44^\circ and ∠SQP=35∘\angle SQP = 35^\circ, find ∠PST\angle PST.

44°35°OPSQRTU
Worked solution (try it first)
  1. ∠SPT\angle SPT is between the tangent PTPT and the chord PSPS, so it equals the angle in the alternate segment: ∠SPT=∠SQP=35∘\angle SPT = \angle SQP = 35^\circ.
  2. The angles of triangle PSTPST add up to 180∘180^\circ: ∠PST=180∘−44∘−35∘\angle PST = 180^\circ - 44^\circ - 35^\circ
    =101∘= 101^\circ, option A.

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