Last updated at Feb. 8, 2025 by Teachoo
Ex 9.2, 1 Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. Given: Circle C1 with radii 5cm & C2 with radii 3cm Intersecting at P & Q. OP = 5cm , XP = 3cm & OX = 4cm To find: Length of common chord i.e., length of PQ Solution: Let the point where OX intersects PQ be R. 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 POR ≅ Δ QOR ⇒ ∠ PRO = ∠ QRO & PR = RQ 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° Let OR = x, So, XR = OX – OR = 4 – x Now, From (4) & (5) 52 – x2 = –7 – x2 + 8x 25 – x2 = –7 – x2 + 8x 25 + 7 – x2 + x2 = 8x 32 = 8x 8x = 32 x = 32/8 x = 4 Putting value of x in (4) PR2 = 25 – x2 PR2 = 25 – 42 PR2 = 25 – 16 PR2 = 9 PR = √9 = 3 ∴ PQ = 2PR = 2 × 3 = 6 Hence, length of common chord = 6 m Note: OR = x = 4 cm & XR = 4 – x = 4 – 4 = 0 cm since XR = 0 this means point X & R coincide Hence actual figure is as follows
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo