The differential equation of the family of ci | vedclass

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(A) The equation of the family of circles passing through $(0,0)$ and having its center on the $X$-axis is given by $(x-r)^2 + y^2 = r^2$,where $r$ is

Vedclass · Accepted Answer

$2 x y \frac{d y}{d x}+x^2-y^2=0$

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(A) The equation of the family of circles passing through $(0,0)$ and having its center on the $X$-axis is given by $(x-r)^2 + y^2 = r^2$,where $r$ is the radius of the circle and $(r, 0)$ is the center.Expanding the equation: $x^2 - 2xr + r^2 + y^2 = r^2$,which simplifies to $x^2 + y^2 - 2xr = 0$ or $r = \frac{x^2 + y^2}{2x}$.Differentiating both sides with respect to $x$:$2x + 2y \frac{dy}{dx} - 2r = 0$$x + y \frac{dy}{dx} = r$Substituting the value of $r$ from the original equation:$x + y \frac{dy}{dx} = \frac{x^2 + y^2}{2x}$$2x^2 + 2xy \frac{dy}{dx} = x^2 + y^2$$2xy \frac{dy}{dx} + x^2 - y^2 = 0$

The differential equation of the family of circles passing through $(0,0)$ and having centre on the $X$-axis is

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