Q.409. Formation of differential equation representing family of circles.

Question 409.
Prove that the differential equation of the family of circles, all touching x-axis is given by
(1 + y'^2)^3 = [yy" + (1 + y'^2)]^2.

Answer 409.
Let (h, k) be the centers of the circles with h and k as arbitrary constants.
As the circles touch the x-axis, radius, r = l k l
=> general equation of the family of circles is
(x - h)^2 + (y - k)^2 = k^2 ... ( 1 )


Differentiating w.r.t. x,
=> 2(x - h) + 2(y - k) y' = 0
=> (x - h) = - (y - k)y'
=> (x - h)^2 - (y - k)^2 y'^2 = 0 ... ( 2 )


( 1 ) - ( 2 )
=> (y - k)^2 (1 + y'^2) = k^2 ... ( 3 )

Differentiating w.r.t. x,
2(y - k) y' (1 + y'^2) + (y - k)^2 * 2y'y" = 0

Dividing by 2y'(y - k)
=> (1 + y'^2) + (y - k) y" = 0
=> y - k = - (1 + y'^2)/y" and k = y + (1 + y'^2)/y"
Plugging y - k and k from above in eqn. ( 3 ),
[ - (1 + y'^2)/y"]^2 (1 + y'^2) = [y + (1 + y'^2)/y"]^2
=> (1 + y'^2)^3 = [yy" + (1 + y'^2)]^2

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