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ICSE Class 9 Mathematics Chapter 17 Circle Digital Edition
For Class 9 Mathematics, this chapter in ICSE Class 9 Maths Chapter 17 Circle provides a detailed overview of important concepts. We highly recommend using this text alongside the ICSE Solutions for Class 9 Mathematics to learn the exercise questions provided at the end of the chapter.
Chapter 17 Circle ICSE Book Class Class 9 PDF (2026-27)
17 Circle
17.1 Introduction
Consider the shape of the wheel of a bicycle, car, etc. Each of it is said to be a circle.
Observe the path traced by the tip of minute-hand of a wall clock, the path traced is a circle.
17.2 Circle
A circle is defined as the figure (closed curve) obtained by joining all those points in a plane which are at the same fixed distance from a fixed point in the same plane.
Infact, a circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point, in the same plane, always remains constant.
The perimeter of the circle is called its circumference.
The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.
The line segment, joining any two points on the circumference of the circle, is called a chord.
A chord which passes through the centre of the circle is called diameter. It is the largest chord of a circle.
Also, diameter is twice the radius.
17.3 More About Circle
Draw a circle with radius r and centre O as shown alongside:
Exterior point: A point lying in the same plane as that of the circle is called an exterior point, if its distance from the centre of the circle is greater than the radius of the circle.
In the given figure, points A, B, C and D are exterior points as each of OA, OB, OC and OD is greater than the radius of the circle.
Interior point: A point lying in the same plane as that of the circle is called an interior point, if its distance from the centre of the circle is less than the radius of the circle.
In the figure, given above, points L, M and N are interior points as each of OL, OM and ON is less than the radius of the circle.
Teacher's Note
The concept of circles is fundamental in engineering and design - from wheel construction to architectural planning, understanding circle properties helps us build stronger and more efficient structures.
Point on the circumference of a circle: A point in the same plane as that of the circle will be on the circumference of the circle, if its distance from the centre of the circle is equal to the radius of the circle.
In the figure, given above, points P, Q and R are on the circumference of the circle as each of OP, OQ and OR is equal to the radius of the circle.
Concentric circles: Two or more circles are said to be concentric, if they have the same centre but different radii.
In the adjoining figure, O is the centre of each circle drawn; so the circles are called concentric circles.
Equal circles: Circles are said to be equal if they have equal radii. Equal circles are also called congruent circles.
Circumscribed circle: A circle that passes through all the vertices of a polygon is called a circumscribed circle. The centre of circumscribed circle is called circumcentre and the polygon is called inscribed polygon. Examples are shown below:
Inscribed circle: A circle that touches all the sides of a polygon is called an inscribed circle (or, in-circle) of the polygon.
The centre of inscribed circle is called incentre and the polygon is called circumscribed polygon. Examples are shown below:
17.4 Arc, Segment And Sector
Arc: A part of the circumference of a circle is called its arc.
The adjoining figure shows a chord AB of a circle with centre O.
The chord AB divides the circumference of the circle into two parts and each of these two parts is an arc.
If both the arcs are unequal in size, smaller arc (arc APB in the adjoining figure) is called minor arc, whereas the bigger arc (arc AQB in the adjoining figure) is called major arc.
If both the arcs are equal, each is called a semi-circle.
A minor arc is always smaller than the semi-circle.
A major arc is always bigger than the semi-circle.
Unless otherwise stated, an arc stands for a minor arc.
Segment: The part of the circle, bounded by an arc and a chord, is called a segment.
In the figure, shown alongside, chord AB divides the circle into two segments. The smaller segment, which is less than semi-circle, is called minor segment and the bigger segment, which is greater than semi-circle, is called major segment.
The centre of the circle lies in the major segment.
Sector: The region bounded by an arc and two radii, joining the centre to the end points of the arc, is called a sector.
The figure, given alongside, shows a circle with centre O. The arc APB, radii OA and OB form the sector AOBP.
Clearly, the region (shaded portion) bounded by arc APB and radii OA and OB is a sector.
The minor arc corresponds to the minor sector and major arc corresponds to the major sector.
When both the arcs are equal, the region enclosed by each arc alongwith its diameter is called a semi-circle.
In the case of a semi-circle, both the segments are equal and both the sectors are also equal.
17.5 Chord Properties:
Theorem 22
A straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is at right angles to the chord.
Given: A circle with centre O and OC bisects the chord AB.
To Prove: OC perpendicular AB.
Construction: Join OA and OB.
Proof:
| Statement | Reason |
|---|---|
| In triangle OAC and triangle OBC; (i) OA = OB | Radii of the same circle |
| (ii) OC = OC | Common |
| (iii) AC = BC | OC bisects AB - Given |
| Therefore triangle OAC is congruent to triangle OBC | S.S.S. |
And angle OCA = angle OCB - Corresponding angles of congruent triangles are congruent.
But, angle OCA + angle OCB = 180° - ABC is a straight line.
Therefore angle OCA = angle OCB = 90° - From (2) and (3).
Therefore OC perpendicular AB.
Hence Proved
Theorem 23
(Converse of Theorem 22)
The perpendicular to a chord, from the centre of the circle, bisects the chord.
Given: A circle with centre O and OP is perpendicular to the chord AB.
To Prove: AP = BP.
Construction: Join OA and OB.
Proof:
| Statement | Reason |
|---|---|
| In triangle OAP and triangle OBP; (i) OA = OB | Radii of the same circle. |
| (ii) OP = OP | Common |
| (iii) angle OPA = angle OPB = 90° | OP perpendicular AB - Given |
| Therefore triangle OAP is congruent to triangle OBP | R.H.S. |
| Therefore AP = BP | Corresponding parts of congruent triangles are congruent. |
Hence Proved
Remember: Greater is the size of a chord, smaller is its distance from the centre and vice-versa.
The adjoining figure shows a circle with centre O. OP is perpendicular to chord AB and OQ is perpendicular to chord CD.
Since, chord AB is greater than chord CD therefore AB is at a smaller distance from the centre as compared to CD i.e. OP is less than OQ.
Conversely, as OP is less than OQ therefore AB is greater than CD.
Theorem 24
Equal chords of a circle are equidistant from the centre.
Given: A circle with centre O in which chord AB = chord CD.
To Prove: Chords AB and CD are equidistant from the centre i.e. if OP perpendicular AB and OQ perpendicular CD, then to prove that OP = OQ.
Construction: Join OB and OD.
Proof:
| Statement | Reason |
|---|---|
| BP = (1/2) AB | Perpendicular from the centre bisects the chord. |
| DQ = (1/2) CD | Perpendicular from the centre bisects the chord. |
| But; AB = CD | Given |
| Therefore BP = DQ | From (1), (2) and (3). |
| In triangle OPB and triangle OQD, (i) BP = DQ | From (4) |
| (ii) OB = OD | Radii of the same circle |
| (iii) angle OPB = angle OQD = 90° | OP perpendicular AB and OQ perpendicular CD |
| Therefore triangle OPB is congruent to triangle OQD | R.H.S. |
| Therefore OP = OQ | Corresponding parts of congruent triangles |
Hence Proved
Theorem 25
(Converse of Theorem 24)
Chords of a circle, equidistant from the centre of the circle, are equal.
Given: A circle with centre O in which chords AB and CD are equidistant from the centre. i.e. if OP perpendicular AB and OQ perpendicular CD, then OP = OQ.
To Prove: Chord AB = chord CD.
Construction: Join OB and OD.
Proof:
| Statement | Reason |
|---|---|
| In triangle OPB and triangle OQD, (i) OP = OQ | Given |
| (ii) OB = OD | Radii of the same circle |
| (iii) angle OPB = angle OQD = 90° | OP perpendicular AB and OQ perpendicular CD |
| Therefore triangle OPB is congruent to triangle OQD | R.H.S. |
| Therefore PB = QD | Corresponding parts of congruent triangles |
| Therefore (1/2) AB = (1/2) CD | Perpendicular from centre bisects the chord. |
| Therefore Chord AB = Chord CD |
Hence Proved
Theorem 26
There is one and only one circle, which passes through three given points not in a straight line.
Given: Three points A, B and C, which are not in a straight line.
To Prove: One and only one circle can be drawn through A, B and C.
Teacher's Note
This theorem is essential in navigation systems and GPS technology - three known reference points are always needed to pinpoint a location accurately on Earth's surface.
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ICSE Book Class 9 Mathematics Chapter 17 Circle
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