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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 $\dpi{100} y=2$ 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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