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(C) The equation of the family of all circles touching the $x-$axis at the origin $(0, 0)$ with center $(0, a)$ and radius $|a|$ is given by:$x^2 + (y

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$2x$

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(C) The equation of the family of all circles touching the $x-$axis at the origin $(0, 0)$ with center $(0, a)$ and radius $|a|$ is given by:$x^2 + (y - a)^2 = a^2$$x^2 + y^2 - 2ay + a^2 = a^2$$x^2 + y^2 - 2ay = 0$ ... $(1)$Differentiating both sides with respect to $x$:$2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$$x + y\frac{dy}{dx} = a\frac{dy}{dx}$$a = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}$Substituting the value of $a$ into equation $(1)$:$x^2 + y^2 - 2y \left( \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} \right) = 0$$(x^2 + y^2)\frac{dy}{dx} - 2y(x + y\frac{dy}{dx}) = 0$$(x^2 + y^2)\frac{dy}{dx} - 2xy - 2y^2\frac{dy}{dx} = 0$$(x^2 - y^2)\frac{dy}{dx} = 2xy$Comparing this with the given equation $(x^2 - y^2)\frac{dy}{dx} = g(x)y$,we get:$g(x)y = 2xy$$g(x) = 2x$

If the differential equation representing the family of all circles touching the $x-$axis at the origin is $(x^2 - y^2)\frac{dy}{dx} = g(x)y$,then $g(x)$ equals

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