Question 1.
\nProve that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-1.png\")
\nSolution:
\nData: Two circles having centres A and B, intersect at C and D.
\nTo Prove: \u2220ACB = \u2220ADB.
\nConstruction: Join A and B.
\nProof: In \u2206ABC and \u2206ABD,
\nAC = AD (\u2235 radii of same circle are equal)
\nBC = DD
\nAB is common.
\n\u2234 \u2206ABC \u2245 \u2206ABD (SSS Postulate.)
\n\u2234 \u2220ACB = \u2220ADB.<\/p>\n
Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. <\/p>\n","protected":false},"excerpt":{"rendered":" Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.6 Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Solution: Data: Two circles having centres A and B, intersect at C and D. To Prove: \u2220ACB = \u2220ADB. Construction: Join A and B. …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2551"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2551"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2551\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2551"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2551"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2551"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nTwo chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm., find the radius of the circle.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-2.png\")
\nSolution:
\nTwo chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre.
\nDistance between AB and CD is 6 cm.
\nTo Prove: Radius of the circle, OP =?
\nConstruction: Join OP and OQ, OB, and OD.
\nProof: Chord AB || Chord CD.
\nAB = 5 cm, and CD =11 cm.
\nOP \u22a5 AB
\n\u2234 BP = AP = \\(\\frac{5}{2}\\) = 2.5 cm.
\nOQ\u22a5CD
\n\u2234 CQ = QD = \\(\\frac{11}{2}\\) = 5.5 cm.
\nPQ = 6 cm. (Data)
\nLet OQ = 2 cm then, OP = (6 – x) cm.
\nIn \u2206BPO, \u2220P = 90\u00b0
\nAs per Pythagoras theorem,
\nOB2<\/sup> = BP2<\/sup> + PO2<\/sup>
\n= (2.5)2<\/sup> + (6 – x)2<\/sup>
\n= 6.25 + 36 – 12x + x2<\/sup>
\nOB2<\/sup> = x2<\/sup>– 12x + 42.25 …………….. (i)
\nIn \u2206OQD, \u2220Q= 90\u00b0
\n\u2234 OD2<\/sup> = OQ2<\/sup> + QD2<\/sup>
\n= (x)2<\/sup> + (5.5)2<\/sup>
\nOD2<\/sup> = x2<\/sup> + 30.25 ……………….. (ii)
\nOB = OD (\u2235 radii of same circle)
\nFrom (i) and (ii).
\nx2<\/sup> – 12x + 42.25 = x2<\/sup> + 30.25
\n-12x = 30.25 – 42.25
\n-12x = -12
\n12x = 12
\n\u2234 x = \\(\\frac{12}{12}\\)
\n\u2234 x = 1 cm.
\nFrom (ii),
\nOD2<\/sup> = x2<\/sup> + 30.25
\n= (1)2<\/sup> + 30.25
\n= 1 + 30.25
\n\u2234 OD2<\/sup> = 31.25
\nOD = \\(\\sqrt{31.25}\\)
\n\u2234 OD = 5.59 cm.
\n\u2234 Radius of circle OP = OD = 5.59 cm.<\/p>\n
\nThe lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm, from the centre, what is the distance of the other chord from the center?
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-3.png\")
\nSolution:
\nData: Chords of a circle are 6 cm. and 8 cm. are parallel. The smaller chord is at distance of 4 cm. from the centre.
\nTo Prove: Distance between bigger chord and centre =?
\nConstruction: Join OA and OC.
\nProof: AB || CD, AB = 6 cm, CD = 8 cm.
\nOP\u22a5CD, OQ\u22a5CD.
\nIn \u2206OPA, \u2220P = 90\u00b0
\n\u2234 OA2<\/sup> = OP2<\/sup> + PA2<\/sup> (According to Pythagoras theorem)
\n= (4)2 <\/sup>+ (3)2<\/sup> = 16 + 9
\nOA2<\/sup> = 25
\n\u2234 OA = 5 cm.
\nOA = OC = 5 cm. (radii of the same circle.)
\nNow, in \u2206OQC,
\nOC2<\/sup> = OQ2<\/sup> + QC2<\/sup>
\n(5)2<\/sup> = x2<\/sup> + (4)2<\/sup>
\n25 = x2<\/sup> + 16
\nx2<\/sup> = 25 – 16 = 9
\n\u2234 x = \\(\\sqrt{9}\\) \u2234 x = 3 cm.
\n\u2234 Bigger chord is at a distance of 3 cm from the centre.<\/p>\n
\nLet the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove \u2220ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-4.png\")
\nSolution:
\nData: The vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle.
\nTo Prove: \u2220ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. OR
\n\u2220ABC= \\(\\frac{1}{2}\\) [\u2220DOE – \u2220AOC].
\nConstruction: OA, OC, OE, OD are joined.
\nProof: In \u2206AOD and \u2206COE,
\nOA = OC, OD = OE radii of same circle.
\nAD = CE (Data)
\n\u2234 \u2206AOD \u2245 \u2206COE (SSS Postulate)
\n\u2220OAD = \u2220OCE ……….. (i)
\n\u2234\u2220ODA = \u2220OEC …………. (ii)
\nOA = OD
\n\u2234 \u2220OAD = \u2220ODA …………. (iii)
\nFrom (i) and (ii),
\n\u2220OAD = \u2220OCE = \u2220ODA = \u2220OEC = x\u00b0.
\nIn \u2206ODE, OD = OE
\n\u2220ODE = \u2220OED = y\u00b0.
\nADEC is a cyclic quadrilateral.
\n\u2234 \u2220CAD + \u2220DEC = 180\u00b0
\nx + a + x + y = 180
\n2x + a + y = 180
\ny = 180 – 2x – a ……….. (iv)
\nBut, \u2220DOE = 180 – 2y
\n\u2220AOC = 180 – 2a
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-5.png\")
<\/p>\n
\nProve that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-6.png\")
\nSolution:
\nData: ABCD is a rhombus. Circle is drawn taking side CD diameter. Let the diagonals AC and BD intersect at ‘O’.
\nTo Prove: Circle passes through the point ‘O’ of the intersection of its diagonals.
\nProof: \u2220DOC = 90\u00b0 (Angle in the semicircle) and diagonals of rhombus bisect at right angles at ‘O’.
\n\u2234 \u2220DOC = \u2220COB = \u2220BOA = \u2220AOD = 90\u00b0
\n\u2234 Circle passes the point of intersection of its diagonal through ‘O’.<\/p>\n
\nABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-7.png\")
\nSolution:
\nData: ABCD is a parallelogram. The circle through A, B, and C intersect CD at E. AE is joined.
\nTo Prove: AE = AD
\nProof: \u2220AEC + \u2220AED = 180\u00b0 …………. (i) (linear pair)
\nABCE is a cyclic quadrilateral.
\n\u2234 \u2220ABC + \u2220AEC = 180\u00b0 ………….(ii) (opposite angles)
\nComparing (i) and (ii),
\n\u2220AEC = \u2220AED = \u2220ABC + \u2220AEC
\n\u2220AED = \u2220ABC ………….. (iii)
\nBut, \u2220ABC = \u2220ADE (Opposite angles of quadrilateral)
\nSubstituting in equation (iii),
\n\u2220AED = \u2220ADE
\n\u2234 AE = AD.<\/p>\n
\nAC and BD are chords of a circle which bisect each other. Prove that
\n(i) AC and BD are diameters,
\n(ii) ABCD is a rectangle.
\nSolution:
\nData : AC and BD are chords of a circle bisect each other at ‘O’.
\nTo Prove:
\ni) AC and BD are diameters.
\nii) ABCD is a rectangle.
\nConstruction: AB, BC, CD and DA are joined.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-8.png\")
\nProof: In \u2206AOB and \u2206COD,
\nAO = OC (Data)
\nBO = OD (Data)
\n\u2220AOB = \u2220COD (vertically opposite angles)
\n\u2234 \u2206AOB \u2245 \u2206COD (SAS postulate)
\n\u2234 \u2220OAB = \u2220OCD
\nThese are pair of alternate angles.
\n\u2234 AB || CD and AB = CD.
\n\u2234 ABCD is a parallelogram.
\n\u2234 \u2220BAD = \u2220BCD (Opposite angles of parallelogram)
\nBut. \u2220BAD + \u2220BCD = 180 (\u2235 Angles of cyclic quadrilateral)
\n\u2220BAD + \u2220BAD = 180
\n2(\u2220BAD) = 180
\n\u2234 \u2220BAD = \\(\\frac{180}{2}\\)
\n\u2234 \u2220BAD = 90\u00b0.
\nIf angles of a quadrilateral are right angles it is rectangle. ABCD is a recrtangle.
\n\u2220BAD = 90\u00b0
\n\u2220BAD is separated from chord BD.
\n\u2234 This is the angl in semicircle.
\n\u2234 Chord BD is a diameter.
\nSimilarly, \u2220ADC = 90\u00b0
\n\u2234 Chord AC is a diameter.<\/p>\n
\nBisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90\u00b0 – \\(\\frac{1}{2}\\) A, 90\u00b0 – \\(\\frac{1}{2}\\) B and 90\u00b0 – \\(\\frac{1}{2}\\) C.
\nSolution:
\nData: AD, BE and CF are angular bisectors of angles A, B and C of \u2206ABC intersects its circumference at D, E and F respectively.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-9.png\")
\nTo prove: Angles of \u2206DEF are 90\u00b0 – \\(\\frac{1}{2}\\) A, 90\u00b0 – \\(\\frac{1}{2}\\) B and 90\u00b0 – \\(\\frac{1}{2}\\) C.
\nProof: AD, BE and CF are angular bisectors of angles A, B and C of \u2206ABC.
\n\u2234 \u2220BAD = \u2220CAD = \\(\\frac{\\angle \\mathrm{A}}{2}\\)
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-10.png\")
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-11.png\")
<\/p>\n
\nTwo congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lies on the two circles.
\nProve that BP = BQ.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-12.png\")
\nSolution:
\nData : Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.
\nTo Prove: BP = BQ
\nConstruction: Join AB.
\nProof: Two congruent triangles with centres O and O’ intersects at A and B. Through A segment PAQ is drawn so that P, Q lie on the two circles.
\nSimilarly, \u2220AQB= 70\u00b0 in circle subtended by chord AB. Because Angles subtended by circumference by same chord.
\n\u2234 \u2220APB = \u2220AQB = 70\u00b0.
\nNow, in \u2206PBQ, \u2220QPB = \u2220PQB.
\n\u2234 Sides opposite to each other are equal.
\n\u2234 BP = BQ.<\/p>\n
\nIn any triangle ABC, if the angle bisector of \u2220A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Maths-Chapter-12-Circles-Ex-12.6-13.png\")
\nSolution:
\nData: In \u2206ABC, if the angle bisector of \u2220A and perpendicular bisector of BC intersect each other. O is the centre of the circle.
\nTo Prove: Angle bisector of \u2220A and perpendicular bisector of BC intersect at D.
\nConstruction: Join OB, OC.
\nProof: Angle subtended at the Centre
\n= 2 \u00d7 angles subtended in the circumference.
\n\u2220BOC = 2 \u00d7 \u2220BAC
\nIn \u2206BOE and \u2206COE,
\n\u2220OEB = \u2220OEC = 90\u00b0 (\u2235 OE\u22a5BC)
\n\u2234 BO = OC (radii)
\nOE is common.
\n\u2234 \u2206BOE \u2245 \u2206COE (RHS postulate)
\nBut, \u2220BOE + \u2220COE = \u2220BOC
\n\u2220BOE + \u2220BOE = \u2220BOC
\n2\u2220BOE = \u2220BOC
\n2\u2220BOE = 2\u2220BAC
\n\u2234 \u2220BOE = \u2220BAC
\nBut, \u2220BOE = \u2220COE = \u2220BAC
\n\u2220BAD = \\(\\frac{1}{2}\\) \u2220BAC
\n\u2220BAD = \\(\\frac{1}{2}\\) \u2220BOE
\n\u2220BAD = \\(\\frac{1}{2}\\) \u2220BOD
\n\u2234 \u2220BOD = 2\u2220BAD
\n\u2234 The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference.
\n\u2234 Angle bisector of \u2220A and perpendicular bisector of BC intersect at D.<\/p>\nKSEEB Solutions for Class 9 Maths<\/a><\/h5>\n