# Question 4:

## Part (a)(i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(-7,\ 9)$, i.e. $x=-7$ **or** $y=9$ | B1 | Special case: if only $(9,-7)$ seen, award B1B0 |
| $x=-7$ **and** $y=9$ | B1 | |

## Part (a)(ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r=\sqrt{(-3-(-7))^2+(12-9)^2}$ or $r=\sqrt{(-11-(-7))^2+(6-9)^2}$ or $r=\frac{1}{2}\sqrt{(-3+11)^2+(12-6)^2}$ | M1 | Correct strategy for radius. Correct method for their centre (allow 1 sign slip within one bracket). Must see $\frac{1}{2}$ if finding length of diameter |
| $r=5$ | A1 | Correct radius |

**(4 marks)**

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+7)^2+(y-9)^2=5^2$ or $x^2+y^2+2\times7x-2\times9y+7^2+9^2-5^2=0$ | M1A1 | M1: correct attempt at circle equation using their values, allow $(x\pm(\text{their}-7))^2+(y\pm(\text{their }9))^2=(\text{their numerical }r)^2$. A1: correct equation in any form |

**(2 marks)**

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m_{\text{radius}}=\frac{12-9}{-3+7}\left(=\frac{3}{4}\right)$ or $m_{\text{tangent}}=-1\div\frac{12-9}{-3+7}\left(=-\frac{4}{3}\right)$ or $m_{\text{tangent}}=-\left(\frac{-3+7}{12-9}\right)\left(=-\frac{4}{3}\right)$ | M1 | Attempt to find radius gradient or tangent gradient. Allow one sign error in numerator or denominator |
| $y-12=-\frac{4}{3}(x+3)$ | M1 | Correct straight line method for tangent using point $Q$. Must be fully correct; if radius gradient found, must apply negative reciprocal. If $y=mx+c$, must reach $c=...$ |
| $4x+3y-24=0$ | A1 | Allow any integer multiple |

**(3 marks) — Total 9**