(Q1) [Cambridge/9794/2007] 一个圆的方程为 $x^2 + y^2 - 8y = 9$。

1. Group the terms and complete the square for the $y$ variable.

$x^2 + (y^2 - 8y) = 9$
$x^2 + (y - 4)^2 - 4^2 = 9$
$x^2 + (y - 4)^2 - 16 = 9$

2. Rearrange into the standard equation of a circle to identify the centre and radius.

$x^2 + (y - 4)^2 = 25$
$x^2 + (y - 4)^2 = 5^2$
Comparing with the standard form $(x - a)^2 + (y - b)^2 = r^2$:
Centre $= (0, 4)$
Radius $= 5$

1. Substitute the linear equation of the line into the equation of the circle to eliminate $x$.

Substitute $x = 3y + k$ into $x^2 + y^2 - 8y = 9$:
$(3y + k)^2 + y^2 - 8y = 9$

2. Expand the squared expression and collect like terms to form the required quadratic equation in $y$.

$(9y^2 + 6ky + k^2) + y^2 - 8y - 9 = 0$
$(9y^2 + y^2) + (6ky - 8y) + (k^2 - 9) = 0$
$10y^2 + (6k - 8)y + (k^2 - 9) = 0$

1. State the condition for the line to be a tangent using the discriminant.

For the line to be a tangent to the circle, the intersection points must be coincident. Thus, the quadratic equation from part (b) must have real and equal roots, meaning its discriminant must be zero:
$\Delta = b^2 - 4ac = 0$

2. Set up the discriminant equation in terms of $k$.

Using $a = 10$, $b = 6k - 8$, and $c = k^2 - 9$ from the equation $10y^2 + (6k - 8)y + (k^2 - 9) = 0$:
$(6k - 8)^2 - 4(10)(k^2 - 9) = 0$

3. Expand and simplify the equation to a standard quadratic form.

$36k^2 - 96k + 64 - 40(k^2 - 9) = 0$
$36k^2 - 96k + 64 - 40k^2 + 360 = 0$
$-4k^2 - 96k + 424 = 0$
Dividing by $-4$:
$k^2 + 24k - 106 = 0$

4. Solve the quadratic equation for the exact values of $k$.

Using the quadratic formula:
$k = \frac{-24 \pm \sqrt{24^2 - 4(1)(-106)}}{2(1)}$
$k = \frac{-24 \pm \sqrt{576 + 424}}{2}$
$k = \frac{-24 \pm \sqrt{1000}}{2}$
$k = \frac{-24 \pm 10\sqrt{10}}{2}$
$k = -12 \pm 5\sqrt{10}$