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Ex 10.3, 3 - If two circles intersect at two points - Circle through 3 points

Ex 10.3, 3 - Chapter 10 Class 9 Circles - Part 2
Ex 10.3, 3 - Chapter 10 Class 9 Circles - Part 3

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Ex 10.3, 3 If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord. Given: Let circle C1 have center O & circle C2 have center X, PQ is the common chord To prove: OX is the perpendicular bisector of PQ i.e. PR = RQ ∠ PRO = ∠ PRX = ∠ QRO = ∠ QRX = 90° Construction: Join PO, PX, QO, QX Proof: In Δ POX & Δ QOX OP = OQ XP = XQ OX = OX ∴ Δ POX ≅ Δ QOX ∠ POX = ∠ QOX Also, In Δ POR & Δ QOR OP = OQ ∠ POR = ∠ QOR OR = OR ∴ Δ OPX ≅ Δ OQX PR = QR & ∠ PRO = ∠ QRO Since PQ is a line ∠ PRO + ∠ QRO = 180° ∠ PRO + ∠ PRO = 180° 2∠ PRO = 180° ∠ PRO = (180°)/2 ∠ PRO = 90° Therefore, ∠ QRO = ∠ PRO = 90° Also, ∠ PRX = ∠ QRO = 90° ∠ QRX = ∠ PRO = 90° Since ∠ PRO = ∠ QRO= ∠ PRX = ∠ QRX = 90° ∴ OX is the perpendicular bisector of PQ.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.