Let k be a positive real number. The line x + y = k and the circle x^2 + y^2 = 2k + 1 are drawn. Find k so that the line is tangent to the circle.
+600 It's obviously tangent at $(k/2, k/2)$. So $\frac{k^2}{2}=2k+1\implies k^2=4k+2\implies k^2-4k-2=0\implies k=2\pm\sqrt{6}$. We take positive root, so $\boxed{k=2+\sqrt{6}}$.
