jmc

The line $x=k$ intersects $y=x^2+6x+5$ at the point $(k,k^2+6k+5)$ and the line $y=mx+b$ at the point $(k,mk+b)$. Since these two points have the same $x$-coordinate, the distance between them is the difference of their $y$-coordinates, so we have $$|(k^2+6k+5)-(mk+b)|=5.$$ Simplifying, this gives us two quadratic equations: $k^2+(6-m)k+5-b=5$ and $k^2+(6-m)k+5-b=-5$. We can express these as $$\begin{array}{rl}{k}^2+(6-m)k-b=0& (1)\\ k^2+(6-m)k+10-b=0.& (2)\end{array}$$ We know that all solutions to both of these equations will be places where the line $y=mx+b$ is a vertical distance of $5$ from the parabola, but we know there can only be one such solution! Thus there must be exactly $1$ solution to one of the equations, and no solutions to the other equation. We find the discriminants ($b^2-4ac$) of the equations, so for equation $(1)$ the discriminant is $(6-m{)}^2-4(1)(-b)=(6-m{)}^2+4b$. For equation $(2)$ the discriminant is $(6-m{)}^2-4(1)(10-b)=(6-m{)}^2+4b-40$. One of these equations must equal zero, and one must be less than zero. Since $-40<0$, adding $(6-m{)}^2+4b$ to both sides doesn't change the inequality and $(6-m{)}^2+4b-40<(6-m{)}^2+4b$, so the greater value must be equal to zero so that the lesser value is always less than zero. Thus we have $(6-m{)}^2+4b=0$.

We are also given that the line $y=mx+b$ passes through the point $(1,6)$, so substituting $x=1$ and $y=6$ gives $6=(1)m+b$ or $m+b=6$. This means that $6-m=b$, so we can substitute in the above equation: $$\begin{array}{rl}(6-m{)}^2+4b& =0\Rightarrow \\ (b{)}^2+4b& =0\Rightarrow \\ b(b+4)& =0.\end{array}$$ We are given that $b\ne 0$, so the only solution is $b=-4$. When we plug this into the equation $m+b=6$, we find $m-4=6$ so $m=10$. Thus the equation of the line is $y=mx+b$ or $\boxed{y=10x-4}$.