1978 AHSME Problems/Problem 11

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), $2$ .

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1978 AHSME Problems/Problem 11