## Question 10(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of relevant radius is $-2$ | B1 | |
| Using $m_1m_2 = -1$, obtain gradient of tangent and form straight line equation through $(-6, 9)$ | M1 | $m_1$ must come from attempt to find gradient of radius using centre and given point |
| Obtain $y = \frac{1}{2}x + 12$ | A1 | OE e.g. $y - 9 = \frac{1}{2}(x+6)$ |

## Question 10(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $(x+4)^2 + (y-5)^2 = 20$ | B1 | If $x^2+y^2-2gx-2fy+c=0$ used correctly with $(-g,-f)=(-4,5)$ and $c=g^2+f^2-r^2$ then M1 |
| Obtain $x^2 + y^2 + 8x - 10y + 21 = 0$ | B1 | A1 if above method used |

## Question 10(c):

| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x=0$ in equation of circle to find $y$-values 3 and 7, or state $C$ to $AB=4$ | B1 | May be implied by $AB=4$ or use of $x$-coordinate of $C$ |
| Attempt value of $\theta$ either using cosine rule or via $\frac{1}{2}\theta$ using right-angled triangle | M1 | Using their $AB$. If $\theta/2$ used, must be multiplied by 2 |
| Obtain $\theta = 0.9273$ | A1 | Or greater accuracy. Correct answer implies M1 |

## Question 10(d):

| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt arc length using $r\theta$ with their $\theta$ (not $\theta/2$) and $r=\sqrt{20}$ | M1 | Expect 4.15 |
| Obtain perimeter $= 8.15$ or greater accuracy | A1 | Condone missing or incorrect units |
| Attempt area using $\frac{1}{2}r^2(\theta - \sin\theta)$ or equivalent with their $\theta$ and $r=\sqrt{20}$ | M1 | If sector–triangle used, both formulae must be correct. If triangle $ACM$ used, area must be multiplied by 2 |
| Obtain area $= 1.27$ or greater accuracy | A1 | Condone missing or incorrect units |