Question

from the differential equation of the family of circle in the second quadrant and touching the coordinates axis​

Answer 1

The equation is of the form (x+a)^2+(y-a)^2=a^2

Differentiating w.r.t x

2(x+a)+2(y-a).y'=0

Therefore a = (x+yy')/(y'-1)

Substituting in the first equation,

(x+(x+yy')/(y'-1))^2+(y-(x+yy')/(y'-1))^2=[(x+yy')/(y'-1)]^2

On expanding and simplifying we get,

(xy'+yy')^2+(2yy'+x-y)^2=(x+yy')^2 is the required solution