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(A) The order of a differential equation is equal to the number of arbitrary constants present in the general equation of the family of curves. For op

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All circles passing through the origin

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(A) The order of a differential equation is equal to the number of arbitrary constants present in the general equation of the family of curves. For option $A$,the equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$. Here,$g$ and $f$ are two independent arbitrary constants. Since there are $2$ arbitrary constants,the differential equation formed will be of order $2$. For option $B$,the equation is $y^2 = 4a(x-h)$,which involves two constants,but the condition of passing through the origin and focus on the $x$-axis restricts it. For option $C$,the equation is $y = mx$,which has only $1$ arbitrary constant. For option $D$,the equation $x^2 - y^2 = k^2$ has only $1$ arbitrary constant $k$. Therefore,the correct option is $A$.

Among the options given below,from which option can a differential equation of order two be formed?

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