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Ex 8.2 , 6 Smaller area enclosed by the circle ๐‘ฅ2+๐‘ฆ2 =4 and the line ๐‘ฅ+๐‘ฆ=2 is (A) 2 (ฯ€ โ€“ 2) (B) ฯ€ โ€“ 2 (C) 2ฯ€ โ€“ 1 (D) 2 (ฯ€ + 2) Step 1: Drawing figure Circle is ๐‘ฅ2+๐‘ฆ2 =4 (๐‘ฅโˆ’0)2+(๐‘ฆโˆ’0)2 = 2๏ทฎ2๏ทฏ So, Center = (0, 0) & Radius = 2 Also, ๐‘ฅ+๐‘ฆ=2 passes through (0, 2) & (2, 0) Hence, intersecting points A = (2, 0) & B = (0, 2) Thus, Area required = Area OACB โˆ’ AREA OAB = ๐œ‹ โˆ’ 2 Thus, Option (B) is correct.

  1. Chapter 8 Class 12 Application of Integrals
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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo