To find the value of \(k\) for which the line is tangent to the curve, we first eliminate \(x\) from the equations. Substitute \(x = k - 2y\) from the line equation into the curve equation:

\(y^2 + 2(k - 2y) = 13\)

Simplify to get:

\(y^2 - 4y + 2k = 13\)

\(y^2 - 4y + 2k - 13 = 0\)

This is a quadratic equation in \(y\). For the line to be tangent to the curve, the discriminant of this quadratic must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 1\), \(b = -4\), and \(c = 2k - 13\). So,

\((-4)^2 - 4(1)(2k - 13) = 0\)

\(16 - 8k + 52 = 0\)

\(68 = 8k\)

\(k = \frac{68}{8} = 8 \frac{1}{2}\)

Thus, the value of \(k\) for which the line is tangent to the curve is \(8 \frac{1}{2}\).