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(C) The general equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + (y-0)^2 = a^2$,where $(a, 0)$ i

Vedclass · Accepted Answer

$y^2=x^2+2xy \frac{dy}{dx}$

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(C) The general equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + (y-0)^2 = a^2$,where $(a, 0)$ is the center and $a$ is the radius. Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0 ... (i)$. Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $2a = 2x + 2y \frac{dy}{dx} ... (ii)$. Substituting the value of $2a$ from equation $(ii)$ into equation $(i)$,we get $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) = 0$. Simplifying this,$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$. Thus,$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$,which gives $y^2 = x^2 + 2xy \frac{dy}{dx}$.

The differential equation of all circles,passing through the origin and having their centres on the $X$-axis,is

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