Answer: \(k=4\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

At points of intersection,

\(x(x+2k)=-2x-6k-1.\)

So

\(x^2+2kx+2x+6k+1=0,\)

or

\(x^2+(2k+2)x+6k+1=0.\)

For the line to be a tangent to the curve, this quadratic in \(x\) must have exactly one repeated root.

Therefore its discriminant is zero:

\((2k+2)^2-4(6k+1)=0.\)

Expanding gives

\(4k^2+8k+4-24k-4=0.\)

So

\(4k^2-16k=0.\)

Factorising,

\(4k(k-4)=0.\)

The question asks for the non-zero value, so

\(k=4.\)

This completes the solution and gives the required result.