# Question 10:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $C = (5, -2)$ | B1 | Correct centre |
| $(x-5)^2 + (y+2)^2 - 25 = 0$ | M1 | $(x \pm 5)^2 - 5^2$ and $(y \pm 2)^2 - 2^2$ seen (or implied by correct answer); or attempt at $r^2 = f^2 + g^2 - c$ |
| Radius $= 5$ | A1 [3] | Correct radius – do not allow A mark from $(x+5)^2$ and/or $(y-2)^2$; $\pm 5$ or $\sqrt{25}$ A0 |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient $PC = \frac{2--2}{8-5} = \frac{4}{3}$ | M1 | Attempt to find gradient of radius (3/4 correct) |
| | A1 | |
| Gradient of tangent $= -\frac{3}{4}$ | B1ft | $\frac{-1}{\text{their gradient}}$ processed |
| $y - 2 = -\frac{3}{4}(x-8)$ | M1 | Equation of straight line through P, using their perpendicular gradient (not from rearrangement); do not allow use of gradient of radius instead of tangent |
| $4y + 3x = 32$ | A1 [5] | Rearrange to required form **www AG**; ignore order of terms |

### Alternative Methods for Part (ii):

**Alternative by rearrangement:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of radius $= \frac{2--2}{8-5} = \frac{4}{3}$ | M1A1 | |
| Attempts to rearrange equation of line to find gradient of line $= -\frac{3}{4}$ and compares with gradient of radius | M1 | |
| Multiply gradients to get $-1$ | B1 | |
| Check $(8, 2)$ lies on line | B1 | |

**Alternative by equating given line to circle:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute for $x/y$ or attempt to get equation in 1 variable only | M1 | |
| $k(x^2 - 16x + 64) = 0$ or $k(y^2 - 4y + 4) = 0$ | A1 | |
| Correct method to solve quadratic | M1 | See Appendix 1 |
| $x = 8,\ y = 2$ found | A1 | |
| States one root implies tangent | B1 | |

**Alternative by implicit differentiation:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at implicit differentiation as evidenced by $2y\frac{dy}{dx}$ term | M1* | |
| $2x + 2y\frac{dy}{dx} - 10 + 4\frac{dy}{dx} = 0$ | A1 | |
| Substitution of $(8, 2)$ to obtain $-\frac{3}{4}$ | A1 | Then as main scheme **OR** attempts to rearrange equation of line to find gradient of line $= -\frac{3}{4}$ **M1dep** |
| Check $(8, 2)$ lies on line | B1 | |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $Q = (0, -2)$ | B1 | $Q$ found correctly; for the M mark, allow splitting into two triangles $\frac{1}{2}\times 6\times 8 + \frac{1}{2}\times 4\times 8$ |
| $R = (0, 8)$ | B1 | $R$ found correctly |
| Area $= \frac{1}{2}\times(8--2)\times 8$ | M1 | Attempt to find area of triangle with their $Q$, $R$ and height 8 i.e. $\frac{1}{2}\times(y_R - y_Q)\times 8$; if using PQ as base then expect to see $\frac{1}{2}\times\sqrt{80}\times\sqrt{80}$ **www** |
| $40$ | A1 [4] | |

---

# Appendix 1 — Solving a Quadratic:

**By factorisation** (example: $3x^2 - 13x - 10 = 0$):

| Attempt | Marks | Guidance |
|---------|-------|----------|
| $(3x+5)(x-2)$ | M1 | $3x^2$ and $-10$ obtained from expansion |
| $(3x-4)(x-3)$ | M1 | $3x^2$ and $-13x$ obtained from expansion |
| $(3x+5)(x+2)$ | M0 | Only $3x^2$ term correct |

**By formula:**

| Attempt | Marks | Guidance |
|---------|-------|----------|
| $\frac{-13 \pm \sqrt{(-13)^2 - 4\times3\times-10}}{2\times3}$ | M1 | Minus sign incorrect at start of formula |
| $\frac{13 \pm \sqrt{(-13)^2 - 4\times3\times10}}{2\times3}$ | M1 | 10 for $c$ instead of $-10$ is the only sign slip |
| $\frac{-13 \pm \sqrt{(-13)^2 - 4\times3\times10}}{2\times3}$ | M0 | 2 sign errors: initial sign and $c$ incorrect |
| $\frac{13 \pm \sqrt{(-13)^2 - 4\times3\times-10}}{2\times-10}$ | M0 | $2c$ on the denominator instead of $2a$ |

**By completing the square** (M1 awarded at $\pm$ stage, provided $x - \frac{13}{6}$ seen or implied; arithmetical errors may be condoned):

$$3x^2 - 13x - 10 = 0 \implies 3\left(x^2 - \frac{13}{3}x\right) - 10 = 0 \implies 3\left[\left(x - \frac{13}{6}\right)^2 - \frac{169}{36}\right] - 10 = 0$$
$$\left(x - \frac{13}{6}\right)^2 = \frac{289}{36} \implies x - \frac{13}{6} = \pm\sqrt{\frac{289}{36}}$$