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Question 4 Form the differential equation of the family of circles touching the x-axis at origin. We know that, Equation of Circle is (๐‘ฅโˆ’๐‘Ž)^2+(๐‘ฆโˆ’๐‘)^2=๐‘Ÿ^2 Center =(๐‘Ž,๐‘) Radius = ๐‘Ÿ 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 โˆด Center = (0 , ๐‘) And, ๐‘Ÿ๐‘Ž๐‘‘๐‘–๐‘ข๐‘ =๐‘ So, Equation of Circle (๐‘ฅโˆ’0)^2+(๐‘ฆโˆ’๐‘)^2=๐‘^2 ๐‘ฅ^2+(๐‘ฆโˆ’๐‘)^2=๐‘^2 ๐‘ฅ^2+๐‘ฆ^2โˆ’2๐‘ฆ๐‘+๐‘^2=๐‘^2 ๐‘ฅ^2+๐‘ฆ^2โˆ’2๐‘ฆ๐‘=0 ๐‘ฅ^2+๐‘ฆ^2=2๐‘ฆ๐‘ Since there is one variable b, we differentiate once Diff. w.r.t. ๐‘ฅ 2๐‘ฅ+2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2๐‘.๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฅ+๐‘ฆ๐‘ฆ^โ€ฒ=๐‘.๐‘ฆโ€ฒ [(๐‘ฅ + ๐‘ฆ๐‘ฆ^โ€ฒ)/๐‘ฆ^โ€ฒ ]=๐‘ Putting Value of ๐‘ in (1) ๐‘ฅ^2+๐‘ฆ^2=2๐‘ฆ[(๐‘ฅ + ๐‘ฆ๐‘ฆ^โ€ฒ)/๐‘ฆ^โ€ฒ ] (๐‘ฅ^2+๐‘ฆ^2 ) ๐‘ฆ^โ€ฒ=2๐‘ฆ(๐‘ฅ+๐‘ฆ๐‘ฆ^โ€ฒ ) (๐‘ฅ^2+๐‘ฆ^2 ) ๐‘ฆ^โ€ฒ=2๐‘ฆ๐‘ฅ+2๐‘ฆ^2 ๐‘ฆ^โ€ฒ ๐‘ฅ^2 ๐‘ฆ^โ€ฒ+๐‘ฆ^2 ๐‘ฆ^โ€ฒ=2๐‘ฆ๐‘ฅ+2๐‘ฆ^2 ๐‘ฆ^โ€ฒ ๐‘ฅ^2 ๐‘ฆ^โ€ฒ+๐‘ฆ^2 ๐‘ฆ^โ€ฒโˆ’2๐‘ฆ^2 ๐‘ฆ^โ€ฒ=2๐‘ฆ๐‘ฅ ๐‘ฅ^2 ๐‘ฆ^โ€ฒโˆ’๐‘ฆ^2 ๐‘ฆ^โ€ฒ=2๐‘ฆ๐‘ฅ ๐‘ฆ^โ€ฒ (๐‘ฅ^2โˆ’๐‘ฆ^2 )=2๐‘ฅ๐‘ฆ ๐ฒโ€ฒ=๐Ÿ๐’™๐’š/(๐’™^๐Ÿ โˆ’ ๐’š^๐Ÿ )

  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

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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