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NCERT Exemplar · Q50

Q.(vi) The differential equation representing the family of circles x2+(y−a)2=a2x^2+(y-a)^2=a^2 will be of order two. (State True or False.)

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The family has only one arbitrary constant, so its differential equation has order one. The statement is False.

The key principle

The order of the differential equation that represents a family of curves equals the number of independent arbitrary constants in the family, because each constant needs one differentiation to eliminate it.

The family

x2+(y−a)2=a2x^2+(y-a)^2=a^2

contains a single arbitrary constant aa (it fixes both the centre (0,a)(0,a) and the radius aa at once). So we expect order one. Let us confirm it.

Eliminate the constant

1. Simplify. Expand:

x2+y2−2ay+a2=a2 ⇒ x2+y2=2ay.x^2+y^2-2ay+a^2=a^2\ \Rightarrow\ x^2+y^2=2ay.

2. Differentiate once with respect to xx:

2x+2ydydx=2adydx.2x+2y\frac{dy}{dx}=2a\frac{dy}{dx}.

3. Remove aa. From step 1, a=x2+y22ya=\dfrac{x^2+y^2}{2y}. Substituting,

x+ydydx=x2+y22ydydx.x+y\frac{dy}{dx}=\frac{x^2+y^2}{2y}\frac{dy}{dx}.

Multiply by 2y2y and collect the derivative terms: …

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