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Distance of a chord from the centre · 3 marks

In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.

Answer: No. Chord length is 2√(r2 − d2), which is not proportional to the distance d. For example, with radius 10 cm, CD at 3 cm is about 19.1 cm, AB at 6 cm is 16 cm, and 19.1 ≠ 2 × 16. All we can say is that AB is shorter than CD.

Step-by-step solution

Given: Distance of AB from the centre = 2 × distance of CD from the centre
To find: Whether CD = 2AB must be true

Idea: Test the claim with the formula chord = 2√(r2 − d2). Doubling the distance does not halve the chord; one counter-example is enough to show the claim is not always true.

  1. Let the radius be r, the distance of CD be d and that of AB be 2d. Then CD = 2√(r2 − d2) and AB = 2√(r2 − 4d2).1 mark
  2. Counter-example: r = 10 cm, d = 3 cm. CD = 2√(100 − 9) = 2√91 ≈ 19.08 cm, and AB = 2√(100 − 36) = 2√64 = 16 cm. Then 2AB = 32 cm ≠ 19.08 cm, so CD ≠ 2AB.1 mark
  3. So we cannot conclude CD = 2AB. (In fact CD is less than a diameter, 20 cm here, so CD = 2AB would need AB below 10 cm.) What we can conclude: AB is farther from the centre, so AB < CD.1 mark
  4. For the curious: CD = 2AB happens only in one special case. Squaring, r2 − d2 = 4(r2 − 4d2), which gives d2 = r2/5. Any other distance breaks it.
No. Since chord length = 2√(r2 − d2) is not proportional to the distance, doubling the distance does not halve the chord. For r = 10 cm and distances 3 cm and 6 cm, CD ≈ 19.08 cm while AB = 16 cm. We can only conclude that AB is shorter than CD.

Check: Special case check: r = 5 and d = √5: CD = 2√20 ≈ 8.94 and AB = 2√(25 − 20) = 2√5 ≈ 4.47, so CD = 2AB there, which confirms it holds only for this particular d.

Answer to write in the exam

Radius r; distance of CD = d, distance of AB = 2d

CD = 2√(r2 − d2), AB = 2√(r2 − 4d2)

Take r = 10 cm, d = 3 cm:

CD = 2√(100 − 9) = 2√91 ≈ 19.08 cm

AB = 2√(100 − 36) = 2√64 = 16 cm ⇒ 2AB = 32 cm ≠ CD

∴ No, CD = 2AB cannot be concluded. (Only AB < CD, as AB is farther from the centre.)

Common mistakes that cost marks

  • Assuming chord length and distance are in simple (inverse) proportion. The relationship involves squares and a square root.
  • Answering “yes” from one lucky example. One counter-example is enough to show a general claim is false.
  • Mixing up which chord is longer: the chord farther from the centre (AB) is the shorter one.

How this can come in the exam

Assertion–Reason (1 mark)

Assertion (A): If chord AB is farther from the centre than chord CD of the same circle, then AB < CD.
Reason (R): Of two chords of a circle, the one nearer to the centre is longer.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
Both are true, and A is exactly R applied to AB and CD.

Try one yourself

In a circle of radius 15 cm, chord CD is 5 cm from the centre and chord AB is 10 cm from the centre. Find CD and AB. Is CD = 2AB?

Show answer

CD = 2√(225 − 25) = 2√200 = 20√2 ≈ 28.3 cm; AB = 2√(225 − 100) = 2√125 = 10√5 ≈ 22.4 cm. 2AB ≈ 44.7 cm ≠ 28.3 cm, so no.

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