Distance of a chord from the centre · 3 marks

Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.

Answer: A chord through A at distance d from O has length 2√(r² − d²), which is shortest when d is largest. The perpendicular OM from O to any chord through A satisfies OM ≤ OA (OA is the hypotenuse of ΔOMA), with equality only when M = A, that is, when the chord is perpendicular to OA.

Step-by-step solution

Given: Circle with centre O and radius r; A point A inside the circle (A ≠ O)
To find: Show that the chord through A perpendicular to OA is the shortest chord through A

Idea: Shorter chords are farther from the centre. Among all chords through A, the one farthest from O is the one whose foot of perpendicular is A itself.

Let RS be any chord through A, and OM ⟂ RS. Its length is RS = 2√(r² − OM²) (the perpendicular from the centre bisects the chord).1 mark
If M ≠ A, ΔOMA is right-angled at M with hypotenuse OA, so OM < OA. If M = A (the chord is perpendicular to OA), OM = OA. So OM ≤ OA always.1 mark
A larger distance gives a shorter chord, so RS ≥ 2√(r² − OA²) = PQ, where PQ is the chord through A perpendicular to OA. Hence PQ is the shortest chord through A.1 mark
(If A is the centre itself, every chord through A is a diameter and all have the same length.)

For any chord through A, its distance from O is at most OA, with equality only for the chord perpendicular to OA. Since a chord farther from the centre is shorter, the chord through A perpendicular to OA is the shortest.

Answer to write in the exam

Let PQ ⟂ OA be the chord through A; let RS be any other chord through A, OM ⟂ RS.

OM < OA (OA is the hypotenuse of right ΔOMA)

RS = 2√(r² − OM²), PQ = 2√(r² − OA²) (perpendicular from the centre bisects the chord)

OM < OA ⇒ r² − OM² > r² − OA² ⇒ RS > PQ

∴ The chord through A perpendicular to OA is the shortest.

Common mistakes that cost marks

Thinking the shortest chord through A is the one through the centre: that is the longest (a diameter).
Not explaining why OM ≤ OA: the hypotenuse is the longest side of a right triangle.
Mixing up “farther from the centre” and “longer”: farther chords are shorter.

How this can come in the exam

MCQ (1 mark)

A point A is 35 cm from the centre of a circle of radius 37 cm. The length of the shortest chord through A is

12 cm
24 cm
72 cm
74 cm

Show answer

(B) 24 cm
The shortest chord is perpendicular to OA: 2√(37² − 35²) = 2√144 = 24 cm.

Try one yourself

A point P is 4 cm from the centre of a circle of radius 8.5 cm. Find the lengths of the shortest and the longest chords through P.