Year 11 Extended · Unit 1 Right-angled triangles and circle geometry · Lesson 7 of 7

Crossing chords and tangents: the lengths that are linked

Two paths cross inside a circular plaza and cut each other into four pieces. This lesson asks what links the four lengths, proves the link with similar triangles, and extends it to a tangent and a secant from a point outside.

Geometry and TrigonometryNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: two triangles are similar with sides in the ratio 3 : 5. A side of 12 cm in the small one matches which side in the large one?
    Answer20 cm.
  2. From Year 9: a right-angled triangle has hypotenuse 20 cm and one side 16 cm. Find the other side.
    Answer12 cm.
  3. From the previous lesson: angles on the same arc of a circle are equal. Angle APB = 33° with P on the major arc. What is angle AQB for another point Q on the same arc?
    Answer33°.
Words used in this lesson
chord
a straight line joining two points on a circle. For example: a path across a round plaza.
tangent
a straight line that touches a circle at exactly one point. For example: a ruler touching a coin at its edge.
secant
a straight line that cuts a circle at two points. For example: a path that crosses the plaza.
similar triangles
triangles with the same angles and sides in the same ratio. For example: a photograph and its enlargement.
contradiction
a result that cannot be true at the same time as what you assumed. For example: a line that is both a tangent and meets the circle twice.

Two paths across a plaza

Two straight paths cross at a point X inside a circular plaza. X cuts the first path into pieces of 3 m and 12 m. It cuts the second path into a piece of 4 m and a longer piece of unknown length.

Predict first: what is the length of the unknown piece? Is there any rule that links the four pieces? Write your guess and your reason before you use the first figure.

6
60
64.0product of the pieces of chord 1
64.0product of the pieces of chord 2

X is 6 from the centre: chord 1 is cut into 4.0 and 16.0, chord 2 into 5.5 and 11.5 (rounded). Both products are 64.0.

Show me the steps
  1. X is 6 from the centre, chord 2 at 60°: chord 1 has pieces 4.0 and 16.0, product 64.0; chord 2 has pieces 5.5 and 11.5, product 64.0 as well.
  2. Turn chord 2 to 120°: its pieces become 11.5 and 5.5, the same pair swapped over, and the product is still 64.0.
  3. Turn chord 2 to 90°, perpendicular to chord 1: the pieces are 8.0 and 8.0 and the product is 64.0. The chord is bisected by the line from the centre, since it is perpendicular to the diameter through X.
  4. Move X to the centre, distance 0: every chord through X is a diameter, with two pieces of 10.0 and a product of 100.0. As X moves out, the product falls to 100 minus d squared.

For two chords crossing at X: AX × XB = CX × XD. Proof: triangles AXD and CXB have angle XAD = angle XCB (same arc DB) and angle AXD = angle CXB (vertically opposite), so they are similar, and AX ÷ CX = XD ÷ XB. Cross-multiplying gives AX × XB = CX × XD.

For a tangent PT and a secant PAB: PT² = PA × PB. Triangles PTA and PBT share the angle at P, and angle PTA = angle PBT (alternate segment), so they are similar: PT ÷ PB = PA ÷ PT.

A tangent is perpendicular to the radius, by contradiction. Suppose OT is not perpendicular to the tangent. Let F be the foot of the perpendicular from O, and T′ the point on the line with FT′ = FT, on the other side of F. Triangles OFT and OFT′ are congruent, so OT′ = OT = r and T′ is a second point on the circle, but a tangent has only one: a contradiction.

12
10
108.0PT squared
108.0PA times PB

P is 12 from the centre: PT² = 12² − 6² = 108.0, and the secant gives PA × PB = 6.2 × 17.4 = 108.0 (rounded).

Show me the steps
  1. P is 12 from the centre, secant at 10°: PT² = 144 − 36 = 108.0. The secant cuts the circle at distances 6.2 and 17.4 from P, and their product is 108.0 to the precision shown.
  2. Turn the secant to 0°, through the centre: PA = 12 − 6 = 6.0 and PB = 12 + 6 = 18.0, and 6.0 × 18.0 = 108.0 again.
  3. Turn the secant to 20°: the distances are 6.9 and 15.7 (rounded), and the product is still 108.0. The secant tilts away from the centre and PA grows while PB shrinks.
  4. Move P to 16 from the centre: PT² = 256 − 36 = 220.0, and every secant from P has PA × PB = 220.0. The constant depends only on how far P is from the centre.

Match the pieces, not the chords. In AX × XB = CX × XD, each side multiplies the two pieces of the same chord. A common slip is to multiply a piece of one chord by a piece of the other, such as AX × CX.

Worked examples

1. Chords AB and CD cross at X, with AX = 5, XB = 8 and CX = 4. Find XD.

  1. AX × XB = CX × XD, so 5 × 8 = 4 × XD.
  2. XD = 40 ÷ 4 = 10.

2. From P, a tangent PT = 14 and a secant PAB with A nearer P. PA = 7. Find AB, and the radius of the circle if P is 25 from the centre.

  1. PT² = PA × PB gives 196 = 7 × PB, so PB = 28 and AB = 28 − 7 = 21.
  2. In right-angled triangle OTP, r² = 25² − 14² = 625 − 196 = 429, so r = √429 ≈ 20.7. This uses the right angle at T. Worked example 3 finds the same number, OP² − r², for the secant through the centre.

3. Show that the equation PA × PB = OP² − r² holds for a secant through the centre O.

  1. Let the secant pass through O. Then PA = OP − r and PB = OP + r.
  2. PA × PB = (OP − r)(OP + r) = OP² − r², a difference of two squares.
  3. This is also PT², so the same number results for the tangent and for every secant, not only the one through the centre.

Your go. The first step is given. Chords AB and CD cross at X, with AX = 3, XB = 10 and CX = 5. Find XD.

  1. Multiply the two pieces of the same chord: AX × XB.
  2. Set it equal to the product of the two pieces of the other chord, CX × XD.
Check the answer

3 × 10 = 5 × XD, so XD = 6. Multiplying AX × CX = 15 and dividing by XB would give 1.5, which pairs pieces of different chords.

Check before you go on

1. Chords AB and CD cross at X, with AX = 6 and XB = 8. Which pair could be CX and XD?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Chords AB and CD of a circle cross at X. AX = 7 cm, XB = 6 cm and CX = 3 cm. How long is XD?

3. From a point P a tangent PT of length 12 cm is drawn to a circle. A secant from P cuts the circle at A and B, with A nearer to P and PA = 8 cm. How long is the chord AB?

Level 2 · Building

4. Chords AB and CD cross at X. AX = x cm, XB = (x + 6) cm, CX = 6 cm and XD = 12 cm. Find the length of the chord AB.

5. A point P is 17 cm from the centre O of a circle, and a tangent from P touches the circle at T with PT = 15 cm. Find the radius of the circle.

6. PT is a tangent to a circle at T. The line PAB cuts the circle at A and B, with A between P and B. Angle TPB = 24° and angle PTA = 36°. Find angle ATB.

Level 3 · Stretch

7. A circle has radius 13 cm. Two parallel chords AB and CD, on the same side of the centre, have lengths 24 cm and 10 cm. How far apart are the chords?

8. Chords AC and BD of a circle cross at X, with the points in the order A, B, C, D round the circle. AX = 6 cm, AB = 4 cm and DC = 10 cm. Find DX.

Something to think about

In the second figure, turn the secant until A and B come together. What does the secant become, and what does the equation PA × PB = PT² say then?

One way to think about it

The secant becomes a tangent: A and B merge into one point T, so PA = PB = PT and PA × PB = PT² says PT × PT = PT². The tangent theorem is the limiting case of the secant theorem, which is why the product is the same for both.

Exit ticket

9. From a point P, a secant cuts a circle at A and B, with A nearer to P. PA = 4 cm and the chord AB = 5 cm. A tangent from P touches the circle at T. How long is PT?