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Misc 5 - Family of circles in first quadrant which touch axes

Misc 5 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 5 - Chapter 9 Class 12 Differential Equations - Part 3


Transcript

Misc 5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes . Let’s first draw the figure Let C be the family of circles in first quadrant touching coordinate axes Let radius be π‘Ž ∴ Center of circle = (βˆ’π‘Ž, π‘Ž) Thus, Equation of a circle is (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’π‘Ž)^2=π‘Ž^2 π‘₯^2+π‘Ž^2βˆ’2π‘Žπ‘₯+𝑦^2+π‘Ž^2βˆ’2π‘Žπ‘¦=π‘Ž^2 π‘₯^2+𝑦^2βˆ’2π‘Žπ‘₯βˆ’2π‘Žπ‘¦+π‘Ž^2=0 Since there is only one variable, we differentiate once Differentiating w.r.t. π‘₯ (π‘₯^2 + 𝑦^2 βˆ’ 2π‘Žπ‘₯ βˆ’ 2π‘Žπ‘¦+π‘Ž^2 )^β€²=(0)^β€² 2π‘₯+2π‘¦π‘¦β€²βˆ’2π‘Žβˆ’2π‘Žπ‘¦β€²+0=0 π‘₯+π‘¦π‘¦β€²βˆ’π‘Žβˆ’π‘Žπ‘¦β€²=0 π‘₯+𝑦𝑦^β€²=π‘Ž+π‘Žπ‘¦^β€² π‘₯+𝑦𝑦^β€²=π‘Ž(1+𝑦^β€²) π‘Ž = (π‘₯ + 𝑦𝑦^β€²)/(1 + 𝑦^β€² ) Putting Value of π‘Ž in (π‘₯βˆ’π‘Ž)^2+(π‘¦βˆ’π‘Ž)^2=π‘Ž^2 (π‘₯βˆ’[(π‘₯ + 𝑦 𝑦^β€²)/(1 + 𝑦^β€² )])^2+(π‘¦βˆ’[(π‘₯ + 𝑦 𝑦^β€²)/(1 + 𝑦^β€² )])^2=((π‘₯ + 𝑦𝑦^β€²)/(1 + 𝑦^β€² ))^2 (π‘₯(1 + 𝑦^β€² ) βˆ’ (π‘₯ + 𝑦𝑦^β€² ))^2/(1 + 𝑦^β€² )^2 +(𝑦(1 + 𝑦^β€² ) βˆ’ (π‘₯ + 𝑦𝑦^β€² ))^2/(1 + 𝑦^β€² )^2 =(π‘₯ + 𝑦𝑦^β€² )^2/(1 + 𝑦^β€² )^2 (π‘₯+π‘₯𝑦^β€²βˆ’π‘₯βˆ’π‘¦π‘¦^β€² )^2+(𝑦+𝑦𝑦^β€²βˆ’π‘₯βˆ’π‘¦π‘¦^β€² )^2=(π‘₯+𝑦𝑦^β€² )^2 (π‘₯𝑦^β€²βˆ’π‘¦π‘¦^β€² )^2+(π‘¦βˆ’π‘₯)^2=(π‘₯+𝑦𝑦^β€² )^2 (π‘₯βˆ’π‘¦)^2 (𝑦^β€² )^2+(π‘₯βˆ’π‘¦)^2=(π‘₯+𝑦𝑦^β€² )^2 (π‘₯βˆ’π‘¦)^2 (1+〖𝑦^β€²γ€—^2 )=(π‘₯+𝑦𝑦^β€² )^2 (𝒙+π’šπ’š^β€² )^𝟐=(π’™βˆ’π’š)^𝟐 (𝟏+γ€–π’š^β€²γ€—^𝟐 )

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.