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

Vedclass · Accepted Answer

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

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(D) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $(x - h)^2 + y^2 = h^2$,where $h$ is the $x$-coordinate of the center.Expanding this,we get $x^2 - 2hx + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 - 2hx = 0$ ..... $(i)$.Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2h = 0$.Dividing by $2$,we get $x + y \frac{dy}{dx} - h = 0$,so $h = x + y \frac{dy}{dx}$.Substituting the value of $h$ into equation $(i)$,we get $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.Rearranging the terms,we get $2xy \frac{dy}{dx} = y^2 - x^2$,which gives $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

The differential equation of all circles passing through the origin and having their centers on the $x$-axis is

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