# If $(x-a)^2+(y-b)^2=c^2,$ for some $c>0,$ prove that $\frac{\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^2\right ]^{\frac{3}{2}}}{\frac{\mathrm{d}^2y }{\mathrm{d} x^2}}$ is a constant independent of a and b.

$(x-a)^2+(y-b)^2=c^2,$  $c>0\: \: \: \: -(i)$

On differentiating equation (i) w.r.t. x we get

$2(x-a)+2(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0$

$\Rightarrow (x-a)+(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-(x-a)}{(y-b)}\: \: \: \: \: -(ii)$

Again, differentiating equation (ii) w.r.t x we get

$\frac{\mathrm{d} ^2y}{\mathrm{d} x^2}=-\left [ \frac{(y+b)\frac{\mathrm{d} }{\mathrm{d} x}(x-a)-(x-a)\frac{\mathrm{d} }{\mathrm{d} x}(y-b)}{(y-b)^2} \right ]$

$=-\left [ \frac{(y-b)-(x-a)\frac{\mathrm{d} y}{\mathrm{d} x}}{(y-b)^2} \right ]$

$\Rightarrow -\left [ \frac{(y-b)+\frac{(x-a)(x-a)}{(y-b)}}{(y-b)^2} \right ]$ $\left [ from \: equation (ii )\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-(x-a)}{(y-b)} \right ]$

$=-\left [ \frac{(y-b)+\frac{(x-a)^2}{y-b}}{(y-b)^2} \right ]$

$=-\left [ \frac{(y-b)^2+(x-a)^2}{(y-b)^3} \right ]$

$=-\left [ \frac{c^2}{(y-b)^3} \right ]$     $\left [ \because from\: equation(ii,(x-a)^2+(y-b)^2=c^2 \right ]$

Now, $\frac{\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^2 \right ]^{\frac{3}{2}}}{\frac{\mathrm{d} ^2y}{\mathrm{d} x^2}}$

$\Rightarrow \frac{-\left [ 1+\frac{(x-a)^2}{(y-b)^2} \right ]^{\frac{3}{2}}}{\frac{c^2}{(y-b)^3}}\Rightarrow \frac{-\left [ (y-b)^2+(x-a)^2 \right ]^{\frac{3}{2}}}{c^2}$

$\Rightarrow \frac{-c^{2\times \frac{3}{2}}}{c^2}\: \: \: \left [ \because c^2=(y-b)^2+(x-a)^2 \right ]$

$\Rightarrow \frac{-c^3}{c^2}\Rightarrow -c\Rightarrow constant$

It shows that $\frac{\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^2 \right ]^{\frac{3}{2}}}{\frac{\mathrm{d} ^2y}{\mathrm{d} x^2}}$ is independent of a and b.

Hence proved.

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