## Question 15:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{(12-5)^2 + (7-6)^2} = \sqrt{50}$ or $5\sqrt{2}$ | M1, A1 | |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x \pm 5)^2 + (y \pm 6)^2 = (\text{their numerical } r)^2$ | M1 | |
| $(x-5)^2 + (y-6)^2 = 50$ | B1, A1 | |

### Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $AP = \frac{1}{7}$ | B1 | |
| Gradient of tangent is $-7$ | M1 | |
| Equation of tangent: $(y-7) = -7(x-12)$ | dM1 A1 | |

### Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB = \sqrt{180} = 6\sqrt{5}$, $BC = \sqrt{160} = 4\sqrt{10}$, $AC = 10$ | M1 A1 A1 | |
| $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{160 + 100 - 180}{20\sqrt{160}}$ | M1 | |
| $C = \text{awrt } 71.6°$ | A1 | |

# Question (Circle/Coordinate Geometry):

## Part (a):
| $\sqrt{(12-5)^2 + (7-6)^2}$ or $r^2 = 7^2 + 1^2$ | M1 | Sight of distance formula; may be implied by $\sqrt{50}$ or $5\sqrt{2}$ or awrt 7.1 |
|---|---|---|
| $\sqrt{50}$ or $5\sqrt{2}$ | A1 | |

## Part (b):
| For the form $(x \pm 5)^2 + (y \pm 6)^2 = (\text{their numerical } r)^2$ | M1 | Accept $x^2 + y^2 \pm 10x \pm 12y + c = 0$ where $c = 61 - (\text{their numerical } r^2)$ |
|---|---|---|
| $(x-5)^2 + (y-6)^2 = \ldots$ or $x^2 + y^2 - 10x - 12y \ldots = 0$ | B1 | Can be awarded from $(x-5)^2 + (y-6)^2 = r^2$ with algebraic $r$ |
| $(x-5)^2 + (y-6)^2 = 50$ | A1 | $(x-5)^2+(y-6)^2=(5\sqrt{2})^2$ and $x^2+y^2-10x-12y+11=0$ are acceptable alternatives |

## Part (c):
| Gradient $AP = \frac{1}{7}$ | B1 | |
|---|---|---|
| Use negative reciprocal of gradient of $AP$ for tangent gradient | M1 | |
| Linear equation through $(12,7)$ with negative reciprocal gradient; find $c$ in $y = mx + c$ | dM1 | Dependent on previous M1 |
| $y - 7 = -7(x-12)$, $y = -7x + 91$ | A1 | Accept any unsimplified form |

## Part (d):
| Attempt to find length of any line using difference between coordinates | M1 | Accept $AB = \sqrt{12^2+6^2}\ (\sqrt{180})$, $BC = \sqrt{12^2+4^2}\ (\sqrt{160})$, $AC = \sqrt{8^2+6^2}\ (\sqrt{100}=10)$ |
|---|---|---|
| Two lengths correct: $AB=\sqrt{180}=13.4$, $BC=\sqrt{160}=12.6$, $AC=10$ | A1 | Either exact or awrt 3 sf |
| All three lengths correct: $AB=\sqrt{180}=13.4$, $BC=\sqrt{160}=12.6$, $AC=10$ | A1 | Either exact or awrt 3 sf |
| $\cos C = \frac{BC^2 + AC^2 - AB^2}{2 \times BC \times AC}$ with lengths in correct positions | M1 | Allow method from $AB^2 = AC^2 + BC^2 - 2 \times AC \times AB\cos C$ |
| $C = 71.6°$ | A1 | Accept awrt $C = 71.6$ |

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