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A Circle features…….the distance around the Circle…… its PERIMETERDiameterRadius … the distance across thecircle, passing through the centreof the circleRadiusDiameter...the distance from the centre ofthe circle to any point on thecircumference
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A Circle features…….Chord -> a line joining two pointson the Circumference.… chord divides circle intotwo segmentsARC -> part of thecircumference of a circleMajorSegmentMinorSegmentTangent -> a line which touchesthe circumference at one pointonly
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A perpendicular from the centre of a circle toa chord bisects the chord.Given: AB is a chord in a circle with centre O. OC ⊥ AB.To prove: The point C bisects the chord AB.Construction: Join OA and OBProof: In ∆ OAC and ∆ OBC,∠OCA = ∠OCB = 90…………(Given)OA = OB …………………..(Radii)OC = OC ……………….(common side)∠OAC = ∠OBC ………………………..(RHS)CA = CB (corresponding sides)The point C bisects the chord AB.Hence the theorem is proved.0A BOC
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The line drawn through the centre of a circle to bisecta chord is perpendicular to the chord.Given: AB be a chord of a circle with centre O and O is joined to the mid-point M of AB.To prove: OM ⊥ ABConstruction:Join OA and OB.Proof:In ∆ OAM & ∆ OBM,OA = OB……………. (Radii of a circle)AM = BM ……………..(Given)OM = OM (Common)Therefore, ∆OAM ≅ ∆OBM ……………………….(SSS Rule)This gives ∠OMA = ∠OMB = 90°0A BOM
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Equal chords of a circle subtend equalangles at the centre.Given: Two equal chords AB and CD of a circle with centre OTo Prove: ∠ AOB = ∠ CODIn ∆ AOB and ∆ COD,OA = OC……. (Radii of a circle)OB = OD………..(Radii of a circle)AB = CD ………...(Given)Therefore,∆ AOB ≅ ∆ COD…………. (SSS rule)This gives ∠ AOB = ∠ COD(Corresponding parts of congruent triangles)OABCD
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The angle subtended by an arc at the centre is double theangle subtended by it at any point on the remaining part ofthe circle.Given: An arc PQ of a circle subtending angles POQ at thecentre O and PAQ at a point A on the remaining part of thecircle.To prove: ∠ POQ = 2 ∠ PAQ.PQOO OPQAQAAP(i) arc PQ is minor (ii)arc PQ is a semicircle iii)arc PQ is major.
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Construction: Let us begin by joining AOand extending it to a point B.Proof: In all the cases,∠ BOQ = ∠ OAQ + ∠ AQObecause an exterior angle of a triangle isequal to the sum of the two interior opposite anglesAlso in ∆ OAQ,OA = OQ …………………… (Radii of acircle)Therefore, ∠ OAQ = ∠ OQA (Theorem )This gives ∠ BOQ = 2 ∠ OAQ ……………………….(i)Similarly, ∠ BOP = 2 ∠ OAP ……………………….(ii)AP QOBTo prove: ∠ POQ = 2 ∠ PAQ.
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From (i) and (ii) we get,∠ BOP + ∠ BOQ = 2 (∠ OAP + ∠ OAQ)Now,∠ POQ = 2 ∠ PAQ …………………..(iii)For the case (iii)where PQ is the major arc(iii) is replaced by reflex anglePOQ = 2 ∠ PAQProvedAP QOBTo prove: ∠ POQ = 2 ∠ PAQ.OQAPB
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Opposite Angles in a Cyclic Quadrilateralare supplementaryRequired to Prove that x + y = 180ºx2xy2yDraw in radiiThe angle at the centre isTWICE the angle at thecircumference2x + 2y = 360º2(x + y) = 360ºx + y = 180ºOpposite Angles in Cyclic Quadrilateral are Supplementary
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Tangent to a CircleThe tangent at any point of a circle is perpendicular to theradius through the point of contactGiven: a circle with centre O and a tangent XY to the circleat a point P.To prove: OP is perpendicular to XY.Construction: Take a point Q on XYother than P and join OQ ,The point Q must lie outside the circle.Proof: OQ is longer than the radiusOP of the circle. That is,OQ > OPOX YP Q
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Tangent to a CircleSince this happens for every point on theline XY except the point P, OP is theshortest of all the distances of the point O to thepoints of XY.So OP is perpendicular to XYOX YP Q
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Tangent to a Circle The lengths of tangents drawn from an external point to acircle are equal.Given: A circle with centre O, a point Plying outside the circle and two tangentsPQ, PR on the circle from PTo prove: PQ = PR.Construction: With centre of circle at O,draw straight lines OA and OB, Draw straight line OP.OQRP
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OQRPTangent to a CircleProof: In right triangles OQP and ORP,OQ = OR .......................(radii of the same circle)∠ OQP = ∠ ORP ………………..(right angles)OP = OP ………………(Common)Therefore,Δ OQP ≅ Δ ORP ……..(RHS)This givesPQ = PR …………..(CPCT)
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