# 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

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Misc 3 Deleted for CBSE Board 2021 Exams only

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

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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.