# if the

Question:

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 $b$.

Solution:

It is given that, $(x-a)^{2}+(y-b)^{2}=c^{2}$

Differentiating both sides with respect to x, we obtain

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

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

$\Rightarrow 2(x-a) \cdot 1+2(y-b) \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{-(x-a)}{y-b}$    ...(1)

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

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

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

$=-\left[\frac{(y-b)-(x-a) \cdot\left\{\frac{-(x-a)}{y-b}\right\}}{(y-b)^{2}}\right]$    [Using (1)]

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

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

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

$=-c$, which is constant and is independent of $a$ and $b$

Hence, proved.