Perpendicular from Centre to Chord
One of the fundamental properties of a circle is: the perpendicular drawn from the centre of a circle to a chord bisects the chord. This is a key theorem in NCERT Class 9 Circles.
The converse is equally important: a line drawn from the centre to the midpoint of a chord is perpendicular to the chord. Together, these two results establish a powerful relationship between the centre, the perpendicular, and the midpoint of any chord.
This theorem is used to find the distance of a chord from the centre, to prove that equal chords are equidistant from the centre, and in numerous construction problems involving circles.
What is Perpendicular from Centre to Chord?
Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.
If OM ⊥ AB, then AM = MB
Converse: The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.
If AM = MB, then OM ⊥ AB
Where:
- O = centre of the circle
- AB = chord of the circle
- M = foot of the perpendicular (or midpoint of chord)
Important:
- The theorem applies to all chords except the diameter (the diameter passes through the centre).
- The perpendicular from the centre to a chord is the shortest distance from the centre to the chord.
- This distance is called the distance of the chord from the centre.
Perpendicular from Centre to Chord Formula
Key Formulas and Results:
1. Perpendicular bisects the chord:
OM ⊥ AB ⇒ AM = MB = AB/2
2. Finding the distance from centre to chord:
In right triangle OMA (using Pythagoras theorem):
OM² + AM² = OA²
Where OA = radius (r), AM = half the chord length, OM = distance from centre.
- OM = √(r² − (AB/2)²)
- AB = 2√(r² − OM²)
3. Equal chords are equidistant from the centre:
- If AB = CD, then the perpendicular distances from O to AB and CD are equal.
4. Converse of equidistant chords:
- Chords equidistant from the centre are equal in length.
Derivation and Proof
Proof: Perpendicular from the centre bisects the chord
Given: Circle with centre O. OM ⊥ AB, where M is the foot of the perpendicular on chord AB.
To prove: AM = MB
Construction: Join OA and OB.
Proof:
- In ▵OMA and ▵OMB:
- OA = OB (Radii of the same circle) …(i)
- ∠OMA = ∠OMB = 90° (OM ⊥ AB) …(ii)
- OM = OM (Common side) …(iii)
- By RHS congruence: ▵OMA ≅ ▵OMB
- By CPCT: AM = MB ■
Proof of Converse: Line from centre to midpoint is perpendicular
Given: Circle with centre O. M is the midpoint of chord AB (AM = MB).
To prove: OM ⊥ AB
Proof:
- Join OA and OB.
- In ▵OMA and ▵OMB:
- OA = OB (Radii) …(i)
- AM = MB (Given, M is midpoint) …(ii)
- OM = OM (Common) …(iii)
- By SSS congruence: ▵OMA ≅ ▵OMB
- By CPCT: ∠OMA = ∠OMB
- But ∠OMA + ∠OMB = 180° (linear pair)
- Therefore: ∠OMA = ∠OMB = 90°
- Hence OM ⊥ AB ■
Proof: Equal chords are equidistant from the centre
Given: AB = CD; OM ⊥ AB at M; ON ⊥ CD at N.
To prove: OM = ON
Proof:
- AM = AB/2 and CN = CD/2 (perpendicular from centre bisects chord)
- AB = CD (Given), so AM = CN …(i)
- OA = OC (Radii) …(ii)
- In right ▵OMA and ▵ONC:
- OA² = OM² + AM² and OC² = ON² + CN²
- Since OA = OC and AM = CN: OM² = ON²
- Therefore OM = ON ■
Types and Properties
Related Theorems and Their Applications:
1. Perpendicular bisector of a chord passes through the centre
- If you draw the perpendicular bisector of any chord, it always passes through the centre.
- This is used to find the centre of a circle: draw perpendicular bisectors of two chords; their intersection is the centre.
2. Equal chords — Equal distances
- Equal chords of a circle (or congruent circles) are equidistant from the centre.
- Converse: chords equidistant from the centre are equal.
3. Longer chord is closer to the centre
- Among two chords, the one closer to the centre is longer.
- The diameter is the closest chord to the centre (distance = 0) and is the longest chord.
4. Distance and chord length relationship
- As the distance from centre increases, chord length decreases.
- At distance 0: chord = diameter (maximum).
- At distance = radius: chord has zero length (point on circle).
5. Common chord of two circles
- The line joining the centres of two intersecting circles is perpendicular to their common chord and bisects it.
Solved Examples
Example 1: Example 1: Finding half the chord length
Problem: A circle has radius 13 cm. The perpendicular distance from the centre to a chord is 5 cm. Find the length of the chord.
Solution:
Given:
- Radius (OA) = 13 cm
- Distance from centre (OM) = 5 cm
Using Pythagoras theorem in ▵OMA:
- OA² = OM² + AM²
- 13² = 5² + AM²
- 169 = 25 + AM²
- AM² = 144
- AM = 12 cm
Chord length:
- AB = 2 × AM = 2 × 12 = 24 cm
Answer: The chord is 24 cm long.
Example 2: Example 2: Finding distance from centre
Problem: A chord of length 30 cm is drawn in a circle of radius 17 cm. Find the distance of the chord from the centre.
Solution:
Given:
- AB = 30 cm, so AM = 15 cm
- Radius (OA) = 17 cm
Using Pythagoras theorem:
- OM² = OA² − AM²
- OM² = 17² − 15²
- OM² = 289 − 225 = 64
- OM = 8 cm
Answer: The distance from the centre is 8 cm.
Example 3: Example 3: Finding radius
Problem: A chord of length 48 cm is at a distance of 7 cm from the centre. Find the radius of the circle.
Solution:
Given:
- AB = 48 cm, so AM = 24 cm
- OM = 7 cm
Using Pythagoras theorem:
- OA² = OM² + AM²
- OA² = 7² + 24²
- OA² = 49 + 576 = 625
- OA = 25 cm
Answer: The radius is 25 cm.
Example 4: Example 4: Equal chords equidistant from centre
Problem: Two chords AB and CD in a circle of radius 10 cm are each 16 cm long. Find the distance of each chord from the centre.
Solution:
Given: AB = CD = 16 cm, r = 10 cm
For chord AB:
- AM = 16/2 = 8 cm
- OM² = 10² − 8² = 100 − 64 = 36
- OM = 6 cm
Since AB = CD (equal chords), by the theorem, they are equidistant from the centre.
Answer: Each chord is 6 cm from the centre.
Example 5: Example 5: Chords at equal distances
Problem: Two chords in a circle of radius 15 cm are each at a distance of 9 cm from the centre. Prove they are equal and find their lengths.
Solution:
Given: r = 15 cm, OM = ON = 9 cm
By the converse theorem: Chords equidistant from the centre are equal.
Finding length:
- AM² = OA² − OM² = 15² − 9² = 225 − 81 = 144
- AM = 12 cm
- AB = 2 × 12 = 24 cm
Answer: Both chords are equal in length = 24 cm.
Example 6: Example 6: Two parallel chords on same side
Problem: In a circle of radius 25 cm, two parallel chords of lengths 48 cm and 14 cm are on the same side of the centre. Find the distance between them.
Solution:
For chord AB = 48 cm:
- AM = 24 cm
- OM² = 25² − 24² = 625 − 576 = 49
- OM = 7 cm
For chord CD = 14 cm:
- CN = 7 cm
- ON² = 25² − 7² = 625 − 49 = 576
- ON = 24 cm
Distance between chords (same side of centre):
- Distance = ON − OM = 24 − 7 = 17 cm
Answer: The distance between the chords is 17 cm.
Example 7: Example 7: Two parallel chords on opposite sides
Problem: Two parallel chords of lengths 30 cm and 16 cm are on opposite sides of the centre in a circle of radius 17 cm. Find the distance between them.
Solution:
For chord AB = 30 cm:
- AM = 15 cm
- OM² = 17² − 15² = 289 − 225 = 64
- OM = 8 cm
For chord CD = 16 cm:
- CN = 8 cm
- ON² = 17² − 8² = 289 − 64 = 225
- ON = 15 cm
Distance between chords (opposite sides of centre):
- Distance = OM + ON = 8 + 15 = 23 cm
Answer: The distance between the chords is 23 cm.
Example 8: Example 8: Finding the centre using perpendicular bisectors
Problem: Two chords AB and CD of a circle are not parallel. Describe how to find the centre of the circle.
Solution:
Method:
- Draw the perpendicular bisector of chord AB.
- Draw the perpendicular bisector of chord CD.
- Both perpendicular bisectors pass through the centre (by the converse theorem).
- The point of intersection of the two perpendicular bisectors is the centre O.
Why this works:
- The perpendicular bisector of any chord passes through the centre.
- Two non-parallel lines intersect at exactly one point.
- That point must be the centre.
Example 9: Example 9: Comparing two chords
Problem: In a circle of radius 20 cm, chord AB is 12 cm from the centre and chord CD is 16 cm from the centre. Which chord is longer?
Solution:
For chord AB (distance = 12 cm):
- AM² = 20² − 12² = 400 − 144 = 256
- AM = 16 cm; AB = 32 cm
For chord CD (distance = 16 cm):
- CN² = 20² − 16² = 400 − 256 = 144
- CN = 12 cm; CD = 24 cm
AB = 32 cm > CD = 24 cm
The chord closer to the centre (AB, at 12 cm) is longer.
Answer: Chord AB is longer (32 cm vs 24 cm).
Example 10: Example 10: Algebraic problem
Problem: A chord is at a distance of x cm from the centre of a circle of radius 2x cm. Find the length of the chord in terms of x.
Solution:
Given:
- Radius (OA) = 2x cm
- Distance from centre (OM) = x cm
Using Pythagoras theorem:
- AM² = (2x)² − x² = 4x² − x² = 3x²
- AM = x√3 cm
- AB = 2 × x√3 = 2x√3 cm
Answer: The chord length is 2x√3 cm.
Real-World Applications
Applications of the Perpendicular from Centre to Chord:
- Finding the centre of a circle: Draw two chords and construct their perpendicular bisectors. They intersect at the centre.
- Engineering and manufacturing: In machining circular parts, perpendicular measurements from the centre to chord surfaces ensure concentricity and accurate tolerances.
- Tunnel and bridge design: Cross-sections of tunnels are often circular arcs. The distance from the centre to the roadway (a chord) determines the clearance height.
- Archaeology: Archaeologists reconstruct complete circles from arc fragments by using the perpendicular bisector method to locate the original centre.
- Pipe and tank measurement: To find the diameter of a cylindrical tank, measure a chord and its distance from the centre, then apply the Pythagorean relationship.
- Lens optics: The distance from the optical centre to a chord of the lens surface determines the curvature and focal properties.
Key Points to Remember
- The perpendicular from the centre to a chord bisects the chord.
- Converse: The line from the centre to the midpoint of a chord is perpendicular to the chord.
- The perpendicular bisector of any chord passes through the centre.
- The perpendicular distance from the centre to a chord is found using Pythagoras theorem: OM = √(r² − (AB/2)²).
- Equal chords are equidistant from the centre.
- Equidistant chords are equal in length (converse).
- A longer chord is closer to the centre than a shorter chord.
- The diameter passes through the centre (distance = 0) and is the longest chord.
- To find the centre of a circle, draw perpendicular bisectors of two non-parallel chords and find their intersection.
- The proof uses RHS congruence (for the theorem) and SSS congruence (for the converse).
Practice Problems
- A chord of length 24 cm is drawn in a circle of radius 13 cm. Find the distance of the chord from the centre.
- The perpendicular distance from the centre of a circle of radius 25 cm to a chord is 7 cm. Find the chord length.
- Two equal chords AB and CD of a circle of radius 10 cm are each 12 cm long. Find their distances from the centre. Verify they are equal.
- Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre in a circle of radius 5 cm. Find the distance between the chords.
- A chord is at a distance equal to half the radius from the centre. Express the chord length in terms of the radius.
- In a circle of radius 20 cm, a chord is 24 cm long. Another chord is 32 cm long. Which chord is closer to the centre? By how much?
- Describe how to find the centre of a circular arc using only a ruler and compass.
- A chord of length 2a is at a distance d from the centre of a circle. Express the radius in terms of a and d.
Frequently Asked Questions
Q1. What does 'perpendicular from the centre to a chord bisects the chord' mean?
If you draw a line from the centre of a circle perpendicular to a chord, it meets the chord at its exact midpoint. The two halves of the chord on either side of this point are equal.
Q2. Is the converse also true?
Yes. If you draw a line from the centre to the midpoint of a chord, that line is perpendicular to the chord. Both the theorem and its converse are proved in NCERT Class 9.
Q3. How do you find the distance of a chord from the centre?
Draw the perpendicular from the centre to the chord. The length of this perpendicular is the distance. Calculate it using Pythagoras theorem: distance = √(radius² − (half chord length)²).
Q4. Are equal chords always equidistant from the centre?
Yes. Equal chords of a circle (or of congruent circles) are equidistant from the centre. The converse is also true: chords equidistant from the centre are equal in length.
Q5. Which chord is longest in a circle?
The diameter is the longest chord. It passes through the centre, so its distance from the centre is 0. All other chords are shorter and have a positive distance from the centre.
Q6. How is this theorem used to find the centre of a circle?
Draw two non-parallel chords. Construct the perpendicular bisector of each chord. These two perpendicular bisectors intersect at the centre of the circle.
Q7. Does this theorem work for a diameter?
A diameter passes through the centre, so the 'perpendicular from the centre' has zero length. The concept applies trivially — the centre itself is on the diameter and bisects it.
Q8. What congruence rule is used in the proof?
The RHS (Right angle-Hypotenuse-Side) congruence rule is used for the main theorem. The SSS congruence rule is used for the converse.
Q9. Can two chords of different lengths be equidistant from the centre?
No. If two chords are equidistant from the centre, they must be equal in length. Different-length chords are at different distances from the centre.
Q10. Is this theorem in the NCERT Class 9 syllabus?
Yes. NCERT Class 9 Chapter 10 (Circles) presents this as Theorem 10.3: 'The perpendicular from the centre of a circle to a chord bisects the chord.' The converse is Theorem 10.4.
Related Topics
- Equal Chords and Equal Angles
- Angle Subtended by a Chord
- Circle Theorems Introduction
- Circle - Basic Concepts
- Angles in the Same Segment
- Angle in a Semicircle
- Cyclic Quadrilateral
- Tangent to a Circle
- Tangent is Perpendicular to Radius
- Number of Tangents from a Point
- Tangents from an External Point
- Tangent-Secant Relationship
- Angle Between Two Tangents
- Important Circle Theorem Problems
