If (x – a)² + (y – b)² = c², for some c > 0, then the expression:$\frac{\begin{bmatrix}1+(\frac{dy}{dx})^2\end{bmatrix}^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}$ equal to:

–c

(x – a)² + (y – b)² = c², c > 0 .......(1)

Differentiating both side.w.r.t.x

$2(x – a) + 2(y – b)\frac{dy}{dx}=0$ ⇒ $\frac{dy}{dx}=\frac{-(x-a)}{y-b}$ ...(2)

Differentiating again

$\frac{d^2y}{dx^2}=\frac{(y-b)(-1)+(x-a).\frac{dy}{dx}}{(y-b)^2}$ ...(3)

Putting value of $\frac{dy}{dx}$ from eq. (2) in eq. (3)

$∴\frac{d^2y}{dx^2}=\frac{(y-b)(-1)+(x-a).\frac{-(x-a)}{y-b}}{(y-b)^2}$

$\frac{d^2y}{dx^2}=-[\frac{(y-b)^2)+(x-a)^2}{(y-b)^3}]=\frac{-c^2}{(y-b)^3}$ ...(4)

Now, $\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{[1+(\frac{-(x-a)^2}{y-b})]^{\frac{3}{2}}}{\frac{-c^2}{(y-b)^3}}$

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

$=\frac{([\frac{c}{(y-b)}]^2)^{\frac{3}{2}}}{\frac{-c^2}{(y-b)^3}}=\frac{c^3}{(y-b)^3}×\frac{(y-b)^3}{-c^2}=-c$