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(C) The general equation of a family of circles with a constant radius $r$ and center $(a, b)$ is given by $(x-a)^2 + (y-b)^2 = r^2$. Since there are

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$8$

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(C) The general equation of a family of circles with a constant radius $r$ and center $(a, b)$ is given by $(x-a)^2 + (y-b)^2 = r^2$. Since there are two arbitrary constants $a$ and $b$, the order of the differential equation is $k = 2$. Differentiating with respect to $x$: $2(x-a) + 2(y-b)y' = 0 \Rightarrow (x-a) + (y-b)y' = 0$. Differentiating again: $1 + (y')^2 + (y-b)y'' = 0 \Rightarrow (y-b) = -\frac{1+(y')^2}{y''}$. Substituting this back into the first derivative equation: $(x-a) = -y' \left(-\frac{1+(y')^2}{y''}\right) = \frac{y'(1+(y')^2)}{y''}$. Substituting $(x-a)$ and $(y-b)$ into the original equation: $\left(\frac{y'(1+(y')^2)}{y''}\right)^2 + \left(-\frac{1+(y')^2}{y''}\right)^2 = r^2$. Simplifying: $\frac{(1+(y')^2)^2}{(y'')^2} ( (y')^2 + 1 ) = r^2 \Rightarrow (1+(y')^2)^3 = r^2(y'')^2$. The highest order derivative is $y''$, so the order $k = 2$. The power of the highest order derivative is $2$, so the degree $l = 2$. Thus, $k^2 + l^2 = 2^2 + 2^2 = 4 + 4 = 8$.

If $k$ and $l$ respectively are the order and degree of the differential equation whose general solution represents the family of circles of constant radius, then $k^2+l^2=$

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