## Question 9(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \sqrt{(6^2 + 3^2)}$ or $r^2 = 45$ | B1 | Sight of $r = 6.7$ implies B1 |
| $(x-5)^2 + (y-1)^2 = r^2$ or $x^2 - 10x + y^2 - 2y = r^2 - 26$ | M1 | Using centre given and *their* radius or $r$ in correct formula |
| $(x-5)^2 + (y-1)^2 = 45$ or $x^2 - 10x + y^2 - 2y = 19$ | A1 | Do not allow $(\sqrt{45})^2$ for $r^2$ |

## Question 9(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ has coordinates $(11, 4)$ | B1 | |
| $0.5$ | B1 | OE, Gradient of $AB$, $BC$ or $AC$ |
| Grad of $CD = -2$ | M1 | Calculation of gradient needs to be shown for this M1 |
| $(\frac{1}{2} \times -2 = -1)$ then states perpendicular $\rightarrow$ hence shown or tangent | A1 | Clear reasoning needed |

**Alternative method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ has coordinates $(11, 4)$ | B1 | |
| $0.5$ | B1 | OE, Gradient of $AB$, $BC$ or $AC$ |
| Gradient of perpendicular is $-2$; equation of perpendicular is $y - 4 = -2(x-11)$ | M1 | Use of $m_1 m_2 = -1$ with *their* gradient of $AB$, $BC$ or $AC$ and correct method for equation of perpendicular. Could use $D(5,16)$ instead of $C(11,4)$ |
| Checks $D(5, 16)$ or checks gradient of $CD$ and states $D$ lies on the line or $CD$ has gradient $-2 \rightarrow$ hence shown or tangent | A1 | Clear check and reasoning needed |

**Alternative method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ has coordinates $(11, 4)$ | B1 | |
| $0.5$ | B1 | OE, Gradient of $AB$, $BC$ or $AC$ |
| Gradient of perpendicular is $-2$; equation: $y - 4 = -2(x-11)$; $(x-5)^2 + (-2x+26-1)^2 = 45 \rightarrow (x^2 - 22x + 121 = 0)$ | M1 | Solving simultaneously with equation of circle |
| $(x-11)^2 = 0$ or $b^2 - 4ac = 0 \rightarrow$ repeated root $\rightarrow$ hence shown or tangent | A1 | Must state repeated root |

**Alternative method 3:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ has coordinates $(11, 4)$ | B1 | |
| Finding $CD = \sqrt{180}$ and $BD = \sqrt{225}$ | B1 | OE. Calculated from co-ordinates of $B$, $C$ & $D$ without using $r$ |
| Checking (their $BD)^2 -$ (their $CD)^2$ is the same as (their $r)^2$ | M1 | |
| $\therefore$ Pythagoras valid $\therefore$ perpendicular $\rightarrow$ hence shown or tangent | A1 | Triangle $ACD$ could be used instead |

**Alternative method 4:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $C$ has coordinates $(11, 4)$ | B1 | |
| Finding vectors $\overrightarrow{AC}$ and $\overrightarrow{CD}$ or $\overrightarrow{BC}$ and $\overrightarrow{CD}$ $\left(= \begin{pmatrix}6\\3\end{pmatrix} \text{ and } \begin{pmatrix}-6\\12\end{pmatrix} \text{ or } \begin{pmatrix}12\\6\end{pmatrix} \text{ and } \begin{pmatrix}-6\\12\end{pmatrix}\right)$ | B1 | Must be correct pairing |
| Applying scalar product to one of these pairs of vectors | M1 | Accept *their* $\overrightarrow{AC}$ and $\overrightarrow{CD}$ or *their* $\overrightarrow{BC}$ and $\overrightarrow{CD}$ |
| Scalar product $= 0$ then states $\therefore$ perpendicular $\rightarrow$ hence shown or tangent | A1 | |

## Question 9(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(-1, 4)$ | B1 B1 | WWW. B1 for each coordinate. Note: Equation of $DE$ which is $y = 2x + 6$ may be used to find $E$ |

---