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Circle (I), Class 9 Mathematics Detailed Chapter Notes PDF Download

CHORD PROPERTIES OF A CIRCLE

CIRCLE

A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant. The fixed point is called the centre and the constant distance is

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

called the radius of the circle. The given figure consists of a circle with centre O and radius equal to r units.

TERMS AND FACTS RELATED TO CIRCLES

Radius : A line segment joining the centre and a point on the circle is called its radius, generally denoted by r. The plural of radius is radii.
In the figure, OA, OB and OC are the radii of a circle.
Circumference : The perimeter of a circle is called its circumference. O
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Circumference = 2πr
Position of a Point With Respect To a Circle Let us consider a circle with centre O and radius r. A point P is said to lie.
(i) inside the circle, if OP < r. (ii) on the circle, if OP = r.
(iii) outside the circle, if OP > r. O r P
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

In the figure of a circle with centre O and radius r, Z
(i) The points A, O, B lie inside the circle; (ii) The points P, Q, R lie on the circle;
(iii) The points X, Y, Z lie outside the circle.
Interior and Exterior of a Circle
The region consisting of all those points which lie inside a circle,
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
is called the interior of the circle.
The region consisting of all those points which lie outside a circle, is called the exterior of the circle.
Circular Region or Circular Disc
The region consisting of all those points which are either on the circle or lie inside the circle, is called the circular region.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Chord : A line segment joining any two points on a circle is called a chord of the circle. In the figure, PQ, RS and AB are the chords of a circle with centre O. O
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Diameter : A chord of the circle passing through the centre of a circle is called its diameter. In the figure, AOB is a diameter of a circle with centre O.
Diameter = 2 × Radius
Properties :(i) Diameter is the largest chord of a circle.
(ii) All diameters of a circle are equal in length.

Secant : A line which intersects a circle in two distinct points is called a secant of the circle. In the figure, the line NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 cuts the circle in two points C and D. So, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 is a secant of the circle.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Tangent : A line that intersects the circle in exactly one point is called a tangent of the circle. The point at which the tangent intersects the circle is called its point of contact. In the figure, SPT is a tangent at the point P of the circle with centre O. Clearly, P is the point of contact of the tangent with the circle. O
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Facts About Tangents :
(i) No tangent can be drawn to a circle through a point inside the circle:
(ii) One and only one tangent can be drawn to a circle at a point on the circle.
(iii) Two tangents can be drawn to a circle from a point outside it.
In the adjoining figure, PT1 and PT2 are the tangents to the circle from point P.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Touching Circles : Two circles are said to touch each other if and only if they have one and only one point in common. Two circles may touch externally [Fig. (i)] or internally T T [Fig. (ii)].
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

The common point is called the point of contact, and the line joining their centres is called the line of centres. A line touching the two circles is called a common tangent. Thus, in the above figure, P is the point of contact, AB is the line of centres and PT is a common tangent.
Direct Common Tangents : A common tangent to two circles is called a direct common tangent if both the circles lie on the same side of it. In the figure, AB and CD are two direct common tangents.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Transverse Common Tangents : A common tangent to two circles is called a transverse common tangent if the circles lie on its opposite sides. In the figure, PQ and RS are two transverse common tangents.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Arc : A continuous piece of a circle is called an arc of the circle. Let P and Q be any two points on a circle with centre O. Then, clearly the whole circle has been divided into two pieces, namely arc PAQ and arc QBP, to be denoted by NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PAQ and NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9QBP respectively. P Q
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
We may denote them by NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ andNCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9QP respectively. Minor and Major Arc : An arc less than one-half of the whole arc of a circle is called a minor arc, and an arc greater than one-half of the whole arc of a circle is called a major arc of the circle.
Thus, in the above figure, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ is a minor arc, while NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9QP is a major arc.

Central Angle : An angle subtended by an arc at the centre of a circle is called its central angle. In the given figure, central angle of NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9POQ.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Degree Measure of An Arc : Let NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ be an arc of a circle with centre O.
If NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9POQ = θ°, we say that the degree measure of NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ is θ° and we write, m(NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ ) = θ°. If m(NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9PQ ) = θ°, then m(NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9QP ) = (360 – θ)°. Degree measure of a circle is 360°.

Congruent Arcs : Two arcs AB and CD are said to be congruent, if they have same degree measure. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AB NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 CD NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 m( NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AB ) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 m( NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CD ) NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AOB = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9COD.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Semi-Circle : A diameter divides a circle into two equal arcs. Each of these two arcs is called a semi-circle. The degree measure of a semi-circle is 180°. In the given figure of a circle with centre O, NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ABC as well as NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ADC is a semi-circle.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Congruent Circles : Two circles of equal radii are said to be congruent.

Concentric Circles : Circles having same centre but different radii are called concentric circles.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Concyclic Points : The points, which lie on the circumference of the same circle, are called concyclic points. In the adjoining figure, points A, B, C and D lie on the same circle and hence, they are concyclic.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Segment : A segment is a part of a circular region bounded by an arc and a chord, including the arc and the chord. The segment containing the minor arc is called a minor segment, while the other one is a major segment. The centre of the circle lies in the major segment.

Alternate Segments of a Circle : The minor and major segments of a circle are called alternate segments of each other.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Sector of a Circle : The part of the plane region enclosed by an arc of a circle and its two bounding radii is called a sector of the circle.
Thus, the region OABO is the sector of a circle with centre O.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Quadrant : One-fourth of a circular disc is called a quadrant.

Cyclic Quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then such a quadrilateral is called a cyclic quadrilateral.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
If four points lie on a circle, they are said to be concyclic. We also say that quad. ABCD is inscribed in a circle with centre O.

Theorem-1 : Equal chords of a circle subtend equal angles at the centre.
Given : A circle with centre O in which chord PQ = chords RS. P
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To prove : NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9POQ = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ROS
Proof : In ΔPOQ and ΔROS,

STATEMENT REASON

OP = OR

OQ = OS

PQ = RS

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔPOQ = ΔROS

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9POQ = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ROS

(Radii of the same circle)

(Radii of the same circle)

(Given)

(By SSS)

(By CPCTC)

Hence proved

Converse of above theorem : If the angles sutended by the chords at the centre (of a circle) are equal, then the chords are equal.
Given : A circle with centre O. Chord PQ and RS subtend equal angles at the centre of the circle.
i.e.,NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9POQ = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ROS P
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To prove : Chord PQ = Chord RS O
Proof : In ΔPOQ and ΔROS,

STATEMENT REASON

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9POQ =NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9ROS

OP = OR

OQ = OS 

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔPOQ = ΔROS

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 chord PQ = chord RS (By CPCTC)

(Given)

(Radii of the same circle)

(Radii of the same circle)

(By SSS)

(By CPCTC)

Hence proved

Theorem 2 : The perpendicular from the centre of a circle to chord bisects the chord.
Given : AB is a chord of a circle with centre O.
To prove : LA = LB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Join OA and OB. L
Proof :

STATEMENT REASON

In a right ΔOLA and ΔOLB,
OA = OB
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLB = 90°

OL = OL
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔOLA NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔOLB
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9LA = LB By C.P.C.T.

Radii of the same circle

Since OL NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB

Common

By RHS congruency

By C.P.C.T.

Hence proved

Converse of above theorem : The straight line drawn from the centre of a circle to bisect a chord, is perpendicular to the chord.
Given : AB is chord of a circle with centre O and OL bisects AB.
To prove: OL NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Join OA and OB. L
Proof :

STATEMENT REASON

1. In ΔOLA and ΔOLB, we have
            OA = OB
             AL = BL
             OL = OL
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔOLA NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 ΔOLB
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLB
2. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLA + NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLB = 180°
3. NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLA = NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OLB = 90°
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9OL  AB

Radii of the same circle

Given, OL bisects AB

Common

By SSS

... (i)By C.P.C.T. 

... (ii) ALB is a straight line

From (i) and (ii).

Hence proved

Theorem 3. Prove that one and only one circle, passing through three non-collinear points.
Given :
Three non-collinear points A, B, C.
To prove : One and only one circle can be drawn, passing through A, B, and C.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Construction : Join AB and BC. Draw the perpendicular bisectors of AB O and BC. Let these perpendicular bisector intersect meeting at a point O.
Proof :

STATEMENT REASON

1. O lies on the perpendicular bisector of AB

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 OA = OB... (i)

2. O lies on the perpendicular bisector of B

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 OB = OC... (ii)

3. OA = OB = OC

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 O is equidistant from A, B and C
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 Any circle drawn with centre O and radius OA will pass through B and C also.

4. O is the only point equidistant from A, B andC.

Every point on perpendicular bisector of a line segment

is equidistant from its end points i.e. A and B.

Each point on perpendicular bisector of line segment is

equidistant from its end points i.e. B and C.

From (i) and (ii)

Perpendicular bisectors of AB and BC cut each other at
point O only

Hence, one and only circle can be drawn through three non-collinear points A, B and C.

Theorem 4. Equal chords of a circle are equidistant from the centre.
Given : A circle with centre O in which chord AB = chord CD; OL NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB and OM NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9CD.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9


To prove : OL = OM.
Proof :

STATEMENT REASON

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Perpendicular from centre bisects the chord

Perpendicular from centre bisects the chord.

Given.

Halves of equals are equal.

From (i) and (ii).

Radii of the same circle.

From (iii).

As OL NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9AB and OM NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 CD.

By RHS

By C.P.C.T.

Hence, the chords AB and CD are equidistant from the centre O.

Converse of above Theorem : Chords of a circle that are equidistant from the centre of the circle, are equal.
Given : AB and CD are two chords of a circle with centre O;
OL NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB, OM NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 CD and OL = OM.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
To prove : AB = CD
Construction : Join OA and OC.
Proof :

STATEMENT REASON

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Given.

Radii of the same circle.

As OL NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB and OM NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 CD.

By RHS

By C.P.C.T.

Perpendicular from centre bisects the chord

Doubles of equals are equal.

Hence proved.

Ex.1 Determine the length of a chord which is at a distance of 6 cm from the centre of a circle of radius 10 cm.
Sol.
Let AB be a chord of a circle with centre O and radius 10 cm.
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9
Draw OC NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB
Join OB.
In ΔBCO, We have
OC = 6 cm

and OB = 10 cm

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB = BC + BC [Since the perpendicular from the centre to a chord bisects the chord NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AC = BC]
NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB = 2BC
Hence, AB = 2 × 8 = 16 cm

Ex.2 A chord of length 30 cm is drawn in a circle of radius 17 cm. Find its distance from the centre of the circle.

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

Ex.3 In figure, O is the centre of the circle of radius 15 cm. OP NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB, OQ NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 CD, AB || CD, AB = 18 cm and CD = 22 cm. Find PQ.
Sol. Given AB and CD be two parallel chords of a circle with centre O. Also OP NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 AB and OQ NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9 CD. Since, AB || CD, so pointsO, Q and P are collinear. Join OA and OC. We have, AB = 18 cm, CD = 22 cm and OA = OC = 15 cm [Radii of the circle] Since the perpendicular drawn from centre of circle to a chord bisect the chord. Therefore

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

NCRT,Question and Answer,Important,Class 9 Mathematics,CBSE Class 9

 

The document Circle (I), Class 9 Mathematics Detailed Chapter Notes is a part of Class 9 category.
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FAQs on Circle (I), Class 9 Mathematics Detailed Chapter Notes

1. What is the equation of a circle?
Ans. The equation of a circle is given by (x-a)^2 + (y-b)^2 = r^2, where (a, b) represents the center of the circle and r represents the radius.
2. How do you find the center and radius of a circle given its equation?
Ans. To find the center and radius of a circle given its equation, we need to rewrite the equation in the standard form. By comparing the equation with the standard form (x-a)^2 + (y-b)^2 = r^2, we can determine the center as (a, b) and the radius as the square root of r^2.
3. How many types of circles are there?
Ans. There are three types of circles: a) A circle with a radius of zero, also known as a point circle. b) A circle with a finite radius, which is the most common type. c) A circle with an infinite radius, also called a line circle.
4. What is the relationship between the radius and diameter of a circle?
Ans. The radius of a circle is half the length of its diameter. In other words, if r represents the radius and d represents the diameter, then the relationship can be expressed as r = d/2.
5. How do you find the circumference of a circle?
Ans. The circumference of a circle can be found using the formula C = 2πr, where C represents the circumference and r represents the radius of the circle.
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