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

Explanation

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)

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