Last updated at Nov. 14, 2019 by

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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 (π₯βπ)^2+(π¦βπ)^2=π^2 Center =(π,π) Radius = π 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 β΄ Center = (0 , π) And, πππππ’π =π So, Equation of Circle (π₯β0)^2+(π¦βπ)^2=π^2 π₯^2+(π¦βπ)^2=π^2 π₯^2+π¦^2β2π¦π+π^2=π^2 π₯^2+π¦^2β2π¦π=0 π₯^2+π¦^2=2π¦π Since there is one variable b, we differentiate once Diff. w.r.t. π₯ 2π₯+2π¦ ππ¦/ππ₯=2π.ππ¦/ππ₯ π₯+π¦π¦^β²=π.π¦β² [(π₯ + π¦π¦^β²)/π¦^β² ]=π Putting Value of π in (1) π₯^2+π¦^2=2π¦[(π₯ + π¦π¦^β²)/π¦^β² ] (π₯^2+π¦^2 ) π¦^β²=2π¦(π₯+π¦π¦^β² ) (π₯^2+π¦^2 ) π¦^β²=2π¦π₯+2π¦^2 π¦^β² π₯^2 π¦^β²+π¦^2 π¦^β²=2π¦π₯+2π¦^2 π¦^β² π₯^2 π¦^β²+π¦^2 π¦^β²β2π¦^2 π¦^β²=2π¦π₯ π₯^2 π¦^β²βπ¦^2 π¦^β²=2π¦π₯ π¦^β² (π₯^2βπ¦^2 )=2π₯π¦ π²β²=πππ/(π^π β π^π )

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Example 4 Deleted for CBSE Board 2022 Exams

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Example 7 Deleted for CBSE Board 2022 Exams You are here

Example 8 Deleted for CBSE Board 2022 Exams

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Example 28 Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 12 Differential Equations (Term 2)

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About the Author

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.