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(C) To find the differential equation for each option, we differentiate with respect to $x$ and eliminate the arbitrary constant.$(a)$ $x^2+y^2+2gx+4y

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$y^2=2c(x+\sqrt{c})$

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(C) To find the differential equation for each option, we differentiate with respect to $x$ and eliminate the arbitrary constant.$(a)$ $x^2+y^2+2gx+4y+2=0$. Differentiating, $2x+2y\frac{dy}{dx}+2g+4\frac{dy}{dx}=0$. This leads to a first-order, first-degree equation.$(b)$ $x^2=a^2(1+y^2)$. Differentiating, $2x=a^2(2y\frac{dy}{dx})$. Substituting $a^2$ back, we get a first-order, first-degree equation.$(c)$ $y^2=2c(x+\sqrt{c})$. Differentiating, $2y\frac{dy}{dx}=2c$, so $c=y\frac{dy}{dx}$. Substituting $c$ into the original equation: $y^2=2(y\frac{dy}{dx})(x+\sqrt{y\frac{dy}{dx}})$. Rearranging: $y^2-2xy\frac{dy}{dx}=2y\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}$. Squaring both sides: $(y^2-2xy\frac{dy}{dx})^2 = 4y^2(\frac{dy}{dx})^2(y\frac{dy}{dx}) = 4y^3(\frac{dy}{dx})^3$. This equation has order $1$ and degree $3$.$(d)$ $y^2=4ax$. Differentiating, $2y\frac{dy}{dx}=4a$. Substituting $a$ gives $y^2=2xy\frac{dy}{dx}$, which is first-order, first-degree.Thus, option $(c)$ is the correct answer.

$A$ family of curves whose equation is the general solution of a differential equation having order $1$ and degree $3$ is ($g, a, c$ are arbitrary constants):

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