| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Normal to circle at point |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question requiring completing the square to find centre/radius, then using perpendicular gradients to find a point on the circle and the normal equation. All techniques are standard P2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Centre is \((-4, 2)\) | B1 | Correct centre; allow written as \(x = -4\), \(y = 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x^2 + y^2 + 8x - 4y = 0 \Rightarrow (x+4)^2 + (y-2)^2 - 16 - 4 = 0 \Rightarrow r = ...\) | M1 | Correct method for radius; award for \((x \pm a)^2 + (y \pm b)^2 \Rightarrow r = \sqrt{"a^2" + "b^2"}\); or using general form \(r = \sqrt{f^2 + g^2 - c}\); may be implied by correct radius |
| \(r = 2\sqrt{5}\) | A1ft | Exact equivalent e.g. \(\sqrt{20}\); do not accept \(\pm\); only follow through with coordinates \((\pm 4, \pm 2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((-2y-10)^2 + y^2 + 8(-2y-10) - 4y = 0\) or \(x^2 + \left(\frac{-x-10}{2}\right)^2 + 8x - 4\left(\frac{-x-10}{2}\right) = 0\) | M1 | Uses both equations to eliminate one variable; condone slips in substituting; allow use of circle in form \((x-a)^2 + (y-b)^2 = r^2\) |
| \(y^2 + 4y + 4 = 0\) or e.g. \(x^2 + 12x + 36 = 0\) | A1 | Correct 3TQ or any equivalent multiple; condone lack of \(= 0\) |
| \((y+2)^2 = 0 \Rightarrow y = ...\) or \((x+6)^2 = 0 \Rightarrow x = ...\) | dM1 | Solves 3TQ to obtain \(x\) or \(y\) coordinate; dependent on first M1 |
| \((-6, -2)\) | A1 | Allow written as \(x = -6\), \(y = -2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x + 2y + 10 = 0 \Rightarrow m_T = -\frac{1}{2} \Rightarrow m_N = 2\) or \(m_N = \frac{"2" - ("-2")}{"-4" - ("-6")}\) | M1 | Attempts to find gradient of normal using tangent equation (gradient = 2) or using centre and point \(P\) |
| \(y - 2 = 2(x+4)\) or \(y + 2 = 2(x+6)\) | dM1 | Correct straight line method using normal gradient and centre or \(P\); dependent on M1 |
| \(y = 2x + 10\) | A1 | cao |
## Question 6:
### Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre is $(-4, 2)$ | B1 | Correct centre; allow written as $x = -4$, $y = 2$ |
### Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x^2 + y^2 + 8x - 4y = 0 \Rightarrow (x+4)^2 + (y-2)^2 - 16 - 4 = 0 \Rightarrow r = ...$ | M1 | Correct method for radius; award for $(x \pm a)^2 + (y \pm b)^2 \Rightarrow r = \sqrt{"a^2" + "b^2"}$; or using general form $r = \sqrt{f^2 + g^2 - c}$; may be implied by correct radius |
| $r = 2\sqrt{5}$ | A1ft | Exact equivalent e.g. $\sqrt{20}$; do not accept $\pm$; only follow through with coordinates $(\pm 4, \pm 2)$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(-2y-10)^2 + y^2 + 8(-2y-10) - 4y = 0$ or $x^2 + \left(\frac{-x-10}{2}\right)^2 + 8x - 4\left(\frac{-x-10}{2}\right) = 0$ | M1 | Uses both equations to eliminate one variable; condone slips in substituting; allow use of circle in form $(x-a)^2 + (y-b)^2 = r^2$ |
| $y^2 + 4y + 4 = 0$ or e.g. $x^2 + 12x + 36 = 0$ | A1 | Correct 3TQ or any equivalent multiple; condone lack of $= 0$ |
| $(y+2)^2 = 0 \Rightarrow y = ...$ or $(x+6)^2 = 0 \Rightarrow x = ...$ | dM1 | Solves 3TQ to obtain $x$ or $y$ coordinate; dependent on first M1 |
| $(-6, -2)$ | A1 | Allow written as $x = -6$, $y = -2$ |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x + 2y + 10 = 0 \Rightarrow m_T = -\frac{1}{2} \Rightarrow m_N = 2$ or $m_N = \frac{"2" - ("-2")}{"-4" - ("-6")}$ | M1 | Attempts to find gradient of normal using tangent equation (gradient = 2) or using centre and point $P$ |
| $y - 2 = 2(x+4)$ or $y + 2 = 2(x+6)$ | dM1 | Correct straight line method using normal gradient and centre or $P$; dependent on M1 |
| $y = 2x + 10$ | A1 | cao |
---\begin{enumerate}
\item The circle $C$ has equation
\end{enumerate}
$$x ^ { 2 } + y ^ { 2 } + 8 x - 4 y = 0$$
(a) Find\\
(i) the coordinates of the centre of $C$,\\
(ii) the exact radius of $C$.
The point $P$ lies on $C$.\\
Given that the tangent to $C$ at $P$ has equation $x + 2 y + 10 = 0$\\
(b) find the coordinates of $P$\\
(c) Find the equation of the normal to $C$ at $P$, giving your answer in the form $y = m x + c$ where $m$ and $c$ are integers to be found.
\hfill \mbox{\textit{Edexcel P2 2023 Q6 [10]}}