# Question 8:

## Part (a)

| Working/Answer | Mark | Guidance |
|---|---|---|
| Major arc of a circle drawn anywhere | B1 | 1.1b — Major arc drawn anywhere |
| End points of arc at $(2, 5)$ and $(8, 5)$, arc drawn above the coordinates | B1 | 1.1b — Correct end points; condone written as complex numbers |

## Part (b)

| Working/Answer | Mark | Guidance |
|---|---|---|
| Centre lies on the **perpendicular bisector/midpoint/equidistant** of $2$ and $8$ | B1 | 2.4 — States perpendicular bisector or midpoint of 2 and 8; condone "in between 2 and 8" if they write $\frac{2+8}{2}=5$. Note: $\frac{2+8}{2}=5$ alone is B0; "in between 2 and 8" alone is B0 |

## Part (c)

| Working/Answer | Mark | Guidance |
|---|---|---|
| Complete method: $\sin\!\left(\frac{\pi}{3}\right) = \frac{3}{r} \Rightarrow r = \ldots$ | M1 | 3.1a — A complete method to find the radius |
| Or: $6^2 = r^2 + r^2 - 2 \times r \times r \times \cos\!\left(\frac{2\pi}{3}\right) \Rightarrow r = \ldots$ | M1 | 3.1a |
| Or: $h = \frac{3}{\tan(\pi/3)} = \sqrt{3} \Rightarrow r = \sqrt{(\sqrt{3})^2 + 3^2} = \ldots$ | M1 | 3.1a |
| $r = \frac{6}{\sqrt{3}}$ or $2\sqrt{3}$ | A1 | 1.1b — Correct radius |

## Part (d)

| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = 5 + h$ where $h = \frac{3}{\tan(\pi/3)}$ or $h = 2\sqrt{3}\cos\!\left(\frac{\pi}{3}\right)$ or $h = \sqrt{(2\sqrt{3})^2 - 3^2}$ | M1 | 3.1a — Any correct complete strategy; if they attempt to find the height (even if incorrect method) in part (c) then $y = 5 +$ their height |
| $y = 5 + \sqrt{3}$ | A1 | 2.2a — Correct answer |

---