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Chapter-12 CIRCLES MATH CLASS-9.

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1 Chapter-12 CIRCLES MATH CLASS-9

2 Module Objectives Define circle.
Define radius,circumference,arc,line segment,chord. Identify and state the property of chord of a circle. Identify central angle and inscribed angle. State the theorem on angle property of the circle. Prove the theorem logically. Solve problems and riders based on the properties of the circle.

3 Introduction Circles are one of the interesting figures in geometry.
The wheels of the carts, automobiles and trains are circular.

4 What is Circle? A closed curve, every point of which is equidistant from a given fixed point. This fixed point (O) is called the centre of the circle. A circle is the locus of a moving point in a plane such that it is at a constant distance from the given fixed point. Locus is the path traced by a moving point

5 Radius The line segment joining the centre to any point on the circle.
OA and OB are the radii of the given circle. The radii of a circle are always equal. i.e: OA == OB Circumference/Perimeter is the Distance around the circle.

6 Chord & Diameter CHORD-A line segment with its end points lying on the circle. Here AB,PQ and XY are the chords. DIAMETER-A line segment passing through the centre of the circle and has its end points on circle. Diameter is the longest chord of a circle.(PQ) Length of a diameter is twice as its radius. Here PQ=OP+OQ Diameter = 2 (Radius) D=2R or Radius = Diameter/2 i.e: R = D/2 Q X Y O A B P

7 Concentric and Congruent Circles
Circles C1,C2,C3 have the same center O and different radiiOA(1.2cm),OB(1.8cm) and OC(2.5cm). Concentric circles are circles with same centre and different radii. Circles C1,C2,C3 have different centres P,Q and R.But they have same radii, PA=QB=RC=1.5 cm. Congruent circles are circles with same radii but different centres.

8 ARC OF A CIRCLE An arc is a `Part’ of a Circle.
The chord divides the circle into two parts,One smaller than the other. The smaller part is called the minor arc which is denoted as AXB. The greater part is called the major arc which is denoted as AYB. AXB and AYB are conjugate arcs.They lie on either sides of chord AB. If a chord is a diameter ,it divides the circle into two arcs of same size.Each is a semi circle. Here CXD and CYD are semicircles.

9 Segment of a Circle The part of the circular region included by an arc and the chord is called a segment. The region bounded by the chord AB and the major arc APB is called the major segment. The region bounded by the chord AB and the minor arc AQB is called the minor segment. The region bounded by the diameter and the arc is called the semi circular region.

10 Do it Yourself 12.1(a) 1.Observe the figure and represent the following : (a) OP (b) SO (c) SQ (d) SPQ (e ) Shaded portion (f) Conjugate of SPQ. 2. What is the length of the biggest chord of a circle of radius 4 cm? 3. Draw a circle with diameter 7cm. 4. Draw a circle with centre O.Draw two diameters and label their end points A,B,C and D. Draw the chords that connect the end points of the diameters.Name the following: (a) Four pairs of congruent triangles. (b) a pair of parallel chords. ( c ) rectangle. (d) semicircle. 5. Draw a circle with centre ‘O’ and radius 2 cm . For this circle, draw a concentric and a congruent circle.

11 Properties of Chord in a Circle
In each of the circles given below,PG is the chord and OA is perpendicular to PQ.In each case PA=QA. Property: In a circle the perpendicular from the centre to the chord , bisects the chord

12 Properties of chord of a circle
Property: Equal chords of a circle are equidistant from the centre Here the two chords AB and RS are equidistant from the centre(OD=OF).Hence AB=RS. In a circle if BC is the chord and OA is perpendicular to BC, OB²=OA²+AB² r²=d²+AB² r² = d²+l² r=radius d= perpendicular distance of chord from centre l= length of the chord/2

13 Do it Yourself 12.1(b) In a circle, with centre O ,the chords and their distances from the centre are given. Arrange them according to the increasing order of their length. 2. The length of chords AB,PQ,MN,DE and XY are 5.1 cm,2.9 cm,6.3 cm,4.5 cm and 5.4 cm.Arrange them in decreasing order of their distance from the centre of the circle. 3. Chord PQ = Chord AB. If PQ is at a distance of 3 cm from the centre of the circle at what distance is chord AB from the centre of the circle. 4 .In the given figure, OA and OB are the radii of the circle ,AB is the chord. OP is perpendicular to AB. Prove that AP = PB. S.No Name of Chord Distance 1 PQ 4.6 cm 2 AB 3.6 cm 3 XY 1 cm 4 CD 2.1 cm 5 MN 0 cm

14 ANGLE PROPERTIES OF A CIRCLE
Angles are of paramount importance in geometry and also in real life. Surveyors,Engineers,Navigators,Astronomers ,Scientists and many other people use the measurements of angles. In Circle 4,the vertex of the angle lies with the centre of the circle.Such an angle is called central angle.The arms of the central angle are its radii. In Circle 1,the vertex of the angle lies on the circle and its arms intersect the circle at two points.Such an angle is called inscribed angle. How many central angles can be drawn to intercept the circleat A and B? In circles 2 and 3 can angle ACB be called an inscribed angle?Why? How many angles can be inscribed in a circle by the same arc? Think! Think! Think!

15 Relation between Central Angle and Inscribed Angle
In the circles given below, O is the centre of the circle and the arc AXB subtends angle AOB at the centre and ACB on the remaining part. Measure angles AOB and ACB seperately .It can be found that , AOB = ACB. The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle.

16 Angles in a Segment Measure each of the angles in the following figures and record them in the table. Angle Segment Measure 1 Major 2 Minor 3 Semicircle 4 5 6 From the above diagram we can conclude that: Angle in the major segment is an acute angle. Angle in the minor segment is an obtuse angle. Angle in the semi-circle is a right angle. Angles in the same segment are equal.

17 Angles in a Segment Solution: X = 2 × 30˚ (angle at the centre is twice the angle at any point on the circle.) x = 60˚ y = 30˚ (angles in the same segment are equal) X = ½ × 105˚ X = 52.5˚ (angle at any point on the circle is half the angle at centre) X = ½ × 80˚ X = 40˚ (angle at any point on the circle is half the angle at the centre)

18 Angles in a Segment Example-2: Find the angles of the triangle ACB. Solution : ABC = 30˚ (given) ACB = 90˚ (angle in a semicircle) Hence CAB = 180˚ - ( 30˚ + 90˚ ) = 180˚ - 120˚ CAB = 60˚

19 Do it yourself 12.1(c) 1.Why angle D is not an inscribed angle? Why angle E is not a central angle? 3. a) Name four central angles with respect to adjoining figure. b) Name two inscribed angles. c) Name two angles that subtend BC. d) What angle subtends CD? e) What kind of triangle is DOC? f) Name three chords. g) Which chord is a diameter?

20 Do it yourself 12.1(c) 4. In the adjoining figure, a) Name the central angle subtended by AE. b) Name the central angle subtended by BC. c) Name the inscribed angle subtended by BC. d) Name the central angle subtended by CDE. e) Name two chords that are not diameters. f) Name a chord that is a diameter. g) Name the arc subtended by BOA. h) Name the arc subtended by DEB. 5. Write the value of x in each of the following cases.

21 Do it yourself 12.1(c) 6. Draw a rough diagram for each of the following: a) AB is a chord of a circle , with centre O.If OAB = 50˚,Find OBA b) RS is a chord of a circle with centre O.If ROS = 15˚,Find ORS 7. a) What kind of triangle is AOB ? b) What can you say about angles A and B ? c) What kind of angles are 1 and 2 ?

22 Angles in a Segment Statement Reason
Data: `O’ is the centre of the circle.AXB is the arc.AOB is the angle subtended by the arc AXB at the centre.ACB is the angle subtended by the arc AXB at a point on the remaining part of the circle. To Prove: AOB = ACB Construction : Join CO and produce it to D. Statement Reason 1. OA = OC Radii of the same circle are equal. 2. OCA = OAC Angles opposite to equal sides are equal. 3.In ∆AOC, AOD = OCA + OAC Exterior angle of a triangle = sum of interior opposite angles. 4. AOD = OCA OCA Substituting OAC by OCA from statement . AOD = OCA By addition 6.Similarly in ∆ BOC , BOD = OCB Adding statements 5 and 6. 7. AOD BOD = 2 OCA + 2 OCB = 2(OCA+OCB) Taking out 2 as common. AOB = ACB Since AOD + BOD = AOB ,OCA +OCB = ACB

23 Angles in a Segment Example-1 : Prove that angle in a semicircle is a right angle. Data: AOB is the diameter. ACB is angle in the semicircle. To prove: ACB = 90° Proof: 1) AOB = 180° ( AOB is a straight line ) 2) ACB = ½ AOB (angle at any point on the circle is half the angle at centre.) 3) ACB = ½ × 180° (from 1) 4) ACB = 90° Example-2 : From the adjoining diagram , prove that ∆APC and ∆DPB are equiangular. To prove : ∆ APC and ∆DPB are equiangular. Proof : In ∆ APC and ∆DPB APC = BPD (Vertically opposite angles) ACP = ABD (Angles in the same segment are equal) PAC = PDB (Angles in the same segment are equal) Hence ∆ APC and ∆DPB are equiangular.

24 Do it yourself 12.1(d) 1. In the given figure , prove that PRQ = PSQ 2. In the figure given below AC and BC are diameters of two circles intersecting at C and D. Show that A,D,B are collinear. 3.Two chords AB and CD of a circle intersect at P.If BP =PD, show that AC ll BD. 4.In the adjoining figure D is a point outside the circle and ACB = 40°.Show that ADB < 40°.

25 Do it yourself 12.1(d) 5. In the given figure if ASC = 160° and ABC = 80°.Prove that ‘S’ is the circumcentre of the ∆ABC. 6. PQ is a diameter of a circle with the centre O, and R is any other point on the circle and RPO = 25°.Calculate OQR. 7. O is the circumcentre of ∆ABC.If ABC = 32°.Calculate AOC. 8. AC and BD are chords of a circle which intersect at X.If ACD = 35° and BCA = 20 °.Calculate (i) ABD and (i) BDA. 9.’O’ is the circumcentre of ∆ABC.If AB = BC and BAC = 50°.Calculate ABC, AOC and OAC. 10.AOB is a diameter of a circle centre O.If,C is a point on the circle and BCO = 60°.Calculate OCA , OAC and AOC. 11.’O’ is the circumcentre of ∆PQR.If PQR = 40° and RPQ = 50 °.Calculate POQ.

26 END OF CHAPTER


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