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  • Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Q5.    Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

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Now, equation of the circle with center at (x,y) and radius r is
(x-a)^2+(y-b)^2 = r^2
Since, it  touch the coordinate axes in first quadrant 
Therefore, x = y = r
(x-a)^2+(y-a)^2 = a^2                     -(i)
Differentiate it w.r.t x
we will get
2(x-a)+2(y-a)y'= 0\\ \\ 2x-2a+2yy'-2ay' = 0\\ a=\frac{x+yy'}{1+y'}          -(ii)
Put value from equation (ii) in equation (i)
(x-\frac{x+yy'}{1+y'})^2+(y-\frac{x+yy'}{1+y'})^2=\left ( \frac{x+yy'}{1+y'} \right )^2\\ \\ (x+xy'-x-yy')^2+(y+yy'-x-yy')^2=(x+yy')^2\\ \\ (y')^2(x-y)^2+(x-y)^2=(x+yy')^2\\ \\ (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2
Therefore,  the differential equation of the family of circles in the first quadrant which touches the coordinate axes is (x-y)^2\left ( (y')^2+1 \right )=(x+yy')^2
 

Posted by

Gautam harsolia

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