Last updated at Dec. 16, 2024 by Teachoo
Question 4 Form the differential equation of the family of circles touching the x-axis at origin. We know that, Equation of Circle is (๐ฅโ๐)^2+(๐ฆโ๐)^2=๐^2 Center =(๐,๐) Radius = ๐ Since the circle touches the x-axis at origin The center will be on the y-axis So, x-coordinate of center is 0 i.e. a = 0 โด Center = (0 , ๐) And, ๐๐๐๐๐ข๐ =๐ So, Equation of Circle (๐ฅโ0)^2+(๐ฆโ๐)^2=๐^2 ๐ฅ^2+(๐ฆโ๐)^2=๐^2 ๐ฅ^2+๐ฆ^2โ2๐ฆ๐+๐^2=๐^2 ๐ฅ^2+๐ฆ^2โ2๐ฆ๐=0 ๐ฅ^2+๐ฆ^2=2๐ฆ๐ Since there is one variable b, we differentiate once Diff. w.r.t. ๐ฅ 2๐ฅ+2๐ฆ ๐๐ฆ/๐๐ฅ=2๐.๐๐ฆ/๐๐ฅ ๐ฅ+๐ฆ๐ฆ^โฒ=๐.๐ฆโฒ [(๐ฅ + ๐ฆ๐ฆ^โฒ)/๐ฆ^โฒ ]=๐ Putting Value of ๐ in (1) ๐ฅ^2+๐ฆ^2=2๐ฆ[(๐ฅ + ๐ฆ๐ฆ^โฒ)/๐ฆ^โฒ ] (๐ฅ^2+๐ฆ^2 ) ๐ฆ^โฒ=2๐ฆ(๐ฅ+๐ฆ๐ฆ^โฒ ) (๐ฅ^2+๐ฆ^2 ) ๐ฆ^โฒ=2๐ฆ๐ฅ+2๐ฆ^2 ๐ฆ^โฒ ๐ฅ^2 ๐ฆ^โฒ+๐ฆ^2 ๐ฆ^โฒ=2๐ฆ๐ฅ+2๐ฆ^2 ๐ฆ^โฒ ๐ฅ^2 ๐ฆ^โฒ+๐ฆ^2 ๐ฆ^โฒโ2๐ฆ^2 ๐ฆ^โฒ=2๐ฆ๐ฅ ๐ฅ^2 ๐ฆ^โฒโ๐ฆ^2 ๐ฆ^โฒ=2๐ฆ๐ฅ ๐ฆ^โฒ (๐ฅ^2โ๐ฆ^2 )=2๐ฅ๐ฆ ๐ฒโฒ=๐๐๐/(๐^๐ โ ๐^๐ )
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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