Let's assume your circle to be the unit circle. Then you have
\begin{align*}
y &= \sqrt{1-x^2} \\
y^2 &= 1-x^2 \\
a(x) &= \sqrt{(1+x)^2+y^2} = \sqrt{1+2x+x^2+1-x^2} = \sqrt{2\left(1+x\right)} \\
b(x) &= \sqrt{(1-x)^2+y^2} = \sqrt{1-2x+x^2+1-x^2} = \sqrt{2\left(1-x\right)} \\
D(x) &= a(x)+b(x) = \sqrt{2\left(1+x\right)}+\sqrt{2\left(1-x\right)}
\end{align*}
I don't see a way to make this expression much easier, and neither does Wolfram Alpha. But you were writing about an infinitesimal change, so let's continue:
\begin{align*}
D'(x) &= \frac{\mathrm d\,D}{\mathrm dx} = \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{2-2x^2}} \\
D'(0) &= 0 \\
D''(x) &= \frac{\mathrm d^2\,D}{\mathrm dx^2} = -\frac1{2\sqrt2}\left(
\left(1+x\right)^{-\frac32}+
\left(1-x\right)^{-\frac32}
\right) \\
D''(0) &= -\frac{1}{\sqrt2} \\
D'''(0) &= 0
\end{align*}
This corresponds to the beginning of a series expansion:
\begin{align*}
D(x) &= D(0) + \frac{D''(0)}{2}x^2 + O\left(x^4\right) = 2\sqrt2 + \frac{1}{2\sqrt2}x^2 + O\left(x^4\right)
\end{align*}
If you want more terms, ask Wolfram Alpha. You can also obtain a series in terms of $\mathrm ds$ instead of $\mathrm dx$, simply by substituting $x=\sin s$. But since the first difference is in the $x^4$ (resp. $s^4$) term, the above approximation won't notice the difference.