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

Answers (1)
G Gautam harsolia


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
 

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