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(B) The equation of a circle with center $(h, k)$ and radius $r = 5$ is $(x - h)^2 + (y - k)^2 = 25$. Since there are two arbitrary constants $h$ and

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$10$

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(B) The equation of a circle with center $(h, k)$ and radius $r = 5$ is $(x - h)^2 + (y - k)^2 = 25$. Since there are two arbitrary constants $h$ and $k$,we differentiate twice. Differentiating with respect to $x$: $2(x - h) + 2(y - k)y' = 0$,which gives $(x - h) = -(y - k)y'$. Differentiating again: $1 = -[(y')^2 + (y - k)y'']$,which gives $(y - k) = -\frac{1 + (y')^2}{y''}$. Substituting $(y - k)$ back into the first derivative equation: $(x - h) = \left(\frac{1 + (y')^2}{y''}\right)y'$. Substituting these into the original circle equation: $\left(\frac{1 + (y')^2}{y''}\right)^2 (y')^2 + \left(\frac{1 + (y')^2}{y''}\right)^2 = 25$. Simplifying,we get $(1 + (y')^2)^2 \left(\frac{(y')^2 + 1}{(y'')^2}\right) = 25$,which leads to $(1 + (y')^2)^3 = 25(y'')^2$. The order $m$ is the highest derivative present,which is $y''$,so $m = 2$. The degree $l$ is the power of the highest derivative,which is $2$,so $l = 2$. Thus,$2l + 3m = 2(2) + 3(2) = 4 + 6 = 10$.

If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all circles in the $XY$ plane with radius $5$ units,then $2l + 3m =$

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