(a) Substitute the points \(A(1, 1)\) and \(B(2, -6)\) into the circle equation:

For \(A(1, 1)\):

\(1 + 1 + a + b - 12 = 0 \Rightarrow a + b = 10\)

For \(B(2, -6)\):

\(4 + 36 + 2a - 6b - 12 = 0 \Rightarrow 2a - 6b = -28\)

Solving these equations:

\(a + b = 10\)

\(2a - 6b = -28 \Rightarrow a - 3b = -14\)

Subtract the first from the second:

\((a - 3b) - (a + b) = -14 - 10\)

\(-4b = -24 \Rightarrow b = 6\)

Substitute \(b = 6\) into \(a + b = 10\):

\(a + 6 = 10 \Rightarrow a = 4\)

The centre of the circle is \(\left(-\frac{a}{2}, -\frac{b}{2}\right) = (-2, -3)\).

(b) The gradient of the radius at \(A(1, 1)\) is:

\(\frac{1 - (-3)}{1 - (-2)} = \frac{4}{3}\)

The gradient of the tangent is the negative reciprocal:

\(-\frac{3}{4}\)

Using the point-slope form of the line equation:

\(y - 1 = -\frac{3}{4}(x - 1)\)

Rearrange to the form \(px + qy = k\):

\(3x + 4y = 7\)