2005 AMC 12B Problem 14

Since the circle is tangent to $y = 6$ and its center $(0, k)$ is above that line, the radius is $r = k - 6.$ The distance from $(0, k)$ to the line $x - y = 0$ is $\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2},$ and this must also equal $r.$ Setting $\dfrac{k}{\sqrt2} = k - 6$ gives $k = \dfrac{6\sqrt2}{\sqrt2 - 1}$ $= 6\sqrt2\,(\sqrt2+1)$ $= 12 + 6\sqrt2.$ Then $r = k - 6 = 6 + 6\sqrt2.$ Thus, the correct answer is E.