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# Circles

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### Circles

1. 1. T- 1-855-694-8886Email- info@iTutor.comBy iTutor.com
2. 2. 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
3. 3. 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
4. 4. 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
5. 5. 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
6. 6. 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
7. 7. 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.
8. 8. 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.
9. 9. 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
10. 10. 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
11. 11. 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
12. 12. 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
13. 13. 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
14. 14. 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)
15. 15. The EndCall us for moreinformation:www.iTutor.com1-855-694-8886Visit
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