I am trying to find the second derivative of a general circle but I can't seem to get the right answer.
My working goes as follows:
$$ (x-a)^2+(y-b)^2=R^2 $$
$$ 2(x-a)+2(y-b)*\frac{dy}{dx}=0 $$
$$ \frac{dy}{dx}=\frac{a-x}{y-b} $$
I then applied the quotient rule: $$ \frac{d^2y}{dx^2}=\frac{-(y-b)-\frac{dy}{dx}(a-x)}{(y-b)^2} $$
$$ \frac{d^2y}{dx^2}=\frac{-(y-b)-\frac{a-x}{y-b}(a-x)}{(y-b)^2} $$
$$ \frac{d^2y}{dx^2}=\frac{-(y-b)^2-(a-x)^2}{(y-b)^3} $$
However, the correct answer was
$$ \frac{d^2y}{dx^2}=-\frac{2}{y-b} $$
I am assuming that that answer comes from
$$ \frac{d^2y}{dx^2}=\frac{-(y-b)-\frac{y-b}{a-x}(a-x)}{(y-b)^2} $$
But I'm unsure why we multiply by $\frac{y-b}{a-x}$ and not $\frac{a-x}{y-b}$.