Do not meet.

We need to find the value of $k$ such that line is tangent to the parabola. Since $k$ is clearly positive, we will focus of $f(x)=\sqrt{x}$ . First, we need to find when $f'(x)=1$ : $f(x)=\sqrt{x}$ $f'(x)=\frac{1}{2\sqrt{x}}$ $\sqrt{x}=\frac{1}{2};~x=\frac{1}{4}$ Now we use the formula for a tangent line: $y=f'(a)(x-a)+f(a)$ $y=\left(x-\frac{1}{4}\right)+\frac{1}{2}$ $y=x+\frac{1}{4}$ $1+4=\boxed{5}$

For the two graphs to not intersect each other, $y=x+k$ is never equal to $y=√x$ .

$x+k≠√x$

$(x+k)^2≠x$

$x^2+2kx+k^2≠x$

$x^2+(2k-1)x+k^2≠0$

For this to never equal 0, it will have to have no real roots; the discriminant is less than 0.

$b^2-4ac<0$

$(2k-1)^2-4k^2<0$

$-4k+1<0$

$k>\frac{1}{4}$

Therefore, $a+b=1+4=5.$