Question

form the differential equation of the family of circle in the second quartant and touching the coordinate 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

Answer 2

Hey !!

Let C denotes the family of circles in the second quadrant and touching the coordinate axes. Let (-a , a) be the coordinate of the centre of any member of this family

Equation representing the family C is

         ( x + a )² + ( y - a )² = a²    ---------------> (1)

(or)    x² + y² + 2ax - 2ay + a² = 0 -------------> (2)

Differentiating equation (1) w.r.t to x

     2x + 2y dy/dx + 2a - 2a dy/dx = 0

=>  x + y dy/dx = a(dy/dx - 1)

  =>             a = x + yy' / y' - 1                              ( y' = dy/dx )

Substituting the value of a in equation (1)

    [ x + x + yy' / y' - 1 ]  + [ y - yy' / y' - 1 ] = [ x + yy' / y' - 1 ]

= > [ xy' - x + x + yy' ] + [ yy' - y - x - yy' ] = [ x + yy' ]²

= > ( x + y)² y² = (x + y)² = (x + yy')²

=> ( x + y)² [(y')² + 1] = [ x + yy']² is the required differential equation representing the given family of circles.

GOOD LUCK !!