The Longer the Chord, the Closer It Sits to the Centre
Learn why equal chords subtend equal angles at the centre, how the perpendicular from the centre bisects a chord, how to find a chord length from the radius and distance by Pythagoras, and why equal chords are equidistant.
Why does a longer chord sit nearer the centre?
Draw a circle of radius cm and put two chords in it, one cm long and one cm long. Measure how far each sits from the centre.
The cm chord is ** cm** from the centre. The cm chord is ** cm** away. The long one is close in and the short one is out near the edge.
That is not a coincidence but arithmetic, and one right triangle explains it. Drop a perpendicular from the centre to a chord: it bisects the chord, making a right triangle whose hypotenuse is the radius, whose base is half the chord, and whose height is the distance from the centre. So
The radius is fixed, so a bigger half-chord forces a smaller . Check it: gives for the long chord, and gives the same total for the short one, with the roles of and swapped.
At the extreme, the diameter of cm sits cm from the centre — the longest chord passes through it. At the other extreme a chord shrinking to nothing has distance cm, the full radius.
This page covers the second part of the CBSE Class 9 Mathematics chapter on circles — the chord theorems and the calculations they make possible.
The cm chord is ** cm** from the centre. The cm chord is ** cm** away. The long one is close in and the short one is out near the edge.
That is not a coincidence but arithmetic, and one right triangle explains it. Drop a perpendicular from the centre to a chord: it bisects the chord, making a right triangle whose hypotenuse is the radius, whose base is half the chord, and whose height is the distance from the centre. So
The radius is fixed, so a bigger half-chord forces a smaller . Check it: gives for the long chord, and gives the same total for the short one, with the roles of and swapped.
At the extreme, the diameter of cm sits cm from the centre — the longest chord passes through it. At the other extreme a chord shrinking to nothing has distance cm, the full radius.
This page covers the second part of the CBSE Class 9 Mathematics chapter on circles — the chord theorems and the calculations they make possible.
Why do equal chords subtend equal angles at the centre?
Because the two triangles formed with the centre are congruent by SSS, and congruent triangles have equal corresponding angles.
The proof. Let and be equal chords of a circle with centre . Compare triangles and :
- , both radii
- , both radii
- , given
Three pairs of equal sides, so by SSS. Therefore the corresponding angles are equal:
The converse also holds. If , then the same two triangles are congruent by SAS — two radii and the included angle — so . Equal chords and equal central angles imply one another, which is why the result is usually stated as a pair.
Worked example 1. In a circle, chord subtends at the centre and chord also measures the same length as . Then .
Worked example 2 — a chord equal to the radius. If , then has all three sides equal to — it is equilateral, so
So a chord as long as the radius always subtends , in any circle. Check: six such chords laid end to end use up , which is why a regular hexagon inscribed in a circle has side equal to the radius.
Worked example 3 — from the angle to the chord. In a circle of radius cm, a chord subtends at the centre. The triangle is right-angled at with both legs cm:
Check it is less than the diameter of cm, as every chord must be.
Both chords must be in the same circle, or in circles of equal radii. A chord of cm subtends a large angle in a small circle and a small angle in a big one — so the theorem's phrase of a circle is doing real work. In two circles of different sizes, equal chords do not subtend equal angles, and quoting the theorem across circles is a genuine error rather than a technicality.
The proof. Let and be equal chords of a circle with centre . Compare triangles and :
- , both radii
- , both radii
- , given
Three pairs of equal sides, so by SSS. Therefore the corresponding angles are equal:
The converse also holds. If , then the same two triangles are congruent by SAS — two radii and the included angle — so . Equal chords and equal central angles imply one another, which is why the result is usually stated as a pair.
Worked example 1. In a circle, chord subtends at the centre and chord also measures the same length as . Then .
Worked example 2 — a chord equal to the radius. If , then has all three sides equal to — it is equilateral, so
So a chord as long as the radius always subtends , in any circle. Check: six such chords laid end to end use up , which is why a regular hexagon inscribed in a circle has side equal to the radius.
Worked example 3 — from the angle to the chord. In a circle of radius cm, a chord subtends at the centre. The triangle is right-angled at with both legs cm:
Check it is less than the diameter of cm, as every chord must be.
Both chords must be in the same circle, or in circles of equal radii. A chord of cm subtends a large angle in a small circle and a small angle in a big one — so the theorem's phrase of a circle is doing real work. In two circles of different sizes, equal chords do not subtend equal angles, and quoting the theorem across circles is a genuine error rather than a technicality.
What does the perpendicular from the centre do to a chord?
It bisects it — and the converse holds too, so the line from the centre to a chord's midpoint is perpendicular to the chord.
The proof that the perpendicular bisects. Let with on . Compare triangles and :
-
- , both radii (the hypotenuses)
- , common
So by RHS, giving . The foot of the perpendicular is the midpoint.
The proof of the converse. Now let be the midpoint of . Compare the same triangles:
- , radii
- , given
- , common
Congruent by SSS, so . These two angles sit on the straight line and add to , so each is :
Worked example 1 — finding the midpoint's distance. A chord of cm lies in a circle of radius cm. The perpendicular from the centre bisects it into halves of cm, so
Worked example 2 — locating the centre. Given only an arc, draw any two chords and bisect each perpendicularly. Since each perpendicular bisector passes through the centre, the two lines meet at the centre. This is the construction from the first part of this chapter, now with a theorem behind it.
Worked example 3 — a chord bisected by a diameter. If a diameter bisects a chord that is not itself a diameter, the diameter must be perpendicular to it, by the converse. And the reverse: a diameter perpendicular to a chord bisects it.
One case the converse does not settle. If the chord is a diameter, then every diameter through the centre bisects it, and they are not all perpendicular to it. The converse needs the chord to be a non-diameter, because for a diameter the midpoint is the centre itself and the line collapses to a point rather than a direction.
That is a small print condition, but it is the reason careful statements of this theorem say a chord other than a diameter — and it is the sort of boundary case an assertion-reason question is built on.
The proof that the perpendicular bisects. Let with on . Compare triangles and :
-
- , both radii (the hypotenuses)
- , common
So by RHS, giving . The foot of the perpendicular is the midpoint.
The proof of the converse. Now let be the midpoint of . Compare the same triangles:
- , radii
- , given
- , common
Congruent by SSS, so . These two angles sit on the straight line and add to , so each is :
Worked example 1 — finding the midpoint's distance. A chord of cm lies in a circle of radius cm. The perpendicular from the centre bisects it into halves of cm, so
Worked example 2 — locating the centre. Given only an arc, draw any two chords and bisect each perpendicularly. Since each perpendicular bisector passes through the centre, the two lines meet at the centre. This is the construction from the first part of this chapter, now with a theorem behind it.
Worked example 3 — a chord bisected by a diameter. If a diameter bisects a chord that is not itself a diameter, the diameter must be perpendicular to it, by the converse. And the reverse: a diameter perpendicular to a chord bisects it.
One case the converse does not settle. If the chord is a diameter, then every diameter through the centre bisects it, and they are not all perpendicular to it. The converse needs the chord to be a non-diameter, because for a diameter the midpoint is the centre itself and the line collapses to a point rather than a direction.
That is a small print condition, but it is the reason careful statements of this theorem say a chord other than a diameter — and it is the sort of boundary case an assertion-reason question is built on.
Formula
How do you calculate a chord length from the radius?
Use the right triangle made by the radius, half the chord and the distance from the centre:
where is the chord length, its distance from the centre, and the radius. Any two of the three give the third.
**Worked example 1 — find from and .** Radius cm, chord cm. Half the chord is cm:
**Worked example 2 — find from and .** Radius cm, distance cm:
The most common slip in the whole chapter is stopping at and reporting the chord as cm. The Pythagoras step gives the half-chord, and the answer needs doubling.
Worked example 3 — a larger circle. Radius cm, chord cm. Half-chord cm:
**Worked example 4 — find .** A chord of cm lies cm from the centre:
Worked example 5 — the two extremes. In a circle of radius cm:
- gives , so cm — the diameter
- gives — the chord has shrunk to a single point where the circle is touched
So runs from to , and runs from down to , in opposite directions. That is the general form of the observation this page opened with.
A check that catches most errors. The chord must never exceed and must never exceed . If a calculation returns a chord of cm in a circle of radius cm, or asks for , the given data are inconsistent — and noticing that is worth more than producing a number.
The formula holds for both halves because the perpendicular bisects. Every step above depends on the previous section's theorem, which is why the bisection is proved before any arithmetic is attempted rather than assumed from the diagram.
where is the chord length, its distance from the centre, and the radius. Any two of the three give the third.
**Worked example 1 — find from and .** Radius cm, chord cm. Half the chord is cm:
**Worked example 2 — find from and .** Radius cm, distance cm:
The most common slip in the whole chapter is stopping at and reporting the chord as cm. The Pythagoras step gives the half-chord, and the answer needs doubling.
Worked example 3 — a larger circle. Radius cm, chord cm. Half-chord cm:
**Worked example 4 — find .** A chord of cm lies cm from the centre:
Worked example 5 — the two extremes. In a circle of radius cm:
- gives , so cm — the diameter
- gives — the chord has shrunk to a single point where the circle is touched
So runs from to , and runs from down to , in opposite directions. That is the general form of the observation this page opened with.
A check that catches most errors. The chord must never exceed and must never exceed . If a calculation returns a chord of cm in a circle of radius cm, or asks for , the given data are inconsistent — and noticing that is worth more than producing a number.
The formula holds for both halves because the perpendicular bisects. Every step above depends on the previous section's theorem, which is why the bisection is proved before any arithmetic is attempted rather than assumed from the diagram.
Why are equal chords the same distance from the centre?
Because equal chords have equal halves, and the same radius then forces the same distance — the right triangles are congruent.
The proof. Let with perpendicular distances and from the centre. The perpendiculars bisect, so
Now compare triangles and : right angles at and , equal hypotenuses (radii), and . Congruent by RHS, so
The converse holds too. If then the same triangles are congruent by RHS, giving , so . Equal chords and equal distances imply one another.
Worked example 1 — comparing two chords. In a circle of radius cm, chord is cm and chord is cm.
- : cm
- : cm
Unequal chords, unequal distances — and the longer chord is nearer, at cm against cm.
Worked example 2 — two equal chords. If a second chord in that circle also measures cm, it must also be cm from the centre, wherever it lies. There are infinitely many such chords, all tangent to a smaller circle of radius cm drawn about the same centre.
Worked example 3 — parallel chords on the same side. Chords of cm and cm in the radius- circle are parallel and on the same side of the centre. Their distance apart is
Worked example 4 — parallel chords on opposite sides. The same two chords on opposite sides of the centre are
apart.
A question giving two parallel chord lengths has two answers unless it says which side. That is not a trick; it is a genuine ambiguity in the figure, and a complete solution states both cases with their diagrams. Answering only one loses half the marks, and it is the most frequent omission in parallel-chord problems.
The general ordering, stated once. Among the chords of a circle, length and distance from the centre move in opposite directions. The diameter is longest and nearest at ; a chord out at has shrunk to nothing. So "which chord is longer" and "which chord is nearer the centre" are the same question asked two ways, and either fact can be deduced from the other without calculation.
The proof. Let with perpendicular distances and from the centre. The perpendiculars bisect, so
Now compare triangles and : right angles at and , equal hypotenuses (radii), and . Congruent by RHS, so
The converse holds too. If then the same triangles are congruent by RHS, giving , so . Equal chords and equal distances imply one another.
Worked example 1 — comparing two chords. In a circle of radius cm, chord is cm and chord is cm.
- : cm
- : cm
Unequal chords, unequal distances — and the longer chord is nearer, at cm against cm.
Worked example 2 — two equal chords. If a second chord in that circle also measures cm, it must also be cm from the centre, wherever it lies. There are infinitely many such chords, all tangent to a smaller circle of radius cm drawn about the same centre.
Worked example 3 — parallel chords on the same side. Chords of cm and cm in the radius- circle are parallel and on the same side of the centre. Their distance apart is
Worked example 4 — parallel chords on opposite sides. The same two chords on opposite sides of the centre are
apart.
A question giving two parallel chord lengths has two answers unless it says which side. That is not a trick; it is a genuine ambiguity in the figure, and a complete solution states both cases with their diagrams. Answering only one loses half the marks, and it is the most frequent omission in parallel-chord problems.
The general ordering, stated once. Among the chords of a circle, length and distance from the centre move in opposite directions. The diameter is longest and nearest at ; a chord out at has shrunk to nothing. So "which chord is longer" and "which chord is nearer the centre" are the same question asked two ways, and either fact can be deduced from the other without calculation.
Exam tip
Exam tip: double the half-chord, and quote the congruence rule by name
Draw the perpendicular from the centre and mark the right angle. The right triangle is the whole method; without it the Pythagoras step has nothing to stand on.
Pythagoras gives the HALF-chord — double it. Radius , distance gives , and the chord is ** cm. This single step loses more marks than any other in the chapter.
Name the congruence rule. Write SSS, SAS or RHS** explicitly, and list the three equal parts with reasons — radii, given, common. Proofs are marked on the reasons, not the conclusion.
Use RHS for perpendicular proofs and SSS or SAS for the equal-chord ones.
State the standard results as you use them: the perpendicular from the centre bisects the chord; equal chords are equidistant from the centre.
Sanity-check the data: no chord exceeds , and no distance exceeds . A negative quantity under the root means the figures are inconsistent.
For parallel chords, give BOTH cases unless the question fixes the side — same side subtracts (), opposite sides add ().
**A chord equal to the radius subtends , since the triangle is equilateral — worth recognising on sight.
Theorems compare chords within one circle or across circles of equal radii; never quote them between circles of different sizes.
And remember the longer chord is always nearer the centre** — it converts many comparison questions into a one-line answer.
Pythagoras gives the HALF-chord — double it. Radius , distance gives , and the chord is ** cm. This single step loses more marks than any other in the chapter.
Name the congruence rule. Write SSS, SAS or RHS** explicitly, and list the three equal parts with reasons — radii, given, common. Proofs are marked on the reasons, not the conclusion.
Use RHS for perpendicular proofs and SSS or SAS for the equal-chord ones.
State the standard results as you use them: the perpendicular from the centre bisects the chord; equal chords are equidistant from the centre.
Sanity-check the data: no chord exceeds , and no distance exceeds . A negative quantity under the root means the figures are inconsistent.
For parallel chords, give BOTH cases unless the question fixes the side — same side subtracts (), opposite sides add ().
**A chord equal to the radius subtends , since the triangle is equilateral — worth recognising on sight.
Theorems compare chords within one circle or across circles of equal radii; never quote them between circles of different sizes.
And remember the longer chord is always nearer the centre** — it converts many comparison questions into a one-line answer.
Did you know
Why every chord of one length is tangent to a hidden inner circle
Take a circle of radius cm and draw every possible chord of length cm. There are infinitely many, pointing in every direction.
Each one is exactly cm from the centre, because and that calculation does not care about direction. So every one of those chords just touches a circle of radius cm drawn about the same centre — the midpoints of all of them lie on that inner circle, and the chords themselves are its tangents.
Spin a cm rod so that both ends stay on the big circle, and the rod sweeps out a ring while never entering the inner cm disc. The midpoint of the rod traces the inner circle exactly once per turn.
So a chord length is really a statement about a radius. Asking for all chords of cm is the same as asking for all tangents to a circle of radius cm — two descriptions of one family of lines.
The extremes make the picture complete. A chord of cm is the diameter, and its "inner circle" has radius : all the diameters pass through the single centre point rather than around a disc. A chord shrinking towards nothing has its inner circle growing out to the full radius of cm, where a line touching there meets the big circle at one point only.
The useful version of all this for a problem is short. Equal chords are equidistant from the centre, so a question about several chords of the same length is a question about several tangents to one inner circle — and the answer often comes from that picture faster than from the algebra.
Each one is exactly cm from the centre, because and that calculation does not care about direction. So every one of those chords just touches a circle of radius cm drawn about the same centre — the midpoints of all of them lie on that inner circle, and the chords themselves are its tangents.
Spin a cm rod so that both ends stay on the big circle, and the rod sweeps out a ring while never entering the inner cm disc. The midpoint of the rod traces the inner circle exactly once per turn.
So a chord length is really a statement about a radius. Asking for all chords of cm is the same as asking for all tangents to a circle of radius cm — two descriptions of one family of lines.
The extremes make the picture complete. A chord of cm is the diameter, and its "inner circle" has radius : all the diameters pass through the single centre point rather than around a disc. A chord shrinking towards nothing has its inner circle growing out to the full radius of cm, where a line touching there meets the big circle at one point only.
The useful version of all this for a problem is short. Equal chords are equidistant from the centre, so a question about several chords of the same length is a question about several tangents to one inner circle — and the answer often comes from that picture faster than from the algebra.
Exam relevance
How are chord theorems tested in JEE Main?
Because the perpendicular distance from a centre to a line is the recurring calculation of coordinate geometry, and these theorems are its geometric content.
This is the foundation for Class 10 Circles and Class 11 Mathematics Conic Sections and Straight Lines, examined in JEE Main. The relation
reappears there as the length of a chord cut by a line on a circle, with computed by the perpendicular-distance formula from the centre to the line. Every step of the derivation — perpendicular from the centre, bisected chord, Pythagoras — is exactly the working on this page, and the only new tool is a way to compute from an equation instead of reading it off a diagram.
The same relation decides how a line meets a circle, a standard JEE Main question type. Comparing with gives three cases: cuts the circle in two points, touches it as a tangent, and misses it. That trichotomy is the algebraic form of this page's extremes, where ran from to and the chord ran from down to nothing.
The equal-chord theorem becomes the parallel-chord calculation. Two parallel lines cutting equal chords from a circle must be equidistant from the centre, which converts a length condition into a distance equation — and the two-case ambiguity flagged here, same side or opposite sides, survives into those problems unchanged.
Where the congruence work leads. The proofs here are a student's main practice at writing a formal geometric argument with a named congruence rule, and that skill is assessed directly in board papers and in NTSE and olympiad geometry. Competitive papers rarely ask for a proof, but they assume the results without restatement.
What the questions look like. For board work, expect prove a chord theorem with the congruence rule named, find a chord, a radius or a distance by Pythagoras, find the distance between two parallel chords, and locate the centre from two chords by construction. For JEE Main, the direct forms are the chord length cut by a given line, the condition for tangency, and the equation of a chord with a stated midpoint.
How board and competitive emphasis differ. A board paper rewards the written proof — the three equal parts, the reasons, the rule. A competitive paper rewards recognising in one line that the perpendicular distance is what the question is really about, and reaching for the distance formula immediately.
The single trap that costs the most marks. Reporting the half-chord as the chord. Pythagoras returns when the radius is and the distance is , and the chord is — the doubling is easy to skip because is a satisfying answer to the calculation just completed. **The defence is to label the diagram rather than ** before computing, so that the symbol itself reminds you a step remains.
This is the foundation for Class 10 Circles and Class 11 Mathematics Conic Sections and Straight Lines, examined in JEE Main. The relation
reappears there as the length of a chord cut by a line on a circle, with computed by the perpendicular-distance formula from the centre to the line. Every step of the derivation — perpendicular from the centre, bisected chord, Pythagoras — is exactly the working on this page, and the only new tool is a way to compute from an equation instead of reading it off a diagram.
The same relation decides how a line meets a circle, a standard JEE Main question type. Comparing with gives three cases: cuts the circle in two points, touches it as a tangent, and misses it. That trichotomy is the algebraic form of this page's extremes, where ran from to and the chord ran from down to nothing.
The equal-chord theorem becomes the parallel-chord calculation. Two parallel lines cutting equal chords from a circle must be equidistant from the centre, which converts a length condition into a distance equation — and the two-case ambiguity flagged here, same side or opposite sides, survives into those problems unchanged.
Where the congruence work leads. The proofs here are a student's main practice at writing a formal geometric argument with a named congruence rule, and that skill is assessed directly in board papers and in NTSE and olympiad geometry. Competitive papers rarely ask for a proof, but they assume the results without restatement.
What the questions look like. For board work, expect prove a chord theorem with the congruence rule named, find a chord, a radius or a distance by Pythagoras, find the distance between two parallel chords, and locate the centre from two chords by construction. For JEE Main, the direct forms are the chord length cut by a given line, the condition for tangency, and the equation of a chord with a stated midpoint.
How board and competitive emphasis differ. A board paper rewards the written proof — the three equal parts, the reasons, the rule. A competitive paper rewards recognising in one line that the perpendicular distance is what the question is really about, and reaching for the distance formula immediately.
The single trap that costs the most marks. Reporting the half-chord as the chord. Pythagoras returns when the radius is and the distance is , and the chord is — the doubling is easy to skip because is a satisfying answer to the calculation just completed. **The defence is to label the diagram rather than ** before computing, so that the symbol itself reminds you a step remains.
Key takeaways
Chord theorems and chord calculations: quick revision
- Equal chords subtend equal angles at the centre, proved by SSS on the two triangles with the centre; the converse follows by SAS.
- A chord equal to the radius makes an equilateral triangle with the centre, so it subtends **** — six of them fill , giving the inscribed regular hexagon.
- A chord subtending in a radius- circle has length cm, less than the diameter of cm.
- The theorems compare chords within one circle, or between circles of equal radii — never across different sizes.
- The perpendicular from the centre bisects the chord, proved by RHS; the converse — the line to the midpoint is perpendicular — follows by SSS, since the two equal angles on a straight line are each.
- The converse needs the chord to be a non-diameter; for a diameter the midpoint is the centre itself.
- The key relation: .
- Radius , chord : cm. Radius , : half-chord , so **chord cm**.
- Radius , chord : cm. Chord at : cm.
- Pythagoras gives the half-chord — double it. This is the chapter's most common lost mark.
- Extremes: gives the diameter ; gives a chord of length .
- **No chord exceeds and no distance exceeds — a negative under the root means inconsistent data.
- Equal chords are equidistant from the centre, proved by RHS; the converse holds too.
- The longer chord is nearer the centre**: in a radius- circle, cm sits at cm and cm sits at cm.
- **Parallel chords of and cm**: cm apart on the same side, cm on opposite sides.
- Give both cases unless the question fixes the side.
- Every chord of a given length is tangent to an inner circle of radius about the same centre, and its midpoint lies on that circle.
- Two chords, each perpendicularly bisected, locate the centre where the bisectors cross.
Draw a circle with a compass, mark any chord, then predict its distance from the centre by calculation before measuring it — the closer your two numbers, the steadier your construction.
- A chord equal to the radius makes an equilateral triangle with the centre, so it subtends **** — six of them fill , giving the inscribed regular hexagon.
- A chord subtending in a radius- circle has length cm, less than the diameter of cm.
- The theorems compare chords within one circle, or between circles of equal radii — never across different sizes.
- The perpendicular from the centre bisects the chord, proved by RHS; the converse — the line to the midpoint is perpendicular — follows by SSS, since the two equal angles on a straight line are each.
- The converse needs the chord to be a non-diameter; for a diameter the midpoint is the centre itself.
- The key relation: .
- Radius , chord : cm. Radius , : half-chord , so **chord cm**.
- Radius , chord : cm. Chord at : cm.
- Pythagoras gives the half-chord — double it. This is the chapter's most common lost mark.
- Extremes: gives the diameter ; gives a chord of length .
- **No chord exceeds and no distance exceeds — a negative under the root means inconsistent data.
- Equal chords are equidistant from the centre, proved by RHS; the converse holds too.
- The longer chord is nearer the centre**: in a radius- circle, cm sits at cm and cm sits at cm.
- **Parallel chords of and cm**: cm apart on the same side, cm on opposite sides.
- Give both cases unless the question fixes the side.
- Every chord of a given length is tangent to an inner circle of radius about the same centre, and its midpoint lies on that circle.
- Two chords, each perpendicularly bisected, locate the centre where the bisectors cross.
Draw a circle with a compass, mark any chord, then predict its distance from the centre by calculation before measuring it — the closer your two numbers, the steadier your construction.