(a) $x^2 + y^2 + 6x - 4y - 14 = 0$ | —

Attempts $(x \pm 3)^2 + (y \pm 2)^2 \ldots (= 0)$ | M1

(i) Centre: $(-3, 2)$ | A1

(ii) Radius: $\sqrt{27}$ or $3\sqrt{3}$ | A1

(3 marks)

(b) Attempts $y/k = \text{``}2\text{''} \pm \text{``}3\sqrt{3}\text{''} $ | M1

Both $k = 2 + 3\sqrt{3}$ and $k = 2 - 3\sqrt{3}$ | A1

**Alt:** $y = k \Rightarrow x^2 + 6x + k^2 - 4k - 14 = 0$

$\Rightarrow b^2 - 4ac = 0 \Rightarrow 6^2 - 4 \times 1 \times (k^2 - 4k - 14) = 0$

$\Rightarrow 4k^2 - 16k - 92 = 0 \Rightarrow k = \ldots$

$k = 2 \pm 3\sqrt{3}$ (oe)

(2 marks)

(c) Attempts Pythagoras to find $h$: $h^2 = r^2 - 2^2$ | M1

Attempts $p = 2 - h$ | dM1

$p = 2 - \sqrt{23}$ (oe) | A1

**Alt1:** Chord touches circle when $x = -1$ (or $-5$) or need $(x + 3) = \pm 2$

$\Rightarrow 2^2 + (y - 2)^2 = 27 \Rightarrow y = \ldots$ or $(-1)^2 + y^2 + 6(-1) - 4y - 14 = 0 \Rightarrow y = \ldots$

$p = 2 - \sqrt{23}$

**Alt2:** Sets $y = p$ and applies difference of roots to be 4

$\Rightarrow x^2 + 6x + p^2 - 4p - 14 = 0 \Rightarrow x = \frac{-6 \pm \sqrt{36 - 4(p^2 - 4p - 14)}}{2}$

$\Rightarrow 2\sqrt{23} + 4p - p^2 = 4$

$\Rightarrow p^2 - 4p - 19 = 0 \Rightarrow p = \ldots$

$p = 2 - \sqrt{23}$

(3 marks)

**Guidance Notes (a):**

M1 Attempts to complete the square. Look for $(x \pm 3)^2 + (y \pm 2)^2 \ldots = 0$. M1 may be awarded for a centre of $(\pm 3, \pm 2)$ (from no, or from incorrect, working).

A1 Centre $(-3, 2)$

A1 Radius $\sqrt{27}$ or $3\sqrt{3}$

**Guidance Notes (b):**

M1 Attempts one value of $k$ by $\text{``}2\text{''} \pm \text{``}r\text{''} $. In the alternative it is for setting $y = k$ in the circle equation, applying $b^2 - 4ac = 0$ and then solving the resulting quadratic in $k$, or equivalent method. But substitution of $x = 0$ is M0.

A1 For both $k = 2 + 3\sqrt{3}$ and $k = 2 - 3\sqrt{3}$. Allow $y \leftrightarrow k$. Accept exact equivalents, e.g. with $\sqrt{27}$.

Alt by differentiation requires: M1 Achieves $ax + by + cx + d = 0$ followed by setting $\frac{dy}{dx} = 0$, finding $x$ and substituting into the circle equation then attempting to solve the quadratic. A1 Correct answers.

**Guidance Notes (c):**

M1 For an attempt to use Pythagoras's theorem with 2 and their 'r' to find 'h' or $h^2$.

dM1 Attempts $p = \text{``}2\text{''} \pm \text{``}h\text{''} $. Condone $p \leftrightarrow y$. Follow through their y coordinate of the centre. Can be implied by the correct decimal answer (awrt $-2.8$).

A1 $p = 2 - \sqrt{23}$ only. Condone $p \leftrightarrow y$. Accept equivalent exact forms e.g. $-\sqrt{23} - 2$.

(8 marks)

---