#### Answer

$$2\pi r $$

#### Work Step by Step

Since
\begin{aligned}
x^{2}+y^{2}&=r^{2} \\ y&=\sqrt{r^{2}-x^{2}} \\
y^{\prime} &=-\frac{x}{\sqrt{r^{2}-x^{2}}}
\end{aligned}
Then
\begin{align*}
\text{Circumference} &=2 \int_{-r}^{r} \sqrt{1+\left(-\frac{x}{\sqrt{r^{2}-x^{2}}}\right)^{2}} d x\\
&=2 \int_{-r}^{r} \sqrt{1+\frac{x^{2}}{r^{2}-x^{2}}} d x \\
&=2 \int_{-r}^{r} \sqrt{\frac{x^{2}+r^{2}-x^{2}}{r^{2}-x^{2}}} d x\\
&=2 \int_{-r}^{r} \frac{r}{\sqrt{r^{2}-x^{2}}} d x\\
&=2 r \sin^{-1}\left(\frac{x}{r}\right) \bigg|_{-r}^{r}\\
&=2r\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\\
&=2\pi r
\end{align*}