## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre $(5, -2)$ | B1 | |
| Radius $= 5$ | M1 | $5$ or $\sqrt{25}$ soi |
| Diameter $= 10$ | A1 [3] | |

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## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Gradient of line $= \frac{2 - {^-2}}{7-5} (= 2)$ | M1 | Uses $\frac{y_2 - y_1}{x_2 - x_1}$ with their centre. 3/4 substitutions correct |
| | A1 | |
| $y - 2 = 2(x-7)$ or $y -{^-2} = 2(x-5)$ | M1 | Correct equation of straight line through $(7, 2)$ or their centre, any non-zero gradient. Allow other points on the line e.g. mid-point is $(6,0)$ |
| $y = 2x - 12$ | A1 [4] | o.e. 3 term equation |

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## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sqrt{(7-5)^2 + (2-{^-2})^2}$ | M1 | Use of $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ with their centre. 3/4 substitutions correct. **Must have** square root as length specifically asked for. |
| $= \sqrt{20}$ | A1 | |
| $\sqrt{20} < 5$ so $P$ lies inside the circle | B1 FT [3] | Compares their length $CP$ with their radius and states consistent conclusion. **Both lengths must be mentioned.** SC If M0, award **B1** for finding $CP^2 = 20$ and stating $20 < 25$ and concluding inside **www** |

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## Question 10(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-5)^2 + (2x+2)^2 (= 25)$ | M1* | Substitute for $x$/$y$ or attempt to eliminate one of the variables |
| $(x-5)^2 + (2x+2)^2 = 25$ | A1 | Correct unsimplified equation ($= 0$ can be implied) |
| $x^2 - 10x + 25 + 4x^2 + 8x + 4 = 25$ | | |
| $5x^2 - 2x + 4 = 0$ | A1 | Obtain correct 3 term quadratic. If $x$ eliminated, $5y^2 - 4y + 16 = 0$ |
| $b^2 - 4ac = 4 - (4 \times 5 \times 4)$ | M1 dep | Attempt to determine whether equation has real roots with consistent conclusion regarding roots/intersection |
| $b^2 - 4ac < 0$ so no real roots | A1 [5] | Fully justified statement that line and circle do not meet **www**. If the discriminant is evaluated, this must be $-76$ (from the quadratic in $x$) or $-304$ (from the quadratic in $y$) for full marks. |

# Allocation of Method Mark for Solving a Quadratic

This page contains **general guidance** for awarding M1 when solving quadratics, not individual question mark schemes. Here is the extracted content:

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## Factorisation Method

| Attempt | Mark | Guidance |
|---------|------|----------|
| $(2x+2)(x-9)=0$ | **M1** | $2x^2$ and $-18$ obtained from expansion |
| $(2x+3)(x-4)=0$ | **M1** | $2x^2$ and $-5x$ obtained from expansion |
| $(2x-9)(x-2)=0$ | **M0** | Only $2x^2$ term correct |

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## Formula Method

| Attempt | Mark | Guidance |
|---------|------|----------|
| $\frac{-5\pm\sqrt{(-5)^2-4\times2\times-18}}{2\times2}$ | **M1** | Minus sign incorrect at start of formula |
| $\frac{5\pm\sqrt{(-5)^2-4\times2\times18}}{2\times2}$ | **M1** | $18$ for $c$ instead of $-18$ |
| $\frac{-5\pm\sqrt{(-5)^2-4\times2\times18}}{2\times2}$ | **M0** | 2 sign errors: initial sign and $c$ incorrect |
| $\frac{5\pm\sqrt{(-5)^2-4\times2\times-18}}{2\times-5}$ | **M0** | $2b$ on the denominator |

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## Completing the Square Method

| Working | Mark | Guidance |
|---------|------|----------|
| $2x^2-5x-18=0$ | | |
| $2\left(x^2-\frac{5}{2}x\right)-18=0$ | | |
| $2\left[\left(x-\frac{5}{4}\right)^2-\frac{25}{16}\right]-18=0$ | | |
| $\left(x-\frac{5}{4}\right)^2=\frac{169}{16}$ | | |
| $x-\frac{5}{4}=\pm\sqrt{\frac{169}{16}}$ | **M1** | Arithmetical errors may be condoned provided $x-\frac{5}{4}$ seen or implied |

> **Note:** For equations such as $2x^2-5x-18=0$, then $b^2=5^2$ would be condoned in the discriminant and would not be counted as a sign error. Repeated sign errors for $a$ in both occurrences scores **M0**.
>
> If the formula is **not quoted**, substitution must be completely correct to earn **M1**.
>
> If candidate makes **repeated attempts**, mark only the last full attempt.