Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE.
Step-by-step solution
To find: Show that CF < CG
Idea: The radius is the same for both chords, and in each right triangle (radius)2 = (distance)2 + (half-chord)2. If one half-chord is bigger, its distance must be smaller to keep the same total.
- The perpendicular from the centre bisects a chord, so F and G are the midpoints of AB and DE: AF = ½AB and GD = ½DE.½ mark
- AC = CD (radii). By the Baudhāyana–Pythagoras theorem in right ΔCFA and ΔCGD: AC2 = CF2 + AF2 and CD2 = CG2 + GD2.1 mark
- So CF2 + AF2 = CG2 + GD2.½ mark
- AB > DE, so AF = ½AB > ½DE = GD, and AF2 > GD2.1 mark
- Subtracting a larger number from the same total leaves less: CF2 < CG2, so CF < CG. The longer chord is nearer the centre.1 mark
- Comment: the chord nearest the centre passes through it (distance 0): the diameter, the longest chord. Pushing a chord towards the edge shrinks it to a point at distance equal to the radius.
Answer to write in the exam
Given: AB > DE, CF ⟂ AB, CG ⟂ DE. To prove: CF < CG.
AF = ½AB, GD = ½DE (perpendicular from the centre bisects the chord)
AC2 = CF2 + AF2, CD2 = CG2 + GD2 (Baudhāyana–Pythagoras theorem)
AC = CD (radii) ⇒ CF2 + AF2 = CG2 + GD2
AB > DE ⇒ AF > GD ⇒ AF2 > GD2
⇒ CF2 < CG2
∴ CF < CG
Common mistakes that cost marks
- Reversing the conclusion: a larger half-chord means a smaller distance, because their squares add up to the same radius squared.
- Comparing AB and DE directly in the equation instead of the half-chords AF and GD.
- Measuring “distance” to a point of the chord other than the foot of the perpendicular.
How this can come in the exam
AB = 12 cm and CD = 9 cm are chords of the same circle, at distances p and q from the centre. Then
- p > q
- p = q
- p < q
- p + q = 0
Show answer
(C) p < q
The longer chord AB is nearer the centre, so p < q.
Assertion (A): A diameter is the longest chord of a circle.
Reason (R): Of two chords of a circle, the longer one is nearer to the centre.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- 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. A diameter is at distance 0, the least possible, so by R it is longer than every other chord; R explains A.
Try one yourself
In a circle of radius 41 cm, chord PQ is 80 cm and chord RS is 18 cm. Find their distances from the centre and check which is nearer.
Show answer
PQ: √(412 − 402) = √81 = 9 cm. RS: √(412 − 92) = √1600 = 40 cm. The longer chord PQ is nearer (9 cm < 40 cm).
More questions like this
- Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
- Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r2 − d2).
- 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.
- A circle with centre O is drawn, and A, B, C, D are points on the circle (see the figure). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.
- Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K between A, B on the circle. Measure the angle subtended at the centre by arc AKB. Take three points P, Q, R on the circle outside arc AKB. Measure the angles subtended by arc AKB at points P, Q, R. What do you notice?
Repeat this activity for a different arc AKB.