Circles-GEOMETRY
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1. The document defines various terms related to circles such as radius, diameter, chord, arc, sector, segment, and circumference. 2. It states several properties of circles including that all radii of a circle are equal, the diameter of a circle is twice the radius, equal chords of a circle subtend equal angles at the centre, and there is one and only one circle passing through three non-collinear points. 3. Examples are provided to illustrate properties such as two arcs being congruent if their corresponding chords are equal, and the perpendicular drawn from the centre of a circle to a chord bisects the chord.
![Mathematics Secondary Course388
Notes
MODULE - 3
Geometry
Circles
(v) Measure ∠ OMP or ∠ OMQ with set square or protractor.
We observe that ∠ OMP = ∠ OMQ = 900
.
The line joining the centre of a circle to the mid point of a chord is perpen-
dicular to the chord.
Activity for you :
Take three non collinear pointsA, B and C. JoinAB and
BC. Draw perpendicular bisectors MN and RS of AB
and BC respectively.
SinceA, B, C are not collinear, MN is not parallel to RS.
TheywillintersectonlyatonepointO.JoinOA,OBand
OC and measure them.
We observe that OA = OB = OC
Now taking O as the centre and OAas radius draw a circle which passes throughA, B
and C.
Repeat the above procedure with another three non-collinear points and observe that
thereisonlyonecirclepassingthroughthreegivennon-collinearpoints.
There is one and only one circle passing through three non-collinear points.
Note. It is important to note that a circle can not be drawn to pass through three collinear
points.
Activity for you :
(i) Draw a circle with centre O [Fig. 15.20a]
(ii) Draw two equal chordsAB and PQ of the circle.
(iii) Draw OM ⊥ PQ and ON ⊥ PQ
(iv) Measure OM and ON and observe that they are equal.
Equal chords of a circle are equidistant from the centre.
In Fig. 15.20 b, OM = ON
Measure and observe thatAB = PQ. Thus,
Chords, that are equidistant from the centre of a
circle, are equal.
The above results hold good in case of congruent circles
also.
We now take a few examples using these properties of
circle.
Fig. 15.20
Fig. 15.20a
Fig. 15.20b](https://image.slidesharecdn.com/circles-150531165605-lva1-app6891/85/Circles-GEOMETRY-8-320.jpg)
Circles-GEOMETRY
- 1. Mathematics Secondary Course 381 Notes MODULE - 3 Geometry Circles 15 CIRCLES Youarealreadyfamiliarwithgeometricalfiguressuchasalinesegment,anangle,atriangle, aquadrilateralandacircle.Commonexamplesofacircleareawheel,abangle,alphabet O, etc. In this lesson we shall study in some detail about the circle and related concepts. OBJECTIVES Afterstudyingthislesson,youwillbeableto • define a circle • give examples of various terms related to a circle • illustrate congruent circles and concentric circles • identify and illustrate terms connected with circles like chord, arc, sector, segment, etc. • verify experimentally results based on arcs and chords of a circle • use the results in solving problems EXPECTED BACKGROUND KNOWLEDGE • Linesegmentanditslength • Angleanditsmeasure • Parallelandperpendicularlines • Closedfiguressuchastriangles,quadrilaterals,polygons,etc. • Perimeterofaclosedfigure • Region bounded by a closed figure • Congruenceofclosedfigures
- 2. Mathematics Secondary Course382 Notes MODULE - 3 Geometry Circles 15.1 CIRCLE AND RELATED TERMS 15.1.1 Circle Acircleisacollectionofallpointsinaplanewhichareat a constant distance from a fixed point in the same plane. Radius:Alinesegmentjoiningthecentreofthecircleto a point on the circle is called its radius. In Fig. 15.1, there is a circle with centre O and one of its radius is OA. OB is another radius of the same circle. Activity for you : Measure the length OA and OB and observe that they are equal. Thus All radii (plural of radius) of a circle are equal Thelengthoftheradiusofacircleisgenerallydenotedby the letter ‘r’. It is customry to write radius instead of the lengthoftheradius. A closed geometric figure in the plane divides the plane into three parts namely, the inner part of the figure, the figure and the outer part. In Fig. 15.2, the shaded portion is the inner part of the circle, the boundary is the circle and the unshaded portion is the outer part of the circle. Activity for you (a) Take a point Q in the inner part of the circle (See Fig. 15.3). Measure OQ and find that OQ < r. The inner part of the circle is called the interior of the circle. (b) Now take a point P in the outer part of the circle (Fig. 15.3). Measure OPand find that OP> r. The outer part of the circle is called the exterior of the circle. 15.1.2 Chord Alinesegmentjoininganytwopointsofacircleiscalled a chord. In Fig. 15.4,AB, PQ and CD are three chords of a circle with centre O and radius r. The chord PQ passes through the centre O of the circle. Such a chord is called a diameter of the circle. Diameter is usually de- noted by ‘d’. Fig. 15.1 Fig. 15.2 Fig. 15.3 Fig. 15.4
- 3. Mathematics Secondary Course 383 Notes MODULE - 3 Geometry Circles Achord passing though the centre of a circle is called its diameter. Activity for you : Measure the length d of PQ, the radius r and find that d is the same as 2r. Thus we have d = 2r i.e. the diameter of a circle = twice the radius of the circle. Measure the length PQ, AB and CD and find that PQ > AB and PQ > CD, we may conclude Diameter is the longest chord of a circle. 15.1.3 Arc A part of a circle is called an arc. In Fig. 15.5(a) ABC is an arc and is denoted by arc ABC (a) (b) Fig. 15.5 15.1.4 Semicircle A diameter of a circle divides a circle into two equal arcs, each of which is known as a semicircle. In Fig. 15.5(b), PQ is a diameter and PRQ is semicircle and so is PBQ. 15.1.5 Sector The region bounded by an arc of a circle and two radii at its end points is called a sector. In Fig. 15.6, the shaded portion is a sector formed by the arc PRQ and the unshaded portion is a sector formed by the arc PTQ. 15.1.6 Segment A chord divides the interior of a circle into two parts, Fig. 15.6
- 4. Mathematics Secondary Course384 Notes MODULE - 3 Geometry Circles eachofwhich iscalledasegment.InFig.15.7,theshaded region PAQP and the unshaded region PBQP are both segments of the circle. PAQPis called a minor segment and PBQP is called a major segment. 15.1.7 Circumference ChooseapointPonacircle.Ifthispointmovesalongthe circle once and comes back to its original position then the distance covered by P is called the circumference of thecircle Fig. 15.8 Activity for you : Take a wheel and mark a point Pon the wheel where it touches the ground. Rotate the wheel along a line till the point P comes back on the ground. Measure the distance be- tween the Ist and last position of P along the line. This distance is equal to the circumfer- ence of the circle. Thus, The length of the boundary of a circle is the circumference of the circle. Activity for you Consider different circles and measures their circumference(s) and diameters. Observe that in each case the ratio of the circumference to the diameter turns out to be the same. The ratio of the circumference of a circle to its diameter is always a con- stant. This constant is universally denoted by Greek letter π . Therefore, π== 2r c d c , where c is the circumference of the circle, d its diameter and r is itsradius. An approximate value of π is 7 22 .Aryabhata -I (476A.D.), a famous Indian Mathema- tician gave a more accurate value of π which is 3.1416. In fact this number is an irrationalnumber. Fig. 15.7
- 5. Mathematics Secondary Course 385 Notes MODULE - 3 Geometry Circles 15.2 MEASUREMENT OF AN ARC OF A CIRCLE Consider an arc PAQ of a circle (Fig. 15.9). To measure its length we put a thread along PAQ and then measure the length of the thread with the help of a scale. Similarly,youmaymeasurethelengthofthearcPBQ. 15.2.1 Minor arc An arc of circle whose length is less than that of a semi- circle of the same circle is called a minor arc. PAQ is a minor arc (See Fig. 15.9) 15.2.2 Major arc An arc of a circle whose length is greater than that of a semicircle of the same circle is called a major arc. In Fig. 15.9, arc PBQ is a major arc. 15.3 CONCENTRIC CIRCLES Circleshavingthesamecentrebutdifferentradiiarecalled concentric circles (See Fig. 15.10). 15.4 CONGRUENT CIRCLES OR ARCS Two cirlces (or arcs) are said to be congruent if we can superimpose (place) one over the other such that they cover each other completely. 15.5 SOME IMPORTANT RULES Activity for you : (i)DrawtwocircleswithcenreO1 andO2 and radius r and s respectively (See Fig. 15.11) Fig. 15.11 Fig. 15.9 Fig. 15.10
- 6. Mathematics Secondary Course386 Notes MODULE - 3 Geometry Circles (ii) Superimpose the circle (i) on the circle (ii) so that O1 coincides with O2 . (iii)We observe that circle (i) will cover circle (ii) if and only if r = s Two circles are congurent if and only if they have equal radii. In Fig. 15.12 if arc PAQ = arc RBS then ∠ POQ = ∠ ROS and conversely if ∠ POQ = ∠ ROS then arc PAQ = arc RBS. Two arcs of a circle are congurent if and only if the angles subtended by them at the centre are equal. In Fig. 15.13, if arc PAQ = arc RBS then PQ = RS and conversely if PQ = RS then arc PAQ = arc RBS. Two arcs of a circle are congurent if and only if their corresponding chords are equal. Activity for you : (i) Draw a circle with centre O (ii) Draw equal chords PQ and RS (See Fig. 15.14) (iii) Join OP, OQ, OR and OS (iv) Measure ∠ POQ and ∠ ROS We observe that ∠ POQ = ∠ ROS Conversely if ∠ POQ = ∠ ROS then PQ = RS Equal chords of a circle subtend equal angles at the centre and conversely if the angles subtended by the chords at the centre of a circle are equal, then the chords are equal. Note : The above results also hold good in case of congruent circles. We take some examples using the above properties : Fig. 15.12 Fig. 15.13 Fig. 15.14
- 7. Mathematics Secondary Course 387 Notes MODULE - 3 Geometry Circles Example 15.1 : In Fig. 15.15, chord PQ = chord RS. Show that chord PR = chord QS. Solution : The arcs corresponding to equal chords PQ and RS are equal. Add to each arc, the arc QR, yielding arc PQR = arc QRS ∴chord PR = chord QS Example 15.2 : In Fig. 15.16, arc AB = arc BC, ∠ AOB = 30o and ∠ AOD = 70o . Find ∠ COD. Solution : Since arcAB = arc BC ∴ ∠ AOB = ∠ BOC (Equals arcs subtend equal angles at the centre) ∴ ∠ BOC = 30o Now ∠ COD = ∠ COB + ∠ BOA + ∠ AOD = 30o + 30o + 70o = 130o . Activity for you : (i) Draw a circle with centre O (See Fig. 15.17). (ii) Draw a chord PQ. (iii) From O draw ON ⊥ PQ (iv) Measure PN and NQ Youwillobservethat PN = NQ. The perpendicular drawn from the centre of a circle to a chord bisects the chord. Activity for you : (i) Draw a circle with centre O (See Fig. 15.18). (ii) Draw a chord PQ. (iii) Find the mid point M of PQ. (iv) Join O and M. Fig. 15.15 Fig. 15.16 Fig. 15.17 Fig. 15.18
- 8. Mathematics Secondary Course388 Notes MODULE - 3 Geometry Circles (v) Measure ∠ OMP or ∠ OMQ with set square or protractor. We observe that ∠ OMP = ∠ OMQ = 900 . The line joining the centre of a circle to the mid point of a chord is perpen- dicular to the chord. Activity for you : Take three non collinear pointsA, B and C. JoinAB and BC. Draw perpendicular bisectors MN and RS of AB and BC respectively. SinceA, B, C are not collinear, MN is not parallel to RS. TheywillintersectonlyatonepointO.JoinOA,OBand OC and measure them. We observe that OA = OB = OC Now taking O as the centre and OAas radius draw a circle which passes throughA, B and C. Repeat the above procedure with another three non-collinear points and observe that thereisonlyonecirclepassingthroughthreegivennon-collinearpoints. There is one and only one circle passing through three non-collinear points. Note. It is important to note that a circle can not be drawn to pass through three collinear points. Activity for you : (i) Draw a circle with centre O [Fig. 15.20a] (ii) Draw two equal chordsAB and PQ of the circle. (iii) Draw OM ⊥ PQ and ON ⊥ PQ (iv) Measure OM and ON and observe that they are equal. Equal chords of a circle are equidistant from the centre. In Fig. 15.20 b, OM = ON Measure and observe thatAB = PQ. Thus, Chords, that are equidistant from the centre of a circle, are equal. The above results hold good in case of congruent circles also. We now take a few examples using these properties of circle. Fig. 15.20 Fig. 15.20a Fig. 15.20b
- 9. Mathematics Secondary Course 389 Notes MODULE - 3 Geometry Circles Examples 15.3 : In Fig. 15.21, O is the centre of the circle and ON ⊥ PQ. If PQ = 8 cm and ON = 3 cm, find OP. Solution: ON ⊥ PQ (given) and since perpendicular drawn from the centre of a circle to a chord bisects the chord. ∴ PN = NQ = 4 cm In a right triangle OPN, ∴ OP2 = PN2 + ON2 or OP2 = 42 + 32 = 25 ∴ OP = 5 cm. Examples 15.4 : In Fig. 15.22, OD is perpendicular to the chordAB of a circle whose centre is O and BC is a diameter. Prove that CA= 2OD. Solution : Since OD ⊥ AB (Given) ∴ D is the mid point ofAB (Perpendicularthroughthecentrebisectsthe chord) Also O is the mid point of CB (Since CB is a diameter) Now in ΔABC, O and D are mid points of the two sides BC and BAof the triangleABC. Sincethelinesegmentjoiningthemidpointsofanytwosidesofatriangleisparalleland halfofthethirdside. ∴ OD = 2 1 CA i.e. CA = 2OD. Example 15.5 :Aregular hexagon is inscribed in a circle. What angle does each side of the hexagon subtend at the centre? Solution:Aregularhexagonhassixsideswhichareequal. Thereforeeachsidesubtendsthesameangleatthecentre. Let us suppose that a side of the hexagon subtends an angle xo at the centre. Then,wehave 6xo = 360o ⇒ x = 60o Hence, each side of the hexagon subtends an angle of 60o at the centre. Fig. 15.21 Fig. 15.22 Fig. 15.23
- 10. Mathematics Secondary Course390 Notes MODULE - 3 Geometry Circles Example 15.6 : In Fig. 15.24, two parallel chords PQ and AB of a circle are of lengths 7 cm and 13 cm respectively.IfthedistancebetweenPQandABis3cm, findtheradiusofthecircle. Solution : Let O be the centre of the circle. Draw perpendicular bisector OLof PQ which also bisectsAB at M. Join OQ and OB (Fig. 15.24) Let OM = x cm and radius of the circle be r cm Then OB2 = OM2 +MB2 and OQ2 = OL2 + LQ2 ∴ 2 22 2 13 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += xr ...(i) and 2 22 2 7 )3( ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++= xr ...(ii) Thereforefrom(i)and(ii), 2 2 2 2 2 7 )3( 2 13 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + xx ∴ 4 49 9 4 169 6 −−=x or 6x = 21 ∴ 2 7 =x ∴ 4 218 4 169 4 49 2 13 2 7 22 2 =+=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =r ∴ 2 218 =r Hence the radius of the circle is 2 218 =r cm. Fig. 15.24
- 11. Mathematics Secondary Course 391 Notes MODULE - 3 Geometry Circles CHECK YOUR PROGRESS 15.1 In questions 1 to 5, fill in the blanks to make each of the statementstrue. 1. In Fig. 15.25, (i)AB is a ... of the circle. (ii) Minor arc corresponding toAB is... . 2. A ... is the longest chord of a circle. 3. The ratio of the circumference to the diameter of a circle is always ... . 4. The value of π as 3.1416 was given by great Indian Mathematician... . 5. Circles having the same centre are called ... circles. 6. Diameterofacircleis30cm.Ifthelengthofachordis20cm,findthedistanceofthe chord from the centre. 7. Findthecircumferenceofacirclewhoseradiusis (i) 7 cm (ii) 11cm. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 7 22 Take π 8. In the Fig. 15.26, RS is a diameter which bisects the chords PQ andAB at the points M and N respectively. Is PQ ||AB ? Given reasons. Fig. 15.26 Fig. 15.27 9. In Fig. 15.27, a line l intersects the two concentric circles with centre O at pointsA, B, C and D. Is AB = CD ? Give reasons. LET US SUM UP • The circumference of a circle of radius r is equal to 2π r. Fig. 15.25
- 12. Mathematics Secondary Course392 Notes MODULE - 3 Geometry Circles • Twoarcsofacirclearecongurentifandonlyifeithertheanglessubtendedbythemat the centre are equal or their corresponding chords are equal. • Equal chords of a circle subtend equal angles at the centre and vice versa. • Perpendicular drawn from the centre of a circle to a chord bisects the chord. • Thelinejoiningthecentreofacircletothemidpointofachordisperpendiculartothe chord. • Thereisoneandonlyonecirclepassngthroughthreenon-collinearpoints. • Equal chords of a circle are equidistant from the centre and the converse. TERMINAL EXERCISE 1. If the length of a chord of a circle is 16 cm and the distance of the chord from the centre is 6 cm, find the radius of the circle. 2. TwocircleswithcentresOandO′ (SeeFig.15.28)arecongurent.Findthelengthof the arc CD. Fig. 15.28 3. A regular pentagon is inscribed in a circle. Find the angle which each side of the pentagon subtends at the centre. 4. In Fig. 15.29, AB = 8 cm and CD = 6 cm are two parallel chords of a circle with centre O. Find the distance between the chords. Fig. 15.29
- 13. Mathematics Secondary Course 393 Notes MODULE - 3 Geometry Circles 5. In Fig.15.30 arc PQ = arc QR, ∠ POQ = 150 and ∠ SOR = 110o . Find ∠ SOP. Fig. 15.30 6. In Fig. 15.31,AB and CD are two equal chords of a circle with centre O. Is chord BD = chord CA? Give reasons. Fig. 15.31 7. If AB and CD are two equal chords of a circle with centre O (Fig. 15.32) and OM ⊥ AB, ON ⊥ CD. Is OM = ON ? Give reasons. Fig. 15.32 8. In Fig. 15.33,AB = 14 cm and CD = 6 cm are two parallel chords of a circle with centre O. Find the distance between the chordsAB and CD. Fig. 15.33
- 14. Mathematics Secondary Course394 Notes MODULE - 3 Geometry Circles 9. In Fig. 15.34,AB and CD are two chords of a circle with centre O, intersecting at a pointPinsidethecircle. Fig. 15.34 OM ⊥ CD, ON ⊥ AB and ∠ OPM = ∠ OPN. Now answer: Is (i) OM = ON, (ii)AB = CD ? Give reasons. 10. C1 andC2 areconcentriccircleswithcentreO(SeeFig.15.35), lisalineintersecting C1 at points P and Q and C2 at points Aand B respectively, ON ⊥ l, is PA= BQ? Givereasons. Fig. 15.35 ANSWERS TO CHECK YOUR PROGRESS 15.1 1. (i) Chord (ii)APB 2.Diameter 3. Constant 4. Aryabhata-I 5.Concentric 6. 5 5 cm. 7. (i) 44 cm (ii) 69.14 cm 8.Yes 9.Yes ANSWERS TO TERMINAL EXERCISE 1. 10 cm 2. 2a cm 3. 72o 4. 1 cm 5. 80o 6. Yes (Equal arcs have corresponding equal chrods of acircle) 7. Yes (equal chords are equidistant from the centre of the circle) 8. 10 2 cm 9. (i)Yes (ii) Yes (ΔOMP ≅ ΔONP) 10. Yes (N is the middle point of chords PQ andAB).