Last updated at Nov. 14, 2019 by Teachoo

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Example 7 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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Chapter 9 Class 12 Differential Equations

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.