# Misc 5 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 11, 2019 by Teachoo

Last updated at Dec. 11, 2019 by Teachoo

Transcript

Misc 5 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes . Letβs first draw the figure Let C be the family of circles in first quadrant touching coordinate axes Let radius be π β΄ Center of circle = (βπ, π) Thus, Equation of a circle is (π₯βπ)^2+(π¦βπ)^2=π^2 π₯^2+π^2β2ππ₯+π¦^2+π^2β2ππ¦=π^2 π₯^2+π¦^2β2ππ₯β2ππ¦+π^2=0 Since there is only one variable, we differentiate once Differentiating w.r.t. π₯ (π₯^2 + π¦^2 β 2ππ₯ β 2ππ¦+π^2 )^β²=(0)^β² 2π₯+2π¦π¦β²β2πβ2ππ¦β²+0=0 π₯+π¦π¦β²βπβππ¦β²=0 π₯+π¦π¦^β²=π+ππ¦^β² π₯+π¦π¦^β²=π(1+π¦^β²) π = (π₯ + π¦π¦^β²)/(1 + π¦^β² ) Putting Value of π in (π₯βπ)^2+(π¦βπ)^2=π^2 (π₯β[(π₯ + π¦ π¦^β²)/(1 + π¦^β² )])^2+(π¦β[(π₯ + π¦ π¦^β²)/(1 + π¦^β² )])^2=((π₯ + π¦π¦^β²)/(1 + π¦^β² ))^2 (π₯(1 + π¦^β² ) β (π₯ + π¦π¦^β² ))^2/(1 + π¦^β² )^2 +(π¦(1 + π¦^β² ) β (π₯ + π¦π¦^β² ))^2/(1 + π¦^β² )^2 =(π₯ + π¦π¦^β² )^2/(1 + π¦^β² )^2 (π₯+π₯π¦^β²βπ₯βπ¦π¦^β² )^2+(π¦+π¦π¦^β²βπ₯βπ¦π¦^β² )^2=(π₯+π¦π¦^β² )^2 (π₯π¦^β²βπ¦π¦^β² )^2+(π¦βπ₯)^2=(π₯+π¦π¦^β² )^2 (π₯βπ¦)^2 (π¦^β² )^2+(π₯βπ¦)^2=(π₯+π¦π¦^β² )^2 (π₯βπ¦)^2 (1+γπ¦^β²γ^2 )=(π₯+π¦π¦^β² )^2 (π+ππ^β² )^π=(πβπ)^π (π+γπ^β²γ^π )

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Deleted for CBSE Board 2021 Exams only

Misc 4 Important

Misc 5 Important Deleted for CBSE Board 2021 Exams only You are here

Misc 6

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10 Important

Misc 11

Misc 12

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16

Misc 17 Important Deleted for CBSE Board 2021 Exams only

Misc 18

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.