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Question

(A) All circles passing through the origin and having their center on the $x$-axis at $(a, 0)$ will have a radius of $a$.The general equation of such

Vedclass · Accepted Answer

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

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(A) All circles passing through the origin and having their center on the $x$-axis at $(a, 0)$ will have a radius of $a$.The general equation of such circles is given by:$(x - a)^2 + y^2 = a^2$$x^2 - 2ax + a^2 + y^2 = a^2$$x^2 + y^2 - 2ax = 0$  --- $(i)$This equation has only one arbitrary constant,$a$. Therefore,we differentiate it once with respect to $x$:$2x + 2y\frac{dy}{dx} - 2a = 0$$x + y\frac{dy}{dx} = a$Substituting the value of $a$ into equation $(i)$:$x^2 + y^2 - 2x(x + y\frac{dy}{dx}) = 0$$x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} = 0$$y^2 - x^2 - 2xy\frac{dy}{dx} = 0$$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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