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If $(x-a)^2+(y-b)^2=c^2$, for some $c>0$, prove that $\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2 y}{d x^2}}$ is a constant independent of a and b.
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Given $(x-a)^2+(y-b)^2=c^2 \quad \ldots(i)$
Differentiating w.r.t. $\mathrm{x}$, we get
$\Rightarrow(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{b}) \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad \ldots (ii)$
Again differentiating w.r.t. $\mathrm{x}$,
$\Rightarrow(y-b)=-\left\{\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}\right\} \quad \ldots(iii)$
Substitute this value of $(\mathrm{y}-\mathrm{b})$ in (ii)
$(x-a)=\left\{\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}\right\}\left(\frac{d y}{d x}\right) \quad \ldots(iv)$
Putting these values of $(y-b)$ and $(x-a)$ from (iii) \& (iv) in (i)
$\Rightarrow \frac{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{3 / 2}}{\frac{d^2 y}{d x^2}}=c$
which is a constant, independent of $\mathrm{a}$ and $\mathrm{b}$.
Differentiating w.r.t. $\mathrm{x}$, we get
$\Rightarrow(\mathrm{x}-\mathrm{a})+(\mathrm{y}-\mathrm{b}) \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad \ldots (ii)$
Again differentiating w.r.t. $\mathrm{x}$,
$\Rightarrow(y-b)=-\left\{\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}\right\} \quad \ldots(iii)$
Substitute this value of $(\mathrm{y}-\mathrm{b})$ in (ii)
$(x-a)=\left\{\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}\right\}\left(\frac{d y}{d x}\right) \quad \ldots(iv)$
Putting these values of $(y-b)$ and $(x-a)$ from (iii) \& (iv) in (i)
$\Rightarrow \frac{\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{3 / 2}}{\frac{d^2 y}{d x^2}}=c$
which is a constant, independent of $\mathrm{a}$ and $\mathrm{b}$.
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