**(i)** Gradient of $AB = \frac{1-3}{7-1} = -\frac{1}{3}$ | M1* | Uses $\frac{y_2 - y_1}{x_2 - x_1}$ for any 2 points |

Gradient of $AC = \frac{-9-3}{-3-1} = 3$ | A1, A1 | One correct gradient (may be for gradient of BC or =1) | Gradients for both AB and AC found correctly |

| M1 | Attempts to show that $m_1 \times m_2 = -1$ oe, accept "negative reciprocal" | Do not allow final mark if vertex A found from wrong working. (Dependent on 1st M 1 A1 A1)

Vertex A | DB1 | | Accept BAC etc for vertex A or "between AB and AC" Allow if marked on diagram.

**OR** Length of $AB = \sqrt{(7-1)^2 + (1-3)^2} = \sqrt{40}$ | M1* | Correct use of Pythagoras, square rooting not needed |

$AC = \sqrt{(-3-1)^2 + (-9-3)^2} = \sqrt{160}$ | | Any length or length squared correct |

$BC = \sqrt{(-3-7)^2 + (-9-1)^2} = \sqrt{200}$ | A1 | All three correct |

Shows that $AB^2 + AC^2 = BC^2$ | A1 | |

Vertex A | M1, DB1, 5 | Correct use of Pythagoras to show $AB^2 + AC^2 = BC^2$ | i.e must add squares of shorter two lengths |

**(ii)** Midpoint of BC is $(\frac{7 + -3}{2}, \frac{1 + -9}{2}) = (2, -4)$ | M1* | Uses $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ o.e. for BC, AB or AC (3 out of 4 subs correct) | Correct centre (cao) |

Length of $BC = \sqrt{(-3-7)^2 + (-9-1)^2} = \sqrt{200} = 10\sqrt{2}$ | A1 | |

Radius $= 5\sqrt{2}$ | | |

$(x-2)^2 + (y+4)^2 = (5\sqrt{2})^2$ | M1** | Correct method to find d or r or d² or r² o.e. for BC, AB or AC (must be consistent with their midpoint if found) | $(x-a)^2 + (y-b)^2$ seen for their centre |

$(x-2)^2 + (y+4)^2 = 50$ | DM1*, DM1**, A1, A1 | Correct method to find d or r or d² or r² o.e. for BC, AB or AC (must be consistent with their midpoint if found) | $(x-a)^2 + (y-b)^2$ = their r² | Correct equation | Correct equation in required form |

$x^2 + y^2 - 4x + 8y - 30 = 0$ | A1 | | Substitution method 1 (into $x^2 + y^2 + ax + by + c = 0$) Substitutes all 3 points to get 3 equations in a,b,c: M1 At least 2 equations correct A1 Correct method to find one variable M1 One of a, b, c correct A1 Correct method to find other values M1 All values correct A1 Correct equation in required form A1 Alternative markscheme for last 4 marks with f.g. c method: $x^2 - 4x + y^2 + 8y$ for their centre DM1** | $c = (±2)^2 + 4^2 - 50$ DM1** | $c = -30$ A1 Correct equation in required form A1 | Ends of diameter method (p, q) to (c, d): Attempts to use $(x-p)(x-c) + (y-q)(y-d) = 0$ for BC,AC or AB M2 | $(x-7)(x+3) + (y-1)(y+9) = 0$ A2 for both x brackets correct, A2 for both y brackets correct | $x^2 + y^2 - 4x + 8y - 30 = 0$ A1 | SC If M2 A0 A0 then B1 if both x brackets correct and B1 if both y brackets correct for AC or AB |