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(B) The family of all circles in the $XY$ plane with radius $5$ units and center $(x_1, y_1)$ is given by:$(x - x_1)^2 + (y - y_1)^2 = 25$ ... $(i)$Di

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The family of all circles in the $XY$ plane with radius $5$ units and center $(x_1, y_1)$ is given by:$(x - x_1)^2 + (y - y_1)^2 = 25$ ... $(i)$Differentiating with respect to $x$:$2(x - x_1) + 2(y - y_1)y' = 0$$(x - x_1) = -(y - y_1)y'$ ... (ii)Differentiating again with respect to $x$:$1 - [(y')^2 + (y - y_1)y''] = 0$$(y - y_1) = \frac{1 - (y')^2}{y''}$ ... (iii)Substituting $(y - y_1)$ from (iii) into (ii):$(x - x_1) = -\left(\frac{1 - (y')^2}{y''}\right)y' = \frac{(y')^3 - y'}{y''}$Substituting $(x - x_1)$ and $(y - y_1)$ into $(i)$:$\left(\frac{(y')^3 - y'}{y''}\right)^2 + \left(\frac{1 - (y')^2}{y''}\right)^2 = 25$$\frac{(y')^2((y')^2 - 1)^2 + (1 - (y')^2)^2}{(y'')^2} = 25$$\frac{(1 - (y')^2)^2 [ (y')^2 + 1 ]}{(y'')^2} = 25$$(1 - (y')^2)^2 (1 + (y')^2) = 25(y'')^2$Here,the highest order derivative is $y''$,so the order $m = 2$.The power of the highest order derivative is $2$,so the degree $l = 2$.Given $l = 2$ and $m = 2$,we have:$2l + 3m = 2(2) + 3(2) = 4 + 6 = 10$.

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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