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(B) Given the family of curves: $(x-2)^2+(y-a)^2=b^2$  $(i)$ Since there are two parameters $a$ and $b$, we differentiate twice. Differentiating with

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$5$

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(B) Given the family of curves: $(x-2)^2+(y-a)^2=b^2$  $(i)$ Since there are two parameters $a$ and $b$, we differentiate twice. Differentiating with respect to $x$: $2(x-2) + 2(y-a)y' = 0$ $(x-2) + (y-a)y' = 0$  (ii) From (ii), $(y-a) = -\frac{x-2}{y'}$ Substitute this into $(i)$: $(x-2)^2 + \left(-\frac{x-2}{y'}\right)^2 = b^2$ $(x-2)^2 \left(1 + \frac{1}{(y')^2}\right) = b^2$ $(x-2)^2 \left(\frac{(y')^2+1}{(y')^2}\right) = b^2$ Differentiating again with respect to $x$: $2(x-2) \left(\frac{(y')^2+1}{(y')^2}\right) + (x-2)^2 \left(\frac{2y'y'' \cdot (y')^2 - ((y')^2+1) \cdot 2y'y''}{(y')^4}\right) = 0$ This differential equation involves the second derivative $y''$, so the order $m = 2$. The highest power of the highest derivative $y''$ is $1$, so the degree $n = 1$. Therefore, $m^2+n = (2)^2 + 1 = 4 + 1 = 5$.

If the order and degree of the differential equation corresponding to the family of curves $(x-2)^2+(y-a)^2=b^2$, (where $a$ and $b$ are parameters) are $m$ and $n$ respectively, then $m^2+n=$

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