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Can someone help me with this, - Differential equations - JEE Main

The differential equation of the family of circles with fixed radius 5 units and centre on the line y=2 is

  • Option 1)

    (x-2)^{2}y'^{2}=25-(y-2)^{2}

  • Option 2)

    (x-2)y'^{2}=25-(y-2)^{2}

  • Option 3)

    (y-2)y'^{2}=25-(y-2)^{2}

  • Option 4)

    (y-2)^{2}y'^{2}=25-(y-2)^{2}

 
Answers (1)
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As we learnt in 

Formation of Differential Equations -

A differential equation can be derived from its equation by the process of differentiation and other algebraical process of elimination

-

 

 Let the center is (a,2) and radius is 5.

\therefore (y-2)^{2}+ (x-a)^{2}=25

\Rightarrow 2(y-2)\times \frac{dy}{dx}+ 2(x-a)=0

\Rightarrow (y-2)y' + x-a=0

\Rightarrow a= x+(y-2)y'

\Rightarrow (y-2)^{2} +[(y-2)y']^{2}= 25

 


Option 1)

(x-2)^{2}y'^{2}=25-(y-2)^{2}

Incorrect option

Option 2)

(x-2)y'^{2}=25-(y-2)^{2}

Incorrect option

Option 3)

(y-2)y'^{2}=25-(y-2)^{2}

Incorrect option

Option 4)

(y-2)^{2}y'^{2}=25-(y-2)^{2}

Correct option

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