Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Step-by-step solution
To find: The shape formed by their midpoints
Idea: The segment from the centre to the midpoint of a chord is perpendicular to it, so its length is the distance of the chord, which is the same for all equal chords. Points at a fixed distance from O form a circle.
- Let AB be any chord of length l and M its midpoint. OM ⟂ AB (line from the centre to the midpoint of a chord), so in right ΔOMA: OM2 = OA2 − AM2 = r2 − (l/2)2.1 mark
- So OM = d = √(r2 − l2/4), the same for every chord of length l. Every midpoint lies on the circle with centre O and radius d.1 mark
- Conversely, take any point M on that circle and draw the chord through M perpendicular to OM. Its half-length is √(r2 − d2) = l/2, so it is a chord of length l with midpoint M. So the midpoints fill the whole concentric circle of radius d.1 mark
Check: r = 10 cm, l = 12 cm: each midpoint is √(100 − 36) = 8 cm from the centre; the midpoints form a circle of radius 8 cm.
Answer to write in the exam
Let AB be a chord of length l, M its midpoint, O the centre, r the radius.
OM ⟂ AB (line from the centre to the midpoint of a chord)
OM2 = OA2 − AM2 = r2 − l2/4 (Baudhāyana–Pythagoras theorem)
OM = √(r2 − l2/4), the same for all such chords
Every point at this distance from O is the midpoint of such a chord.
∴ The midpoints form a circle with centre O and radius √(r2 − l2/4).
Common mistakes that cost marks
- Answering “a line” or “a chord”: equal chords point in all directions around the centre.
- Giving the radius of the new circle as r − l/2 instead of √(r2 − l2/4).
- Forgetting the special case of diameters, whose midpoints are all the centre.
How this can come in the exam
In a circle of radius 26 cm, the midpoints of all chords of length 20 cm lie on a circle of radius
- 6 cm
- 16 cm
- 24 cm
- 10 cm
Show answer
(C) 24 cm
√(262 − 102) = √576 = 24 cm.
Try one yourself
A circle has radius 65 cm. Describe exactly the set of midpoints of all its chords of length 66 cm.
Show answer
A circle with the same centre and radius √(652 − 332) = √3136 = 56 cm.
More questions like this
- In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.
- Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
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- A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
- Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).