Suppose the graphs $ x^2 + y^2 = k^2 $ and $ xy = k $ intersect, which means the system \begin{align*} x^2 + y^2 &= k^2, \\ xy &= k \end{align*}has a solution. Then \[(x - y)^2 \ge 0.\]Expanding, we get $ x^2 - 2xy + y^2 \ge 0, $ so \[k^2 - 2k \ge 0.\]This is satisfied by all integers except $ k = 1 $. By the same token, $ (x + y)^2 \ge 0, $ or $ x^2 + 2xy + y^2 \ge 0 $. Hence, \[k^2 + 2k \ge 0.\]This is satisfied by all integers except $ k = -1 $. We have found that $ k = 1 $ and $ k = -1 $ do not work. If $ k = 0, $ then $ (x,y) = (0,0) $ is a solution. Now, assume $ k \ge 2 $. The point $ (\sqrt{k},\sqrt{k}) $ lies on the hyperbola $ xy = k, $ and its distance from the origin is \[\sqrt{k + k} = \sqrt{2k} \le \sqrt{k \cdot k} = k.\]Since there are points arbitrary far from the origin on the hyperbola $ xy = k, $ there must be a point on the hyperbola whose distance from the origin is exactly $ k, $ which means it lies on the circle $ x^2 + y^2 = k^2 $. For $ k \le -2, $ the graph of $ xy = k $ is the graph of $ xy = -k $ rotated $ 90^\circ $ about the origin, so the case where $ k \ge 2 $ applies, i.e. the hyperbola $ xy = k $ and circle $ x^2 + y^2 = k^2 $ intersect. Hence, the there are $ \boxed{2} $ integer values of $ k $ for which the graphs do not intersect, namely 1 and $ -1 $.