jmc

Note that the graphs of $y=x^2+k$ and $x=y^2+k$ are reflections of each other in the line $y=x,$ so if they are tangent to each other, then the point of tangency must lie on the line $y=x.$ Furthermore, both graphs will be tangent to the line $y=x.$



This means that the quadratic $x^2+k=x$ will have a double root. We can arrange the equation to get $$x^2-x+k=0.$$We want the discriminant of this quadratic to be 0, giving us $1-4k=0,$ or $k=\boxed{\frac{1}{4}}.$