The differential equation of the family of ci | vedclass

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Question

(A) The equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + y^2 = a^2$,where $a$ is a parameter. Ex

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

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(A) The equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + y^2 = a^2$,where $a$ is a parameter. Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$. Differentiating both sides with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 2a$. Substituting the value of $a = x + y \frac{dy}{dx}$ into the equation $x^2 + y^2 = 2ax$,we get $x^2 + y^2 = 2x(x + y \frac{dy}{dx})$. This simplifies to $x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$. Rearranging the terms,we get $y^2 - x^2 = 2xy \frac{dy}{dx}$,which can be written as $(x^2 - y^2) dx + 2xy dy = 0$.

The differential equation of the family of circles passing through the origin and having centre on the $X$-axis is:

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