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