Answer
$2\pi{r}$
Work Step by Step
$y$ = $\sqrt {r^{2}-x^{2}}$
$\frac{dy}{dx}$ = $\frac{-2x}{2\sqrt{r^{2}-x^{2}}}$
$\frac{dy}{dx}$ = $\frac{-x}{\sqrt{r^{2}-x^{2}}}$
$L$ = $\int_{{\,o}}^{{\,\frac{r}{\sqrt 2}}}$$\sqrt {1+(\frac{-x}{\sqrt{r^{2}-x^{2}}})^{2}}$ $dx$
$L$ = $\int_{{\,o}}^{{\,\frac{r}{\sqrt 2}}}$$\sqrt {(\frac{{r^{2}-x^{2}+x^{2}}}{{r^{2}-x^{2}}})}$ $dx$
$L$ = $\int_{{\,o}}^{{\,\frac{r}{\sqrt 2}}}$$\sqrt {(\frac{{r^{2}}}{{r^{2}-x^{2}}})}$ $dx$
$L$ = $\int_{{\,o}}^{{\,\frac{r}{\sqrt 2}}}$$r\frac{1}{\sqrt {1-(\frac{x}{r})^{2}} }$ $dx$
$L$ = $r[sin^{-1}(\frac{x}{r})]$ $|_{{\,0}}^{{\,\frac{r}{\sqrt 2}}}$
$L$ = $r[sin^{-1}(\frac{1}{\sqrt 2})-sin^{-1}(0)]$
$L$ = $r(\frac{\pi}{4}-0)$
$L$ = $\frac{\pi{r}}{4}$
$circumference$ = $8L$ = $8(\frac{\pi{r}}{4})$ = $2\pi{r}$
