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 $b.$

(N/A) Given equation: $(x-a)^{2}+(y-b)^{2}=c^{2}$
Differentiating both sides with respect to $x$:
$\frac{d}{d x}[(x-a)^{2}]+\frac{d}{d x}[(y-b)^{2}]=\frac{d}{d x}(c^{2})$
$2(x-a) + 2(y-b) \cdot \frac{d y}{d x} = 0$
$\frac{d y}{d x} = -\frac{x-a}{y-b}$ --- $(1)$
Differentiating again with respect to $x$:
$\frac{d^{2} y}{d x^{2}} = -\frac{d}{d x} \left[ \frac{x-a}{y-b} \right] = -\frac{(y-b) \cdot 1 - (x-a) \cdot \frac{d y}{d x}}{(y-b)^{2}}$
Substituting $\frac{d y}{d x}$ from $(1)$:
$\frac{d^{2} y}{d x^{2}} = -\frac{(y-b) - (x-a) \cdot \left( -\frac{x-a}{y-b} \right)}{(y-b)^{2}} = -\frac{(y-b)^{2} + (x-a)^{2}}{(y-b)^{3}} = -\frac{c^{2}}{(y-b)^{3}}$
Now,evaluate the expression:
$\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = \frac{\left[1 + \frac{(x-a)^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}} = \frac{\left[\frac{(y-b)^{2} + (x-a)^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}} = \frac{\left[\frac{c^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}}$
$= \frac{\frac{c^{3}}{|y-b|^{3}}}{-\frac{c^{2}}{(y-b)^{3}}} = -c$
Since $-c$ is a constant independent of $a$ and $b$,the proof is complete.