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Exercise 3 · Q2

Q.Find the differential equation of the family of circles having centre at origin.

Puducherry CbseNCERTSubjective· 2mImportance★★★★★
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✓ Free question

The one-parameter family x2+y2=a2x^2+y^2=a^2 has a single constant, so one differentiation removes it, giving x+ydydx=0x+y\dfrac{dy}{dx}=0.

Family of circles, centre origin: x2+y2=a2x^2+y^2=a^2, where aa = radius (the only arbitrary constant). Number of constants = order of the required differential equation, so we differentiate once.

Steps

  1. Write the family with the origin as centre and radius aa:

x2+y2=a2.x^2+y^2=a^2.

  1. This contains one arbitrary constant aa, so the differential equation is of order 1.

  2. Differentiate both sides with respect to xx:

ddx(x2)+ddx(y2)=ddx(a2).\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(a^2).

2x+2ydydx=0.2x+2y\frac{dy}{dx}=0.

  1. Divide by 22:

x+ydydx=0.x+y\frac{dy}{dx}=0.

  1. The constant aa has been eliminated, giving the required differential equation.
✓Final answer

x+ydydx=0x+y\dfrac{dy}{dx}=0 (order 1, degree 1).

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