(a) To find the y-coordinates of points \(A\) and \(B\), substitute \(x = 0\) into the circle's equation:

\((0-3)^2 + (y-5)^2 = 40\)

\(9 + (y-5)^2 = 40\)

\((y-5)^2 = 31\)

\(y - 5 = \pm \sqrt{31}\)

\(y = 5 \pm \sqrt{31}\)

(b) The midpoint of \(A\) and \(B\) is the center of the new circle. The y-coordinates of \(A\) and \(B\) are \(5 + \sqrt{31}\) and \(5 - \sqrt{31}\), so the midpoint is:

\(\left(0, \frac{(5 + \sqrt{31}) + (5 - \sqrt{31})}{2}\right) = (0, 5)\)

The diameter of the circle is the distance between \(A\) and \(B\), which is:

\(2\sqrt{31}\)

The radius is half of the diameter:

\(\sqrt{31}\)

The equation of the circle with center \((0, 5)\) and radius \(\sqrt{31}\) is:

\(x^2 + (y-5)^2 = 31\)

(a) Substitute the equation of the tangent \(y = mx + 10\) into the circle's equation \(x^2 + y^2 = 20\):

\(x^2 + (mx + 10)^2 = 20\)

\(x^2 + (mx)^2 + 20mx + 100 = 20\)

\((1 + m^2)x^2 + 20mx + 80 = 0\)

Use the discriminant \(b^2 - 4ac = 0\) for tangency:

\((20m)^2 - 4(1 + m^2)80 = 0\)

\(400m^2 - 320 - 80m^2 = 0\)

\(320m^2 = 320\)

\(m^2 = 4\)

\(m = \pm 2\)

(b) For \(m = 2\), the tangent is \(y = 2x + 10\). Substitute into the circle's equation:

\(x^2 + (2x + 10)^2 = 20\)

\(x^2 + 4x^2 + 40x + 100 = 20\)

\(5x^2 + 40x + 80 = 0\)

\(x = -4\) or \(x = 4\)

For \(x = -4\), \(y = 2(-4) + 10 = 2\)

For \(x = 4\), \(y = 2(4) + 10 = 2\)

Coordinates are \((-4, 2)\) and \((4, 2)\).

(c) Use the cosine rule in triangle \(BCD\):

\(BC = 8, BD = \sqrt{(\sqrt{20} + 4)^2 + 2^2}, CD = \sqrt{(\sqrt{20} - 4)^2 + 2^2}\)

\(64 = 80 - 16\sqrt{5} \cos BDC\)

\(\cos BDC = \frac{5}{5}\)

\(\angle BDC = 63.4^\circ\)

(a) To find the equation of line \(BC\), we use the point-slope form. The slope \(m\) of \(BC\) is calculated using the coordinates of \(B(0, 2)\) and the center \(C(2, -4)\):

\(m = \frac{-4 - 2}{2 - 0} = -3\)

Using the point \(B(0, 2)\), the equation of the line is:

\(y - 2 = -3(x - 0)\)

Simplifying gives:

\(y = 2 - 3x\)

(b) To find the coordinates of \(P\), substitute the equation of \(BC\) into the circle's equation:

\((x-2)^2 + (2 - 3x + 4)^2 = 20\)

\((x-2)^2 + (6 - 3x)^2 = 20\)

Expanding and simplifying:

\((x-2)^2 + (6 - 3x)^2 = 20\)

\((x-2)^2 + (36 - 36x + 9x^2) = 20\)

\(10x^2 - 40x + 20 = 0\)

Solving the quadratic equation:

\(x = 2 \pm \sqrt{2}\)

For \(x = 2 - \sqrt{2}\), substitute back into \(y = 2 - 3x\):

\(y = 2 - 3(2 - \sqrt{2}) = 3\sqrt{2} - 4\)

Thus, the coordinates of \(P\) are \((2 - \sqrt{2}, 3\sqrt{2} - 4)\).

(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 \(px + qy = k\):

\(3x + 4y = 7\)

(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 \(k > \frac{4}{3}\).