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Ex 9.3, 6 - Form differential equation of family of circles

Ex 9.3, 6 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.3, 6 - Chapter 9 Class 12 Differential Equations - Part 3

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Question 6 Form the differential equation of the family of circle touching the π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 at origin. General Equation of Circle (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’π‘)^2=π‘Ÿ^2 where Centre at (π‘Ž , 𝑏) and Radius is r If circle touches y-axis at origin, Center will be at x-axis So, Center = (a, 0) & Radius = a Thus, equation of circle becomes (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’0)^2=π‘Ž^2 (π‘₯βˆ’π‘Ž)^2+𝑦^2=π‘Ž^2 π‘₯^2+π‘Ž^2βˆ’2π‘Žπ‘₯+𝑦^2=π‘Ž^2 π‘₯^2βˆ’2π‘Žπ‘₯+𝑦^2=π‘Ž^2βˆ’π‘Ž^2 π‘₯^2βˆ’2π‘Žπ‘₯+𝑦^2=0 2π‘Žπ‘₯=π‘₯^2+𝑦^2 Differentiating Both Sides w.r.t. π‘₯ (𝑑(2π‘Žπ‘₯))/𝑑π‘₯=𝑑(π‘₯^2 )/𝑑π‘₯+𝑑(𝑦^2 )/𝑑π‘₯ 2a = 2x + 2y 𝑑𝑦/𝑑π‘₯ a = x + yy’ …(1) …(2) From (1) 2π‘Žπ‘₯=π‘₯^2+𝑦^2 Putting value of a from (2) 2π‘₯(π‘₯+𝑦𝑦^β€²)=π‘₯^2+𝑦^2 2π‘₯^2+2π‘₯𝑦𝑦^β€²=π‘₯^2+𝑦^2 2π‘₯^2βˆ’π‘₯^2+2π‘₯𝑦𝑦^β€²=+𝑦^2 πŸπ’™π’šπ’š^β€²+𝒙^𝟐=π’š^𝟐 is the required differential equation.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.