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Form the differential equation of the...

Form the differential equation of the family of circles touching the y-axis at origin.

A

`(x^(2)+y^(2))(dy)/(dx)-2xy=0`

B

`(x^(2)-y^(2))+2xy(dy)/(dx)=0`

C

` (x^(2)-y^(2)) (dy)/(dx) - 2xy = 0`

D

`(x^(2)+y^(2))(dy)/(dx)+2xy=0`

Text Solution

Verified by Experts

The correct Answer is:
b
Let centre of circle on X - axis be (h,0)
The radius of circle will be h .

` :.` The equation of circle having centre (h,0) and radius h is
` (x-h)^(2)+ (y-0)^(2) = h^(2)`
` rArr x^(2) +h^(2) - 2hx + y^(2) = h^(2)`
` rArr x^(2) - 2hx +y^(2) = 0`
On differentiating both sides w.r.t x, we get
` 2x - 2h +2y(dy)/(dx) = 0 rArr h = x+y (dy)/(dx)`
On putting `h = x+y (dy)/(dx)` in Eq. (i), we get
` x^(2) - 2 ( x + y (dy)/(dx)) x +y^(2) = 0`
` rArr -x^(2) +y^(2) - 2xy (dy)/(dx) =0`
` rArr (x^(2) -y^(2)) +2xy (dy)/(dx) =0`
` rArr (x^(2) -y^(2)) +2xy (dy)/(dx)=0 `
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Knowledge Check

  • The differential equation of the system of circles touching the y axis at origin is

    A
    `x^(2) + y^(2) - 2xy "" (dy)/(dx) = 0`
    B
    `x^(2) + y^(2) + 2xy "" (dy)/(dx) = 0`
    C
    `x^(2) - y^(2) + 2xy "" (dy)/(dx) = 0`
    D
    `x^(2) - y^(2) - 2xy "" (dy)/(dx) = 0`

  • The differential equation of the system of circles touching the y-axis at the origin is

    A
    ` x^(2) +y^(2) -2xy (dy)/(dx) =0`
    B
    ` x^(2)+y^(2)+2xy (dy)/(dx)=0`
    C
    `x^(2) -y^(2) +2xy (dy)/(dx)=0`
    D
    `x^(2) -y^(2) -2xy(dy)/(dx) =0`

  • The differential equation of system of circles touching the Y-axis at origin is

    A
    `(dy)/(dx) =(x^2+y^2)/(2xy)`
    B
    `(dy)/(dx) =(y^2-x^2)/(xy)`
    C
    `(dy)/(dx) =(y^2-x^2)/(2xy)`
    D
    `(dy)/(dx) =(x^2-y^2)/(2xy)`

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