**Part (a)**

$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$ | M1 A1, A1 |

$(x - 3)^2 + (y + 2)^2 = 12 + "9" + "4"$ | M1 A1 | Centre is $(3, -2)$ |

$r = \sqrt{12 + "9" + "4"} = 5$ (or $\sqrt{25}$) | M1 A1 |

**Part (b)**

$PQ = \sqrt{(7 - 1)^2 + (-5 - 1)^2}$ or $\sqrt{8^2 + 6^2}$ | M1 |

$= 10 = 2 \times \text{radius}$, $\therefore \text{diam.}$ | A1 | (N.B. For A1, need a comment or conclusion) |

[ALT: midpt. of $PQ$ $\frac{7 + (-1)}{2}$, $\frac{1 + (-5)}{2}$: M1, $= (3, -2) = $ centre: A1]

[ALT: eqn. of $PQ$ $3x + 4y - 1 = 0$: M1, verify $(3, -2)$ lies on this: A1]

[ALT: find two grads, e.g. $PQ$ and $P$ to centre: M1, equal $\therefore$ diameter: A1]

[ALT: show that point $S(-1, -5)$ or $(7, 1)$ lies on circle: M1 because $\angle PSQ = 90°$, semicircle $\therefore$ diameter: A1]

**Part (c)**

$R$ must lie on the circle (angle in a semicircle theorem)... often implied by a diagram with $R$ on the circle or by subsequent working) | B1 |

$x = 0 \Rightarrow y^2 + 4y - 12 = 0$ | M1 |

$(y - 2)(y + 6) = 0$ | cM1 |

$y = ...$ (M is dependent on previous M) | A1 |

$y = -6$ or $2$ (ignoring $y = -6$ if seen, and 'coordinates' are not required)) | A1 |

**Guidance:**

**Part (a):**

$1^{\text{st}}$ M1 for attempt to complete square. Allow $(x \pm 3)^2 \pm k$, or $(y \pm 2)^2 \pm k$, $k \ne 0$.

$1^{\text{st}}$ A1 $x$-coordinate 3, $2^{\text{nd}}$ A1 $y$-coordinate $-2$

$2^{\text{nd}}$ M1 for a full method leading to $r = ...$, with their 9 and their 4, $3^{\text{rd}}$ A1 5 or $\sqrt{25}$

The $1^{\text{st}}$ M can be implied by $(\pm 3, \pm 2)$ but a full method must be seen for the $2^{\text{nd}}$ M.

Where the 'diameter' in part (b) has clearly been used to answer part (a), no marks in (a), but in this case the M1 (not the A1) for part (b) can be given for work seen in (a).

**Alternative:**

$1^{\text{st}}$ M1 for comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$ to write down centre $(-g, -f)$ directly. Condone sign errors for this M mark.

$2^{\text{nd}}$ M1 for using $r = \sqrt{g^2 + f^2 - c}$. Condone sign errors for this M mark.

**Part (c):**

$1^{\text{st}}$ M1 for setting $x = 0$ and getting a 3TQ in $y$ by using eqn. of circle.

$2^{\text{nd}}$ M1 (dep.) for attempt to solve a 3TQ leading to at least one solution for $y$.

**Alternative 1:** (Requires the B mark as in the main scheme)

$1^{\text{st}}$ M for using $(3, 4, 5)$ triangle with vertices $(3, -2), (0, -2), (0, y)$ to get a linear or quadratic equation in $y$ (e.g. $3^2 + (y + 2)^2 = 25$).

$2^{\text{nd}}$ M (dep.) as in main scheme, but may be scored by simply solving a linear equation.

**Alternative 2:** (Not requiring realisation that $R$ is on the circle)

B1 for attempt at $m_{PQ} \times m_{QR} = -1$, (NOT $m_{PQ}$) or for attempt at Pythag. in triangle $PQR$.

$1^{\text{st}}$ M1 for setting $x = 0$, i.e. $(0, y)$, and proceeding to get a 3TQ in $y$. Then main scheme.

**Alternative 2 by 'verification':**

B1 for attempt at $m_{PQ} \times m_{QR} = -1$, (NOT $m_{PQ}$) or for attempt at Pythag. in triangle $PQR$.

$1^{\text{st}}$ M1 for trying $(0, 2)$.

$2^{\text{nd}}$ M1 (dep.) for performing all required calculations.

A1 for fully correct working and conclusion.

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