Circle | ICSE Class 9 Maths Notes

This note covers circle terminology, chord midpoints and perpendiculars, distances of chords from the centre, the circle through three non-collinear points, equal arcs and central angles, equal chords and their corresponding arcs, and applications using triangle congruence and Pythagoras’ theorem.

What are the basic parts of a circle?

Definition: A circle is the set of points in a plane at a fixed positive distance from a fixed point. The fixed point is its centre; the fixed distance is its radius.

A plane is a flat surface extending in every direction. A line segment joins two endpoints along a straight line. A radius also means a segment joining the centre to a point on the circle. All radii of one circle have equal lengths.

A chord joins two points on the circle. A diameter is a chord passing through the centre. It consists of two radii lying along the same straight line. Therefore its length is twice the radius. A chord need not pass through the centre.

How should the notation be read?

Let O denote a centre and A and B denote points on its circle. AB names their joining segment and, in a length equation, its length. The symbol = means equality. The notation ∠AOB denotes the angle between OA and OB, with vertex O.

A vertex is the point where the sides of an angle meet. The symbol ° denotes degrees, the unit used here for angles. A right angle measures 90°. Lines meeting at a right angle are perpendicular; the symbol ⊥ means “is perpendicular to”.

A midpoint divides a segment into two equal lengths. To bisect a segment means to divide it at its midpoint. A perpendicular bisector is a line that does both: it passes through the midpoint and is perpendicular to the segment.

Worked example 1. Can a circle of radius 3 cm pass through points A and B separated by 6 cm? Here cm means centimetres.

Answer: Yes. Choose O as the midpoint of AB. Then OA = OB = 6 ÷ 2 = 3 cm, where ÷ means division. The circle with centre O and radius 3 cm passes through both points, with AB as a diameter.

This construction checks the required distances directly. It does not claim that two given points alone determine a unique circle. The specified radius is additional information. The theorem giving a unique circle uses three points that do not lie on one straight line.

Why does the line from the centre to a chord’s midpoint meet it at right angles?

Theorem: The centre-to-midpoint line is perpendicular to a non-diameter chord

Let AB be a chord of a circle with centre O, and let M be its midpoint. Assume that AB is not a diameter. The statement to prove is OM ⊥ AB. Join OA and OB, forming the triangles OMA and OMB.

A triangle has three straight sides. Congruent triangles have the same shape and size, with equal corresponding sides and angles. Corresponding parts occupy matching positions. The SSS criterion, meaning side-side-side, establishes congruence when three sides of one triangle equal the three corresponding sides of another.

1. OA = OB because both segments are radii of the same circle.
2. AM = BM because M is the midpoint of AB.
3. OM is a common side, so the triangles OMA and OMB are congruent by SSS.
4. The matching angles OMA and OMB are equal. Since A, M and B lie on a straight line, these adjacent angles total 180°.
5. Each angle is therefore 90°, proving that OM is perpendicular to AB.

Adjacent angles share a vertex and a side without overlapping interiors. Here their non-common sides point in opposite directions along AB. Their equality and straight-line sum together establish the right angle; neither fact alone completes the proof.

Why is the diameter exception necessary?

If AB were a diameter, its midpoint would be O itself. There would be no separate segment OM defining a direction. The exclusion “not a diameter” is therefore part of the theorem’s condition and must stay in its statement.

Draw a circle with centre O and a chord AB away from O. Mark M halfway along AB. Join OA, OB and OM. Mark AM and BM equal, then mark the right angle at M established by the proof.

The proof begins with a midpoint and ends with a perpendicular. When solving a problem, check which of these facts is given. A statement and its converse exchange the starting condition and conclusion; the converse requires its own justification or an established theorem.

How does a perpendicular from the centre help calculate chord lengths?

Theorem: A perpendicular from the centre bisects a chord

If a perpendicular from the centre O meets a chord AB at M, then AM = MB. This result is used without proof here. The point where a perpendicular meets a line is called its foot. Thus M is both the foot and the midpoint.

Join OA. The triangle OMA is right-angled at M. Its hypotenuse, the side opposite the right angle, is OA. Pythagoras’ theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides.

The superscript ² means the square of a quantity, obtained by multiplying it by itself. Consequently, OA² = OM² + AM². The sign + means addition. This relation uses half the chord, AM, rather than the complete chord AB.

Worked example 2. In a circle, AD is a diameter of length 34 cm and AB is a chord of length 30 cm. Find the perpendicular distance of AB from the centre O.

Answer: Draw OM perpendicular to AB, with M on AB. The radius OA = 34 ÷ 2 = 17 cm and the half-chord AM = 30 ÷ 2 = 15 cm. Pythagoras gives OM² = 17² − 15² = 289 − 225 = 64. Here − means subtraction. Since OM is a length, OM = 8 cm.

How can a diagram contain two different distances on the same line?

In the next example, C is the foot on the chord, while D is on the circle beyond C. Both points lie along the same perpendicular from O. The distance to the chord is OC. The smaller gap between the chord and the circumference, the circle’s boundary, is CD.

The circle has centre O, chord AB across its lower part and radius OA. The line from O meets AB at C and continues to the circle at D. The accompanying data give OA = 5 cm and AB = 8 cm, with OD perpendicular to AB.

See Fig. 10.3 in your NCERT textbook

Worked example 3. A circle has centre O and radius OA = 5 cm. Its chord AB is 8 cm long. The perpendicular from O meets AB at C and then the circle at D, with C between O and D. Find CD.

Answer: The perpendicular bisects the chord, so AC = 8 ÷ 2 = 4 cm. In the right triangle OCA, OC² = OA² − AC² = 25 − 16 = 9, giving OC = 3 cm. Since OD is a radius, OD = 5 cm. Therefore CD = OD − OC = 5 − 3 = 2 cm.

For either calculation, begin by identifying the radius, the perpendicular and the half-chord. Then identify the requested segment. The final subtraction in the second example is justified by the stated order O, C, D along the line, rather than by their appearance alone.

The units also distinguish lengths from squared lengths. For example, the equation involving squares compares squared lengths; after taking the positive square root, the answer is a length in centimetres. The positive square root is the positive number whose square equals the given value.

Why are equal chords equidistant from the centre?

The distance from a point to a line is the length of the perpendicular drawn from that point to the line. Two chords are equidistant from the centre when these perpendicular lengths are equal. An oblique segment, meeting the chord without a right angle, does not measure this distance.

Theorem: Equal chords have equal perpendicular distances

Let AB and CD be equal chords in a circle with centre O. Draw OM perpendicular to AB at M, and ON perpendicular to CD at N. Join OA and OC. We will compare the right triangles OMA and ONC.

The RHS criterion means right angle-hypotenuse-side. Two right triangles are congruent if their hypotenuses and one pair of corresponding other sides are equal. It is particularly useful when radii form the hypotenuses of triangles drawn to chords.

1. The perpendiculars bisect the chords, so AM is half of AB and CN is half of CD.
2. Since AB = CD, their halves are equal: AM = CN.
3. OA = OC because they are radii of the same circle. Both triangles have right angles at the perpendicular feet.
4. The triangles OMA and ONC are congruent by RHS. Their corresponding sides OM and ON are therefore equal.

This establishes the required distances. The proof does not require the chords to be parallel, meaning that their lines never meet in the plane. Their positions around the centre may differ; equal lengths and a common circle supply the needed conditions.

Theorem: Chords at equal distances from the centre are equal

The converse is used without proof here. If OM and ON are perpendicular distances from one centre to chords AB and CD, and OM = ON, then AB = CD. The equality of distances must refer to perpendiculars, not arbitrary joining segments.

Worked example 4. Chords AB and CD of the same circle are each at a distance of 4 cm from its centre. Must their lengths be equal?

Answer: Yes. Their perpendicular distances are both 4 cm, so the chords are equidistant from the centre. By the converse chord-distance theorem, AB = CD. The data establish equality but do not supply a numerical length for either chord.

The two results give a useful pair of routes: start with equal chords to obtain equal distances, or start with equal distances to obtain equal chords. State the route that matches the information actually provided.

Why is there exactly one circle through three non-collinear points?

Collinear points lie on a single straight line. Non-collinear points do not. Let A, B and C be three distinct, non-collinear points. A circle through them must have a centre equally distant from A, B and C.

Theorem: Three non-collinear points determine a unique circle

A point on the perpendicular bisector of a segment is equally distant from its endpoints. To see this, join the point to both endpoints and to the midpoint. The resulting right triangles have equal halves of the segment and a common perpendicular side.

The SAS criterion, meaning side-angle-side, proves congruence when two corresponding sides and the angles included between them are equal. It gives equality of the distances to the endpoints in this construction. Conversely, join a point equally distant from the endpoints to the midpoint. SSS makes the two triangles congruent, so their equal adjacent angles are right angles. The point therefore lies on the perpendicular bisector; if it is the midpoint itself, it already lies on that line.

1. Construct the perpendicular bisectors of AB and BC. Since A, B and C are non-collinear, AB and BC are not parallel, and their perpendicular bisectors intersect at a single point O.
2. Because O is on the perpendicular bisector of AB, OA = OB.
3. Because O is on the perpendicular bisector of BC, OB = OC. Hence OA = OB = OC.
4. Draw the circle with centre O and radius OA. All three given points lie on it, proving existence.
5. Any centre of a circle through A, B and C must lie on both of those perpendicular bisectors. Their sole intersection is O, and the radius must be OA. This proves uniqueness.

What do the construction and proof establish separately?

Existence means that such a circle can be drawn. Uniqueness means that no different circle satisfies the same conditions. Drawing a candidate circle establishes the first claim only after its equal radii are justified. The final step rules out a second possible centre or radius.

The circle through the three vertices of a triangle is its circumcircle, and its centre is the circumcentre. The perpendicular bisector of CA also passes through O, since OC = OA. Thus the three perpendicular bisectors are concurrent, meaning they pass through one common point.

Mark non-collinear points A, B and C. Draw perpendicular bisectors of AB and BC, labelling their intersection O. Draw the circle with centre O through A and check that it passes through B and C.

The condition of non-collinearity cannot be dropped. For three distinct collinear points, the perpendicular bisectors of two consecutive joining segments are distinct parallel lines. They have no common point that can serve as the centre of a circle through all three.

How do equal arcs relate to angles at the centre?

An arc is a connected part of a circle’s circumference, its boundary. Two distinct endpoints divide the circle into two arcs. Unless the endpoints form a diameter, the shorter is the minor arc and the longer is the major arc. Diameter endpoints divide the circle into semicircles.

An arc subtends an angle at the centre when radii joining its endpoints form that angle. The selected arc determines which angle is meant. A minor arc uses the smaller central angle; a major arc uses the reflex angle, which is greater than 180° and less than 360°.

Theorem: Equal central angles give equal arcs, and conversely

In the same circle, or in congruent circles with equal radii, arcs subtending equal central angles are equal. Conversely, equal corresponding arcs subtend equal angles at the centre. Here equality compares arcs of the same kind, such as two minor arcs.

For a proof, place the centres of equal circles together and align one bounding radius of each arc. Equal central angles make the other bounding radii coincide. Because the radii and selected angular regions agree, the arcs coincide and are congruent, meaning exactly superposable.

Conversely, superpose equal corresponding arcs and their centres. Their endpoints coincide, so the radii through those endpoints coincide. The central angles bounded by those radii must therefore be equal. A rotation, which turns a figure about a fixed point without changing its size, gives the same argument within one circle.

Worked example 5. Two congruent circles have centres O and O′, read “O prime”. Arc AXB subtends 75° at O, and arc A′YB′ subtends 25° at O′. The letters X and Y name points on the respective arcs. Find the ratio of their lengths.

Answer: Split the 75° central angle into three consecutive angles of 25° each. Their corresponding arcs equal the 25° arc in the congruent circle. Thus the first arc contains three equal copies of the second, and the ratio is 3 : 1. The colon denotes a ratio, comparing quantities by division.

The equal-radii condition matters: the same angle can bound different arc lengths in circles of different sizes. Here the circles are explicitly congruent. The argument uses equal central-angle pieces, so no assumption about unequal circles enters the comparison.

How do equal chords and corresponding arcs determine one another?

Theorem: Equal chords cut off equal corresponding arcs, and conversely

In the same circle, equal chords cut off equal corresponding arcs. Conversely, equal corresponding arcs have equal chords. These results are used without proof here. The word corresponding prevents an incorrect comparison between the minor arc of one chord and the major arc of another.

If AB and CD are equal non-diameter chords, their minor arcs are equal to each other, and their major arcs are equal to each other. Each pair of endpoints must be associated with the intended arc. A third letter on the arc makes the selection explicit.

Worked example 6. Arcs AXB and CYD of one circle are congruent. Here X lies on the selected arc from A to B, and Y lies on the selected arc from C to D. Find the ratio of chords AB and CD.

Answer: Congruent corresponding arcs have equal chords, so AB = CD. The ratio AB : CD is therefore 1 : 1. No radius or numerical chord length is required because the question asks for a comparison of the two lengths.

How can adjacent arcs on a semicircle be compared?

When a diameter divides the circle, the radii along that diameter point in opposite directions. Central angles spanning one semicircle total 180°. If one of its arcs is half the other, the larger arc can be divided into two parts equal to the smaller.

Worked example 7. AC is a diameter of a circle with centre O. Point B lies on one semicircle between A and C. Along that semicircle, X lies between A and B and Y between B and C. Arc AXB is half of arc BYC. Find ∠BOC.

Answer: The arc condition gives ∠AOB = half of ∠BOC. Let x denote the degree measure of ∠BOC. The straight angle AOC gives x/2 + x = 180, where / means division. Thus 3x/2 = 180 and x = 120. Therefore ∠BOC = 120°.

A and C are the ends of a horizontal diameter through O. Point B is on the upper semicircle. X marks the arc from A to B, and Y marks the arc from B to C. The radii OA, OB and OC show the two adjacent central angles.

See Fig. 10.11 in your NCERT textbook

These examples use two different conclusions. Congruent arcs give equal chords directly. Adjacent arcs with a specified ratio give a relation between central angles. State which relationship is needed before setting up an equation.

How can chord properties be combined in a longer calculation?

The perpendicular-bisector result also connects chord lengths to the common radius through algebra. Let r denote the radius, d the perpendicular distance from the centre to a chord, and h half the chord’s length. These are lengths measured in the same unit.

Property: Radius, perpendicular distance and half-chord form a right triangle

Pythagoras gives r² = d² + h². Rearranging, h² = r² − d². Since the complete chord has length 2h, its squared length is 4h². Keeping the half-chord visible avoids introducing the complete chord as a side of the right triangle.

Worked example 8. In a circle of radius r, chords AB and AC satisfy AB = 2AC. Their perpendicular distances from the centre are p and q respectively. Prove that 4q² = p² + 3r².

Answer: Let O be the centre, and let M and N be the perpendicular feet on AB and AC. Then OM = p, ON = q, AM = AB/2 and AN = AC/2. Hence AM = 2AN. Pythagoras gives AM² = r² − p² and AN² = r² − q². Squaring AM = 2AN gives r² − p² = 4(r² − q²). Expanding and rearranging produces 4q² = p² + 3r².

How should the reasoning be checked?

First check the correspondence: p belongs to AB, while q belongs to AC. Next check that each perpendicular halves its own chord. Finally, when a length is doubled, its square is multiplied by four. This explains the coefficient 4 in the equation.

A complete solution links every equation to its geometric reason. A diagram can organise the given data, but a chord’s apparent length or position is not a substitute for a stated equality, perpendicular condition or theorem.

Glossary

• Circle — The set of points in a plane at a fixed positive distance from a fixed point.
• Centre — The fixed point from which every point of a circle is equally distant.
• Radius — A segment from the centre to the circle, or the length of that segment.
• Chord — A straight line segment whose two endpoints lie on the same circle.
• Diameter — A chord passing through the centre, with length equal to twice the radius.
• Midpoint — The point on a line segment that divides it into two equal lengths.
• Perpendicular bisector — A line passing through a segment’s midpoint and meeting it at a right angle.
• Perpendicular distance — The length of the perpendicular segment from a point to a given line.
• Arc — A connected portion of a circle’s circumference selected between two endpoints.
• Central angle — An angle with its vertex at the circle’s centre and sides along radii.
• Congruent circles — Circles having equal radii and therefore the same size and shape.
• Non-collinear points — Points that do not all lie on one straight line.
• Circumcentre — The centre of the circle passing through all three vertices of a triangle.
• Converse — A statement formed by exchanging the condition and conclusion of an original conditional statement.

Common errors and misconceptions

• Misconception: Every chord is a diameter. Correct: A diameter must pass through the centre; a general chord only needs its endpoints on the circle.
• Misconception: Any line from the centre to a chord bisects it. Correct: The line must be perpendicular to the chord, or its intersection must already be known to be the midpoint.
• Misconception: The midpoint-to-centre theorem needs no exception. Correct: The chord must not be a diameter, whose midpoint is the centre itself.
• Misconception: Any segment from the centre to a chord measures its distance. Correct: The distance is measured along the perpendicular.
• Misconception: Pythagoras uses the whole chord beside the perpendicular. Correct: The right triangle formed by one radius contains half the chord.
• Misconception: Equal chords let us equate any two arcs they bound. Correct: Compare corresponding arcs, matching minor with minor or major with major.
• Misconception: Three distinct points determine a circle even when collinear. Correct: The existence and uniqueness theorem requires non-collinear points.
• Misconception: Equal central angles give equal arc lengths in all circles. Correct: This comparison requires the same circle or congruent circles.

Exam-style questions with model answers

Q1. Define a chord and distinguish a diameter from a general chord. [2 marks]

1. A chord is a straight line segment joining two points on a circle.
2. A diameter is a chord that passes through the centre; a general chord need not do so.

Q2. A circle has diameter AD = 34 cm and chord AB = 30 cm. Find the perpendicular distance of AB from the centre O. [3 marks]

1. Draw OM perpendicular to AB, with M on AB. The perpendicular from the centre bisects the chord, so AM = 30 ÷ 2 = 15 cm.
2. The radius OA is half the diameter: OA = 34 ÷ 2 = 17 cm. In right triangle OMA, Pythagoras gives OM² = 17² − 15² = 64.
3. Taking the positive square root, the perpendicular distance from the centre to the chord is OM = 8 cm.

Q3. In a circle with centre O, AB is a non-diameter chord and M is its midpoint. Prove that OM is perpendicular to AB. [5 marks]

1. Join the centre O to the chord endpoints A and B. The two radii have equal lengths, so OA = OB.
2. Because M is the given midpoint of AB, its two parts are equal: AM = BM.
3. The side OM belongs to both triangles OMA and OMB. Their three pairs of corresponding sides are therefore equal, proving congruence by SSS.
4. Corresponding angles OMA and OMB are equal. Since A, M and B lie on one straight line, these adjacent angles have sum 180°.
5. Each of two equal angles totalling 180° measures 90°. Therefore OM meets AB at a right angle, which proves OM ⊥ AB.

Q4. Equal chords AB and CD lie in a circle with centre O. The perpendiculars from O meet AB at M and CD at N. Prove OM = ON. [4 marks]

1. The perpendiculars from the centre bisect their chords, so AM = AB/2 and CN = CD/2. Since AB = CD, AM = CN.
2. Join OA and OC. These are radii of the same circle, so their lengths are equal.
3. Triangles OMA and ONC have right angles, equal hypotenuses OA and OC, and equal sides AM and CN. They are congruent by RHS.
4. The corresponding sides OM and ON are therefore equal. These sides are precisely the perpendicular distances of the chords from the centre.

Q5. Explain why exactly one circle passes through three distinct non-collinear points A, B and C. Include both existence and uniqueness. [5 marks]

1. Draw the perpendicular bisectors of AB and BC. Since A, B and C are non-collinear, these bisectors intersect at one point, labelled O.
2. Every point on a perpendicular bisector is equally distant from its segment’s endpoints. Thus OA = OB from the first bisector, and OB = OC from the second.
3. It follows that OA = OB = OC. The circle with centre O and radius OA passes through A, B and C, establishing existence.
4. The centre of any circle passing through all three points must be equally distant from them, and must therefore lie on both perpendicular bisectors.
5. The two bisectors have only one intersection, O. The radius is then fixed as OA. Hence no different circle passes through the three points, establishing uniqueness.

Q6. AC is a diameter of a circle with centre O. B lies on one semicircle between A and C. The arc from A to B along that semicircle is half the adjacent arc from B to C. Find ∠BOC. [3 marks]

1. Divide the larger arc into two arcs equal to the smaller. Equal arcs give equal central angles, so ∠AOB is half of ∠BOC.
2. Let x be the degree measure of ∠BOC. Since AOC is a straight angle and B lies on the stated semicircle, x/2 + x = 180.
3. Thus 3x/2 = 180, giving x = 120. The required central angle is ∠BOC = 120°.

Q7. Chords AB and AC of a circle of radius r satisfy AB = 2AC. Their perpendicular distances from the centre are p and q respectively. Prove 4q² = p² + 3r². [5 marks]

1. Label the centre O. Draw perpendiculars OM to AB and ON to AC, with feet M and N. Then OM = p and ON = q.
2. Each perpendicular bisects its chord. Therefore AM = AB/2 and AN = AC/2. The given relation AB = 2AC implies AM = 2AN.
3. In right triangle OMA, the radius OA is the hypotenuse. Pythagoras gives AM² = r² − p².
4. In right triangle ONA, the same radius OA is the hypotenuse, giving AN² = r² − q². Squaring AM = 2AN gives AM² = 4AN².
5. Substitute the two squared lengths: r² − p² = 4(r² − q²). Expanding and rearranging yields 4q² = p² + 3r², as required.

Q8. Two chords of the same circle are each at a perpendicular distance of 4 cm from the centre. What can be concluded about their lengths, and can either numerical length be found from these data alone? [2 marks]

1. The chords are equal because they are equidistant from the centre of the same circle.
2. Their numerical length cannot be determined from the stated distance alone; the radius or equivalent additional information is needed.

Key takeaways

• A diameter is a chord through the centre, and its length equals twice the radius of the circle.
• The line joining the centre to the midpoint of a non-diameter chord is perpendicular to that chord.
• A perpendicular from the centre bisects a chord, creating a right triangle with one radius and half the chord.
• Equal chords are equidistant from the centre, and chords at equal perpendicular distances from the centre are equal.
• Three distinct non-collinear points determine exactly one circle; its centre is found using perpendicular bisectors.
• Equal central angles give equal corresponding arcs in the same circle or congruent circles, and the converse holds.
• Equal chords cut off equal corresponding arcs; equal corresponding arcs also give equal chords in the same circle.
• In chord calculations, identify the half-chord and perpendicular before applying Pythagoras, and keep every symbol tied to its segment.

Test yourself

What makes a diameter a special chord?

It passes through the centre and has length equal to twice the radius.

Why must the centre-to-midpoint theorem exclude a diameter?

A diameter’s midpoint is the centre itself, so there is no separate joining segment defining a perpendicular direction.

Which length measures the distance from the centre to a chord?

The length of the perpendicular from the centre to the line containing the chord.

What must be used with a radius and a chord’s perpendicular distance in Pythagoras’ theorem?

Use half the chord, because the perpendicular from the centre bisects the complete chord.

Where is the centre of a circle through three distinct non-collinear points?

It is the intersection of the perpendicular bisectors of the segments joining pairs of those points.

Congruent arcs AXB and CYD belong to one circle, with X and Y on the respective arcs. What is AB : CD?

The ratio is 1 : 1 because congruent corresponding arcs have equal chords.

Can equal central angles alone establish equal arc lengths in circles of different radii?

No. Equal arc lengths follow from equal central angles when the circles have equal radii.

Why is proving existence insufficient when a theorem asserts a unique circle?

Existence supplies one circle; uniqueness additionally rules out a different centre or radius satisfying the same conditions.