A Line from the Centre That Halves a Chord Must Be Perpendicular
Prove that a line from the centre bisecting a chord is perpendicular to it, find chord lengths and distances with the Pythagoras theorem, show equal chords are equidistant, and handle two parallel chords.
Why must a line from the centre that halves a chord meet it at a right angle?
Draw a circle, draw any chord that is not a diameter, mark its mid-point, and join that mid-point to the centre. The join always comes out perpendicular to the chord — and the reason is an isosceles triangle.
The proof. Let be the centre, a chord, and the mid-point of .
In and :
- (radii of the same circle)
- (given, is the mid-point)
- (common)
So by SSS, and by CPCTC . These two angles lie on the straight line , so they add to :
**Notice that was isosceles all along**, because any two radii are equal. So is the median to the base of an isosceles triangle, and from the isosceles chapter you already know that the median to the base is also the perpendicular. Every chord property in this chapter is an isosceles-triangle property wearing a circle around it.
The converse is equally useful and proved the same way: the perpendicular from the centre to a chord bisects it, this time by RHS.
Why the diameter is excluded. For a diameter the centre is the mid-point, so there is no separate line to draw and nothing to prove. The theorem is about chords that miss the centre.
This page covers the first part of the ICSE Class 9 Mathematics chapter on the circle: the perpendicular-from-the-centre property, chord and distance calculations, equal chords, and pairs of parallel chords.
The proof. Let be the centre, a chord, and the mid-point of .
In and :
- (radii of the same circle)
- (given, is the mid-point)
- (common)
So by SSS, and by CPCTC . These two angles lie on the straight line , so they add to :
**Notice that was isosceles all along**, because any two radii are equal. So is the median to the base of an isosceles triangle, and from the isosceles chapter you already know that the median to the base is also the perpendicular. Every chord property in this chapter is an isosceles-triangle property wearing a circle around it.
The converse is equally useful and proved the same way: the perpendicular from the centre to a chord bisects it, this time by RHS.
Why the diameter is excluded. For a diameter the centre is the mid-point, so there is no separate line to draw and nothing to prove. The theorem is about chords that miss the centre.
This page covers the first part of the ICSE Class 9 Mathematics chapter on the circle: the perpendicular-from-the-centre property, chord and distance calculations, equal chords, and pairs of parallel chords.
How do you find a chord's length or its distance from the centre?
Drop the perpendicular from the centre to the chord and you get a right-angled triangle whose hypotenuse is the radius and whose legs are the distance and half the chord.
If is the radius, the perpendicular distance from the centre, and the chord,
Three quantities, one equation — so any two give you the third.
Worked example 1 — chord known, find the distance. A chord of cm is drawn in a circle of radius cm. How far is it from the centre?
Half the chord is cm, so
Worked example 2 — distance known, find the chord. A chord of a circle of radius cm is cm from the centre. Find its length.
The step everyone forgets is the last one. The Pythagoras theorem gives you half the chord. Writing cm as the answer is the single most common error in this chapter.
Worked example 3 — find the radius. A chord of cm lies cm from the centre. Find the radius.
Worked example 4 — watching the pattern. In a circle of radius cm, find the chords at distances cm, cm and cm from the centre.
- : cm
- : cm
- : cm
So the further a chord is from the centre, the shorter it is — and at it is the diameter, cm. The diameter is the longest chord of a circle, and this calculation is the proof rather than an assertion.
One boundary case. A distance of cm gives : the line touches the circle at a single point instead of cutting it. That line is a tangent, and it is where the Class 10 circle chapter begins.
If is the radius, the perpendicular distance from the centre, and the chord,
Three quantities, one equation — so any two give you the third.
Worked example 1 — chord known, find the distance. A chord of cm is drawn in a circle of radius cm. How far is it from the centre?
Half the chord is cm, so
Worked example 2 — distance known, find the chord. A chord of a circle of radius cm is cm from the centre. Find its length.
The step everyone forgets is the last one. The Pythagoras theorem gives you half the chord. Writing cm as the answer is the single most common error in this chapter.
Worked example 3 — find the radius. A chord of cm lies cm from the centre. Find the radius.
Worked example 4 — watching the pattern. In a circle of radius cm, find the chords at distances cm, cm and cm from the centre.
- : cm
- : cm
- : cm
So the further a chord is from the centre, the shorter it is — and at it is the diameter, cm. The diameter is the longest chord of a circle, and this calculation is the proof rather than an assertion.
One boundary case. A distance of cm gives : the line touches the circle at a single point instead of cutting it. That line is a tangent, and it is where the Class 10 circle chapter begins.
Why are equal chords the same distance from the centre?
Because the two right-angled triangles formed by the perpendiculars are congruent by RHS.
The proof. Let and be equal chords of a circle with centre , with and .
Since the perpendicular from the centre bisects a chord,
and because , these halves are equal: .
Now in and :
- (by construction)
- (radii, and these are the hypotenuses)
- (just shown)
So by RHS, giving by CPCTC.
The converse is true and proved from the same figure: if two chords are equidistant from the centre, they are equal. Start from and the same RHS congruence delivers , hence .
Worked example 1. In a circle of radius cm, two equal chords of cm are drawn. Find the distance of each from the centre.
Both are cm from the centre, as the theorem requires.
Worked example 2 — using the converse. Two chords of a circle of radius cm are each cm from the centre. Show they are equal and find their length.
Equal distances mean equal chords, and
Worked example 3 — a rider. and are two equal chords of a circle with centre . Prove that bisects .
In and : (radii), (given) and (common). So by SSS, and by CPCTC, which is what was required.
The link back to the equal-distance theorem makes this rider shorter still: since , they are equidistant from , so lies on the bisector of the angle between them — because the set of points equidistant from two lines is the bisector of their angle. That sentence is the locus idea again, and it returns in coordinate geometry.
The proof. Let and be equal chords of a circle with centre , with and .
Since the perpendicular from the centre bisects a chord,
and because , these halves are equal: .
Now in and :
- (by construction)
- (radii, and these are the hypotenuses)
- (just shown)
So by RHS, giving by CPCTC.
The converse is true and proved from the same figure: if two chords are equidistant from the centre, they are equal. Start from and the same RHS congruence delivers , hence .
Worked example 1. In a circle of radius cm, two equal chords of cm are drawn. Find the distance of each from the centre.
Both are cm from the centre, as the theorem requires.
Worked example 2 — using the converse. Two chords of a circle of radius cm are each cm from the centre. Show they are equal and find their length.
Equal distances mean equal chords, and
Worked example 3 — a rider. and are two equal chords of a circle with centre . Prove that bisects .
In and : (radii), (given) and (common). So by SSS, and by CPCTC, which is what was required.
The link back to the equal-distance theorem makes this rider shorter still: since , they are equidistant from , so lies on the bisector of the angle between them — because the set of points equidistant from two lines is the bisector of their angle. That sentence is the locus idea again, and it returns in coordinate geometry.
How do you handle two parallel chords in the same circle?
Find each chord's distance from the centre separately, then add the two distances if the chords are on opposite sides and subtract if they are on the same side.
That one sentence contains the only difficulty in this section: a question with two parallel chords has two answers unless it tells you which side of the centre each one lies.
Worked example 1. A circle of radius cm has two parallel chords of cm and cm. Find the distance between them.
First, each distance from the centre:
- On the same side of the centre: cm
- On opposite sides: cm
Both are valid answers, and a complete solution gives both unless the question rules one out.
Worked example 2. In a circle of radius cm, two parallel chords measure cm and cm. Find the distance between them in each case.
Same side: cm. Opposite sides: cm.
Worked example 3 — finding the radius instead. Two parallel chords of a circle are cm and cm, and they lie on opposite sides of the centre, cm apart. Find the radius.
Let the distance from the centre to the cm chord be , so the distance to the other is . Both must give the same radius:
Setting them equal:
Then , so cm.
Check: the distances are cm and cm, which add to cm, and both give a radius of cm, as required.
Worked example 4 — equal parallel chords. If two parallel chords of a circle are equal, where is the centre?
Equal chords are equidistant from the centre, and they are on opposite sides of it, so the centre lies exactly midway between them — on the common perpendicular bisector of both. That perpendicular is a diameter, and it is the fastest way to locate a centre from two parallel equal chords.
The habit that makes all four examples routine. Draw the radius to one end of each chord and the perpendicular from the centre, and label three things every time: the radius, the distance, and half the chord. Every question in this chapter is that one right-angled triangle, used twice.
That one sentence contains the only difficulty in this section: a question with two parallel chords has two answers unless it tells you which side of the centre each one lies.
Worked example 1. A circle of radius cm has two parallel chords of cm and cm. Find the distance between them.
First, each distance from the centre:
- On the same side of the centre: cm
- On opposite sides: cm
Both are valid answers, and a complete solution gives both unless the question rules one out.
Worked example 2. In a circle of radius cm, two parallel chords measure cm and cm. Find the distance between them in each case.
Same side: cm. Opposite sides: cm.
Worked example 3 — finding the radius instead. Two parallel chords of a circle are cm and cm, and they lie on opposite sides of the centre, cm apart. Find the radius.
Let the distance from the centre to the cm chord be , so the distance to the other is . Both must give the same radius:
Setting them equal:
Then , so cm.
Check: the distances are cm and cm, which add to cm, and both give a radius of cm, as required.
Worked example 4 — equal parallel chords. If two parallel chords of a circle are equal, where is the centre?
Equal chords are equidistant from the centre, and they are on opposite sides of it, so the centre lies exactly midway between them — on the common perpendicular bisector of both. That perpendicular is a diameter, and it is the fastest way to locate a centre from two parallel equal chords.
The habit that makes all four examples routine. Draw the radius to one end of each chord and the perpendicular from the centre, and label three things every time: the radius, the distance, and half the chord. Every question in this chapter is that one right-angled triangle, used twice.
Exam tip
What is the safest layout for a chord calculation?
**Draw the perpendicular from the centre, mark the right angle, and write half the chord as its own labelled step. The arithmetic here is easy; the marks are lost in the labelling.
- State the construction**: *draw ; then (perpendicular from the centre bisects the chord).* That reason is worth a mark on its own
- Write the three quantities before substituting: radius, distance, half-chord. Knowing which two you have tells you what to rearrange
- Double at the end. If the Pythagoras step produced cm, the chord is cm. Underline the doubling so you cannot forget it
- For two parallel chords, give both answers — same side and opposite sides — unless the question fixes the arrangement. A single answer is half an answer
- Keep the radius as the hypotenuse always. It is the longest of the three, so if your distance comes out larger than the radius you have mis-assigned them
- **Quote equal chords are equidistant from the centre by name rather than re-proving it, once it has been established
- Carry units and check plausibility: a chord can never exceed the diameter
The misconception to name. The distance from the centre to a chord means the perpendicular** distance, not the distance to an endpoint — that is the radius. Reading *the chord is cm from the centre as the endpoint is cm away makes every following step wrong. *The word distance in geometry always means the perpendicular one**, as the inequalities chapter proved.
- State the construction**: *draw ; then (perpendicular from the centre bisects the chord).* That reason is worth a mark on its own
- Write the three quantities before substituting: radius, distance, half-chord. Knowing which two you have tells you what to rearrange
- Double at the end. If the Pythagoras step produced cm, the chord is cm. Underline the doubling so you cannot forget it
- For two parallel chords, give both answers — same side and opposite sides — unless the question fixes the arrangement. A single answer is half an answer
- Keep the radius as the hypotenuse always. It is the longest of the three, so if your distance comes out larger than the radius you have mis-assigned them
- **Quote equal chords are equidistant from the centre by name rather than re-proving it, once it has been established
- Carry units and check plausibility: a chord can never exceed the diameter
The misconception to name. The distance from the centre to a chord means the perpendicular** distance, not the distance to an endpoint — that is the radius. Reading *the chord is cm from the centre as the endpoint is cm away makes every following step wrong. *The word distance in geometry always means the perpendicular one**, as the inequalities chapter proved.
Did you know
How can you find the centre of a broken circular plate?
A carpenter needs the centre of a circular tabletop with no centre mark, or you have a piece of a round tawa and want to know the radius of the whole thing. The chord theorems give both, with only a ruler.
To find the centre: draw any two chords across the circle and construct the perpendicular bisector of each. Because the perpendicular from the centre bisects a chord, the centre must lie on both of those bisectors — so it is the point where they cross. Two chords are enough, and any two will do.
To find the radius from a fragment, you do not even need the centre. Lay a straight edge across the broken arc to make a chord, measure the chord and the greatest depth of the arc below it. That depth is called the sagitta, and with a chord and a depth ,
Worked example. A fragment gives a chord of cm with a depth of cm. Then
Check it against the chord theorem: with cm and half the chord cm, the centre is cm from the chord, so the depth from the chord out to the arc is cm, exactly as measured.
And that formula is not a separate fact — it comes straight out of the right-angled triangle you have been using. Since the distance from the centre to the chord is ,
and expanding and cancelling leaves the result. One triangle, rearranged for a different unknown, which is how most practical formulas are made.
The same measurement is how the curve of an arch, a railway bend or a bicycle rim is checked on site: measure a chord, measure the sag, and the radius of a circle you cannot see falls out in one line.
To find the centre: draw any two chords across the circle and construct the perpendicular bisector of each. Because the perpendicular from the centre bisects a chord, the centre must lie on both of those bisectors — so it is the point where they cross. Two chords are enough, and any two will do.
To find the radius from a fragment, you do not even need the centre. Lay a straight edge across the broken arc to make a chord, measure the chord and the greatest depth of the arc below it. That depth is called the sagitta, and with a chord and a depth ,
Worked example. A fragment gives a chord of cm with a depth of cm. Then
Check it against the chord theorem: with cm and half the chord cm, the centre is cm from the chord, so the depth from the chord out to the arc is cm, exactly as measured.
And that formula is not a separate fact — it comes straight out of the right-angled triangle you have been using. Since the distance from the centre to the chord is ,
and expanding and cancelling leaves the result. One triangle, rearranged for a different unknown, which is how most practical formulas are made.
The same measurement is how the curve of an arch, a railway bend or a bicycle rim is checked on site: measure a chord, measure the sag, and the radius of a circle you cannot see falls out in one line.
Exam relevance
How are chord properties tested in JEE-level questions?
This is foundation work that turns into one of the most heavily used formulas in coordinate geometry.
Where it leads. The relation becomes the length of a chord formula in Class 11 Circles:
where is now computed by the distance-from-a-point-to-a-line formula. JEE Main asks this constantly — the length of the chord a given line cuts on a given circle, or the condition for a line to be a tangent, which is the boundary case that you met above as the chord shrinking to zero. Class 10 adds the tangent and cyclic-quadrilateral theorems on top of these.
Where the equal-chord result leads. Equal chords are equidistant from the centre is used to prove that the perpendicular bisector of a chord passes through the centre, which becomes the standard method for finding the centre and radius of a circle through given points — and that is the Class 11 version of the next part of this chapter.
Question types to expect. At this level: find a chord, a distance or a radius, and the two-parallel-chord problem. In competitive papers: chord length cut by a line, tangency conditions, and the shortest chord through a given point inside a circle — which is the one perpendicular to the line joining that point to the centre, since a greater distance means a shorter chord.
That last result comes free from worked example 4 above. The chord shortens as the distance grows, so the chord through an interior point is shortest when that point is the foot of the perpendicular, that is, when the chord is perpendicular to the radius through the point. JEE Main sets exactly this question, and the geometry you have just done answers it without any calculation.
The single trap that costs marks. Using the whole chord instead of half in the Pythagoras step, and — in the parallel-chord problems — giving only one of the two possible separations. Both survive into the coordinate version, where the second appears as a quadratic with two admissible roots.
Board versus competitive emphasis. ICSE marks the construction line and the quoted reason; a competitive paper marks the number. **The transferable sentence is further from the centre means shorter chord** — it answers several JEE questions on sight.
Where it leads. The relation becomes the length of a chord formula in Class 11 Circles:
where is now computed by the distance-from-a-point-to-a-line formula. JEE Main asks this constantly — the length of the chord a given line cuts on a given circle, or the condition for a line to be a tangent, which is the boundary case that you met above as the chord shrinking to zero. Class 10 adds the tangent and cyclic-quadrilateral theorems on top of these.
Where the equal-chord result leads. Equal chords are equidistant from the centre is used to prove that the perpendicular bisector of a chord passes through the centre, which becomes the standard method for finding the centre and radius of a circle through given points — and that is the Class 11 version of the next part of this chapter.
Question types to expect. At this level: find a chord, a distance or a radius, and the two-parallel-chord problem. In competitive papers: chord length cut by a line, tangency conditions, and the shortest chord through a given point inside a circle — which is the one perpendicular to the line joining that point to the centre, since a greater distance means a shorter chord.
That last result comes free from worked example 4 above. The chord shortens as the distance grows, so the chord through an interior point is shortest when that point is the foot of the perpendicular, that is, when the chord is perpendicular to the radius through the point. JEE Main sets exactly this question, and the geometry you have just done answers it without any calculation.
The single trap that costs marks. Using the whole chord instead of half in the Pythagoras step, and — in the parallel-chord problems — giving only one of the two possible separations. Both survive into the coordinate version, where the second appears as a quadratic with two admissible roots.
Board versus competitive emphasis. ICSE marks the construction line and the quoted reason; a competitive paper marks the number. **The transferable sentence is further from the centre means shorter chord** — it answers several JEE questions on sight.
Key takeaways
What should you know about chords before Part 2?
One right-angled triangle does the whole of this chapter, and two congruences prove it.
- A line from the centre bisecting a chord is perpendicular to it (SSS), and the perpendicular from the centre bisects the chord (RHS)
- Both are isosceles-triangle results, since any two radii are equal
- The working relation: — radius as hypotenuse, distance and half the chord as legs
- Double the half-chord at the end of every calculation
- Greater distance from the centre means a shorter chord, so the diameter is the longest chord and gives a tangent
- Equal chords are equidistant from the centre, and the converse holds — both by RHS
- Two parallel chords: subtract the distances if they are on the same side, add them if on opposite sides, and give both answers when the question does not say
- Two chords and their perpendicular bisectors locate the centre of any circle
The fastest self-test is the parallel-chord problem run backwards. Two parallel chords of cm and cm lie cm apart — find the radius from scratch, and check that your two distances add back to cm.
- A line from the centre bisecting a chord is perpendicular to it (SSS), and the perpendicular from the centre bisects the chord (RHS)
- Both are isosceles-triangle results, since any two radii are equal
- The working relation: — radius as hypotenuse, distance and half the chord as legs
- Double the half-chord at the end of every calculation
- Greater distance from the centre means a shorter chord, so the diameter is the longest chord and gives a tangent
- Equal chords are equidistant from the centre, and the converse holds — both by RHS
- Two parallel chords: subtract the distances if they are on the same side, add them if on opposite sides, and give both answers when the question does not say
- Two chords and their perpendicular bisectors locate the centre of any circle
The fastest self-test is the parallel-chord problem run backwards. Two parallel chords of cm and cm lie cm apart — find the radius from scratch, and check that your two distances add back to cm.