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  • From the differential equation of the family of circles in the second quadrant and touching the coordinates axes.        

From the differential equation of the family of circles in the second quadrant and touching the coordinates axes.

 

 

 

 
 
 
 
 

Answers (1)

The equation of circle in the second quadrant which touch the coordinates axes are \left ( x+a \right )^{2}+\left ( y-a \right )^{2}= a^{2}\, a\varepsilon R---(i)
where a is perimetric this equation contains ine arbitary constant so we shall differentiate it quce only and we shall get differentiate equation of first order.

Differentiating (i) w.r.t x, we get
2\left ( x+a \right )+2\left ( y-a \right )\frac{dy}{dx}= 0
\Rightarrow a= -\left [ \frac{x+y\frac{dy}{dx}}{1-\frac{dy}{dx}} \right ]\Rightarrow a= \frac{x+p{y}}{p-1}\, where\, p= \frac{dy}{dx}
substituting the value of a in (i), we get
\left ( x+ \frac{x+py}{p-1} \right )^2+\left ( y-\frac{x+p{y}}{p-1} \right )^{2}= \left ( \frac{x+p{y}}{p-1} \right )^{2}
\Rightarrow \left ( xp-x+x+yp \right )^{2}+\left ( yp-y-x-yp \right )^{2}= \left ( x+yp \right )^{2}
\Rightarrow \left ( x+y \right )^{2}p^{2}+\left ( x+y \right )^{2}= \left ( x+yp \right )^{2}
\Rightarrow \left ( x+y \right )^{2}\left ( p^{2}+1 \right )= \left ( x+yp \right )^{2}
\Rightarrow \left ( x+y \right )^{2}\left [ \left ( \frac{dy}{dx} \right )^{2}+1 \right ]= \left ( x+\frac{ydy}{dx} \right )^{2}
This is the required differential equation representing the given family of circles.

Posted by

Ravindra Pindel

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