(a) To find the centre and radius, rewrite the circle equation by completing the square:

\((x+3)^2 + (y-1)^2 = 36\)

This gives the centre \((-3, 1)\) and radius \(6\).

The lowest point on the circle is directly below the centre by the radius, so it is \((-3, 1-6) = (-3, -5)\).

(b) Substitute \(y = kx - 5\) into the circle equation:

\(x^2 + (kx-5)^2 + 6x - 2(kx-5) - 26 = 0\)

Rearrange to form a quadratic in \(x\):

\((k^2+1)x^2 + (6-12k)x + 9 = 0\)

For two distinct points, the discriminant must be positive:

\((6-12k)^2 - 4(k^2+1) \times 9 > 0\)

Simplify to:

\(144k^2 - 144k + 36 - 36k^2 - 36 > 0\)

\(108k^2 - 144k > 0\)

\(k(108k - 144) > 0\)

\(k = 0\) or \(k = \frac{4}{3}\)

Thus, \(k < 0\) or \(k > \frac{4}{3}\).