smc

The circle $x^2+y^2=r$ has center $(0,0)$ and radius $\sqrt{r}$. Therefore, if the line $x+y=r$ is tangent to the circle $x^2+y^2=r$, then the distance between $(0,0)$ and the line $x+y=r$ is $\sqrt{r}$. The distance between $(0,0)$ and the line $x+y=r$ is $$\frac{|0+0-r|}{\sqrt{1^2+1^2}}=\frac{r}{\sqrt{2}}.$$ Hence, $$\frac{r}{\sqrt{2}}=\sqrt{r}.$$ Then $r=\sqrt{r}\cdot \sqrt{2}$, so $\sqrt{r}=\sqrt{2}$, which means $r=\boxed{2}$ or (B), $2$.