The differential equation corresponding to th | vedclass

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(B) Given the family of circles: $(x-a)^2+(y-b)^2=4$. Differentiating with respect to $x$: $2(x-a)+2(y-b)y'=0 \implies (x-a)+(y-b)y'=0$. Differentiati

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$4\left(\frac{d^2 y}{d x^2}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3$

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(B) Given the family of circles: $(x-a)^2+(y-b)^2=4$. Differentiating with respect to $x$: $2(x-a)+2(y-b)y'=0 \implies (x-a)+(y-b)y'=0$. Differentiating again with respect to $x$: $1+(y')^2+(y-b)y''=0 \implies (y-b) = -\frac{1+(y')^2}{y''}$. Substituting $(y-b)$ back into the first derivative equation: $(x-a) = -y'(y-b) = y' \cdot \frac{1+(y')^2}{y''}$. Now substitute $(x-a)$ and $(y-b)$ into the original equation: $\left(y' \cdot \frac{1+(y')^2}{y''}\right)^2 + \left(-\frac{1+(y')^2}{y''}\right)^2 = 4$. This simplifies to: $\frac{(1+(y')^2)^2}{(y'')^2} \cdot ((y')^2+1) = 4$. Thus,$(1+(y')^2)^3 = 4(y'')^2$,which is $4\left(\frac{d^2 y}{d x^2}\right)^2 = \left[1+\left(\frac{d y}{d x}\right)^2\right]^3$.

The differential equation corresponding to the family of circles given by $(x-a)^2+(y-b)^2=4$,where $a$ and $b$ are parameters,is

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