# Question 9:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Midpoint $PQ = \left(\frac{-9+15}{2}, \frac{8-10}{2}\right)$ | M1 | Uses correct method to find midpoint of line segment $PQ$ |
| $= (3,-1)$, which is centre point $A$, so $PQ$ is the diameter | A1 | Completes proof obtaining $(3,-1)$ and gives correct conclusion |

**Alt 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m_{PQ} = \frac{-10-8}{15--9} = -\frac{3}{4} \Rightarrow PQ: y-8=-\frac{3}{4}(x--9)$ | M1 | Full attempt to find equation of line $PQ$ |
| $PQ: y = -\frac{3}{4}x + \frac{5}{4}$. So $x=3 \Rightarrow y = -\frac{3}{4}(3)+\frac{5}{4}=-1$, so $PQ$ is the diameter | A1 | Completes proof showing $(3,-1)$ lies on $PQ$ and gives correct conclusion |

**Alt 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $PQ = \sqrt{(-9-15)^2+(8--10)^2} \{=\sqrt{900}=30\}$ and either $AP=\sqrt{(3--9)^2+(-1-8)^2}\{=\sqrt{225}=15\}$ or $AQ=\sqrt{(3-15)^2+(-1--10)^2}\{=\sqrt{225}=15\}$ | M1 | Attempts to find distance $PQ$ and either $AP$ or $AQ$ |
| e.g. $PQ=2AP$, then $PQ$ is the diameter | A1 | Correctly shows $PQ=2AP$ (with $PQ=30$, $AP=15$) or $PQ=2AQ$ (with $PQ=30$, $AQ=15$) and gives correct conclusion |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses Pythagoras in correct method to find either radius or diameter | M1 | Uses Pythagoras correctly to find **radius**: e.g. $r^2=(-9-3)^2+(8+1)^2$ or $r^2=(15-3)^2+(-10+1)^2$; **or diameter**: e.g. $d^2=(15+9)^2+(-10-8)^2$. Note: implied by 30 as diameter or 15 as radius |
| $(x-3)^2 + (y+1)^2 = 225$ | M1 | Writes circle equation in form $(x\pm"3")^2+(y\pm"-1")^2=(\text{their }r)^2$ |
| $(x-3)^2 + (y+1)^2 = 225$ (or $15^2$) | A1 | Correct equation |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= \sqrt{("15")^2-(10)^2}$ or $= \frac{1}{2}\sqrt{(2("15"))^2-(2(10))^2}$ | M1 | Attempts using circle property "perpendicular from centre to chord bisects the chord" and applies Pythagoras in form $\sqrt{(\text{their "15"})^2-(10)^2}$ |
| $\{=\sqrt{125}\} = 5\sqrt{5}$ | A1 | $5\sqrt{5}$ by correct solution only |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sin(A\hat{R}Q) = \frac{20}{2("15")}$ or $A\hat{R}Q = 90 - \cos^{-1}\left(\frac{10}{"15"}\right)$ | M1 | Attempts using circle property "angle in a semicircle is a right angle"; writes $\sin(A\hat{R}Q)=\frac{20}{2(\text{their "15"})}$ or $A\hat{R}Q = 90-\cos^{-1}\left(\frac{10}{\text{their "15"}}\right)$. Note: also allow $\cos(A\hat{R}Q)=\frac{15^2+(2(5\sqrt{5}))^2-15^2}{2(15)(2(5\sqrt{5}))}\left\{=\frac{\sqrt{5}}{3}\right\}$ |
| $A\hat{R}Q = 41.8103... = 41.8°$ (to 0.1 of a degree) | A1 | 41.8° by correct solution only |

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