## Question 5:

### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Centre is at $(10, 12)$ | B1 B1 | B1: $x = 10$; B1: $y = 12$ |

### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $(x-10)^2 + (y-12)^2 = -195 + 100 + 144 \Rightarrow r = \ldots$ | M1 | Completes the square for both $x$ and $y$ in an attempt to find $r$. $(x \pm$"$10$"$)^2 \pm a$ and $(y \pm$"$12$"$)^2 \pm b$ and $+195 = 0,\ (a,b \neq 0)$. Allow errors in obtaining $r^2$ but must find square root |
| $r = \sqrt{10^2 + 12^2 - 195}$ | A1 | A correct numerical expression for $r$ including the square root; can be implied by a correct value for $r$ |
| $r = 7$ | A1 | Not $r = \pm 7$ unless $-7$ is rejected |

**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Compares with $x^2 + y^2 + 2gx + 2fy + c = 0$ to write down centre $(-g, -f)$ i.e. $(10, 12)$ | B1B1 | B1: $x=10$; B1: $y=12$ |
| Uses $r = \sqrt{(\pm\text{"10"})^2 + (\pm\text{"12"})^2 - c}$ | M1 | |
| $r = \sqrt{10^2 + 12^2 - 195}$ | A1 | A correct numerical expression for $r$ |
| $r = 7$ | A1 | |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $MN = \sqrt{(25 - \text{"10"})^2 + (32 - \text{"12"})^2}$ | M1 | Correct use of Pythagoras |
| $MN\ (= \sqrt{625}) = 25$ | A1 | |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $NP = \sqrt{(\text{"25"}^2 - \text{"7"}^2)}$ | M1 | $NP = \sqrt{(MN^2 - r^2)}$ |
| $NP\ (= \sqrt{576}) = 24$ | A1 | |

**Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos(NMP) = \dfrac{7}{\text{"25"}} \Rightarrow NP = \text{"25"}\sin(NMP)$ | M1 | Correct strategy for finding $NP$ |
| $NP = 24$ | A1 | |

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