Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.

Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.

Answers (1)

To find: Differential equations of all circles which pass though origin and centre lies on x axis

Assume a point (0,k) on y-axis

Radius of the circle is

\sqrt{0^{2}+k^{2}}=k

General form of the equation of circle is,

(x-a)^{2}+(y-b)^{2}=r^{2}

Here a, c is the center and r is the radius.

Substituting the values in the above equation,

(x-0)^{2}+(y-b)^{2}=k^{2}\\ x^{2}+y^{2}-2yk=0.......(i)

Differentiate the equation with respect to x

2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0\\ Formula:\frac{d}{dx}(x^{n})=nx^{n-1}\\ k=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}

Substituting the value of k in (i)

x^{2}+y^{2}-2y\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}=0\\ \left ( x^{2}+y^{2} \right )\frac{dy}{dx}-2yx+2y^{2}\frac{dy}{dx}=0\\ \left ( x^{2}-y^{2} \right )\frac{dy}{dx}-2yx=0

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question