The differential equation of the family of ci | vedclass

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Question

(A) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $x^{2} + y^{2} - 2hx = 0$,where $h$ is a p

Vedclass · Accepted Answer

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

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(A) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $x^{2} + y^{2} - 2hx = 0$,where $h$ is a parameter. From this equation,we have $2h = \frac{x^{2} + y^{2}}{x}$. Differentiating the equation $x^{2} + y^{2} - 2hx = 0$ with respect to $x$,we get: $2x + 2y \frac{dy}{dx} - 2h = 0$. Substituting the value of $2h$ from the first equation: $2x + 2y \frac{dy}{dx} - \left( \frac{x^{2} + y^{2}}{x} \right) = 0$. Multiplying the entire equation by $x$: $2x^{2} + 2xy \frac{dy}{dx} - x^{2} - y^{2} = 0$. Simplifying the expression: $x^{2} - y^{2} + 2xy \frac{dy}{dx} = 0$,which can be rewritten as $y^{2} = x^{2} + 2xy \frac{dy}{dx}$.

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

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