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

We have, ${(x - a)^2} + {(y - b)^2} = {c^2}$ …(i)

Differentiating (i) with respect to x, we get
$2(x - a) + 2(y - b)\cfrac{{dy}}{{dx}} = 0$

$\Rightarrow$ $(x - a) + (y - b)\cfrac{{dy}}{{dx}} = 0$

Differentiating (ii) with respect to x, we get
$1 + (y - b)\cfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\cfrac{{dy}}{{dx}}} \right)^2} = 0$

$\Rightarrow$ $(y - b)\cfrac{{{d^2}y}}{{d{x^2}}} = - \left[ {1 + {{\left( {\cfrac{{dy}}{{dx}}} \right)}^2}} \right]$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = - \cfrac{{\left[ {1 + {{\left( {\cfrac{{dy}}{{dx}}} \right)}^2}} \right]}}{{(y - b)}}$ …(iii)

From (ii), we have
$\cfrac{{dy}}{{dx}} = - \left( {\cfrac{{x - a}}{{y - b}}} \right)$ $\Rightarrow$

${\left( {\cfrac{{dy}}{{dx}}} \right)^2} = {\left( { - \left( {\cfrac{{x - a}}{{y - b}}} \right)} \right)^2}$

Adding 1 on both sides, we get

$1 + {\left( {\cfrac{{dy}}{{dx}}} \right)^2} = 1 + {\left( { - \left( {\cfrac{{x - a}}{{y - b}}} \right)} \right)^2} = 1 + {\left( {\cfrac{{x - a}}{{y - b}}} \right)^2} = \cfrac{{{c^2}}}{{{{(y - b)}^2}}}$ …(iv)

Also, $\cfrac{{{d^2}y}}{{d{x^2}}} = - \cfrac{{{c^2}}}{{{{(y - b)}^3}}}$ …(v)

[From (iii) \& (iv)]
From (iv) and (v), we have

$\cfrac{{{{\left[ {1 + {{\left( {\cfrac{{dy}}{{dx}}} \right)}^2}} \right]}^{3/2}}}}{{\cfrac{{{d^2}y}}{{d{x^2}}}}} = \cfrac{{\cfrac{{{c^3}}}{{{{(y - b)}^3}}}}}{{\cfrac{{ - {c^2}}}{{{{(y - b)}^3}}}}} = - c$

which is independent of a and b.