## Question 10:

### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+2)^2 + (y-4)^2 = 25$ | M1 | $(x+2)^2$ and $(y-4)^2$ seen (or implied by $x^2 + 4x + y^2 - 8y$) |
| $x^2 + 4x + 4 + y^2 - 8y + 16 - 25 = 0$ | M1 | $(x \pm 2)^2 + (y \pm 4)^2 = 25$ |
| $x^2 + y^2 + 4x - 8y - 5 = 0$ | A1 | Correct equation in correct form (terms can be in any order but must have "$= 0$"); Alt: $x^2 + 4x + y^2 - 8y$ **B1**; $c = 2^2 + (\pm 4)^2 - 25$ **M1**; correct equation in correct form **A1** |

**Total: [3]**

### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of radius $= \frac{8-4}{-5+2}$ | M1 | Uses $\frac{y_2 - y_1}{x_2 - x_1}$ (3/4 substitutions correct) |
| $= -\frac{4}{3}$ | A1 | Allow $\frac{4}{-3}$ |
| Gradient of tangent $= \frac{3}{4}$ | B1FT | |
| $y - 8 = \frac{3}{4}(x + 5)$ | M1 | Correct equation of straight line through $(-5, 8)$, any non-zero gradient |
| $3x - 4y + 47 = 0$ | A1 | Shows rearrangement to given equation **AG**; **CWO** throughout for A1 |

**Alt by rearrangement:** Gradient of radius $= \frac{8-4}{-5+2} = \frac{-4}{3}$ **M1\* A1**; attempts to rearrange equation of line to find gradient of line $= \frac{3}{4}$ **M1dep**; multiply gradients to get $-1$ **B1**; check $(-5, 8)$ lies on line **B1 (dep on both M1s)**

**Alt implicit differentiation:** M1 attempt at implicit diff as evidenced by $2y\frac{dy}{dx}$ term; **A1ft** $2x + 2y\frac{dy}{dx} + 4 - 8\frac{dy}{dx} = 0$ ft from their equation in (i); **A1** substitution of $(-5, 8)$ to obtain $\frac{3}{4}$, then final 2 marks as main scheme

**Total: [5]**

### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3 \times 3) - (4 \times 14) + 47 = 0$ | B1 | Sufficient correct working to verify statement e.g. verifying co-ordinate as shown; Alt: showing line joining $(-5, 8)$ to $(3, 14)$ has same gradient etc. |

**Total: [1]**

### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sqrt{(3-{^-}5)^2 + (14-8)^2} = 10$ | M1 | Use of $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ for $TP$ |
| | A1 | |
| Area of triangle $= \frac{1}{2} \times 10 \times 5 = 25$ | M1 | Must use their $TP$ and their $CP$; correct use of area of triangle formula **M1 dep**; all four values correct **A1** |
| $= 25$ | A1 | Final answer correct **A1**; (use the same principle for any enclosing shape) |

**Alt:** Attempt to find area of enclosing rectangle and subtract areas of other three triangles **M1\***

**Total: [4]**

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