| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Tangent equation at a known point on circle |
| Difficulty | Moderate -0.3 This is a straightforward circle question requiring completion of the square to find centre and radius, then using the perpendicular radius property to find the tangent equation. All steps are standard techniques with no novel problem-solving required, making it slightly easier than average but still requiring multiple competent applications of method. |
| 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 |
|---|---|---|
| (a) Attempts \((x \pm 2)^2 + (y \pm 5)^2 ..... = 0\) | M1 | Attempts to complete the square on both terms or states the centre as \((\pm 2, \pm 5)\). For completing the square look for \((x \pm 2)^2 + (y \pm 5)^2 ..... = ...\) |
| Centre \((-2, 5)\) | A1 | Allow \(x = -2, y = 5\) This alone can score both marks even following incorrect lines eg \((x + 2)^2 .. (y - 5)^2 = ...\) where ... could be, for example a minus sign or blank |
| Radius \(\sqrt{50}\) or \(5\sqrt{2}\) | B1 | You may isw after a correct answer. If a candidate attempts to use \(x^2 + y^2 + 2fx + 2gy + c = 0\) then M1 may be awarded for a centre of \((\pm 2, \pm 5)\) |
| (3 marks) | ||
| (b) Gradient of radius \(= \frac{(5) - 4}{(-2) - 5} = -\frac{1}{7}\) which needs to be in simplest form | B1ft | Correct answer for the gradient of the line joining \(P(5,4)\) to their centre. You may ft on their centre but the value must be fully simplified. |
| Uses \(m_2 = -\frac{1}{m_1}\) to find gradient of tangent | M1 | Awarded for using \(m_2 = -\frac{1}{m_1}\) to find gradient of tangent. Do be aware that some good candidates may do the first two marks at once so you may need to look at what value they are using for the gradient of the tangent. |
| Equation of tangent \(y - 4 = "7"(x - 5) \Rightarrow y = 7x - 31\) | M1 A1 | For an attempt to find the equation of the tangent using \(P(5,4)\) and a changed gradient. Condone bracketing slips only. If the candidate uses the form \(y = mx + c\) they must use \(x\) and \(y\) the correct way around and proceed as far as \(c = ...\). A1: \(y = 7x - 31\) stated. It must be written in this form. (It cannot be awarded from \(y = mx + c\) by just stating \(c = -31\)) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + y^2 + 4x - 10y - 21 = 0 \Rightarrow 2x + 2y\frac{dy}{dx} + 4 - 10\frac{dy}{dx} = 0\) | B1 | |
| Substitutes \(P(5,4)\) into an expression of the form \(ax + by\frac{dy}{dx} + c + d\frac{dy}{dx} = 0\) AND finds the value of \(\frac{dy}{dx} = (7)\). The values of \(a, b, c\) and \(d\) must be non-zero. | M1 | |
| Uses \(m = \frac{dy}{dx}\bigg | _{x=5}\) with \(P(5,4)\) to find equation of tangent | M1 |
| \(y = 7x - 31\) | A1 |
**(a)** Attempts $(x \pm 2)^2 + (y \pm 5)^2 ..... = 0$ | M1 | Attempts to complete the square on both terms or states the centre as $(\pm 2, \pm 5)$. For completing the square look for $(x \pm 2)^2 + (y \pm 5)^2 ..... = ...$
Centre $(-2, 5)$ | A1 | Allow $x = -2, y = 5$ This alone can score both marks even following incorrect lines eg $(x + 2)^2 .. (y - 5)^2 = ...$ where ... could be, for example a minus sign or blank
Radius $\sqrt{50}$ or $5\sqrt{2}$ | B1 | You may isw after a correct answer. If a candidate attempts to use $x^2 + y^2 + 2fx + 2gy + c = 0$ then M1 may be awarded for a centre of $(\pm 2, \pm 5)$
| (3 marks)
**(b)** Gradient of radius $= \frac{(5) - 4}{(-2) - 5} = -\frac{1}{7}$ which needs to be in simplest form | B1ft | Correct answer for the gradient of the line joining $P(5,4)$ to their centre. You may ft on their centre but the value must be fully simplified.
Uses $m_2 = -\frac{1}{m_1}$ to find gradient of tangent | M1 | Awarded for using $m_2 = -\frac{1}{m_1}$ to find gradient of tangent. Do be aware that some good candidates may do the first two marks at once so you may need to look at what value they are using for the gradient of the tangent.
Equation of tangent $y - 4 = "7"(x - 5) \Rightarrow y = 7x - 31$ | M1 A1 | For an attempt to find the equation of the tangent using $P(5,4)$ and a changed gradient. Condone bracketing slips only. If the candidate uses the form $y = mx + c$ they must use $x$ and $y$ the correct way around and proceed as far as $c = ...$. A1: $y = 7x - 31$ stated. It must be written in this form. (It cannot be awarded from $y = mx + c$ by just stating $c = -31$) | (4 marks, 7 marks total)
**Additional notes for (b):** Attempts at (b) using differentiation:
$x^2 + y^2 + 4x - 10y - 21 = 0 \Rightarrow 2x + 2y\frac{dy}{dx} + 4 - 10\frac{dy}{dx} = 0$ | B1 |
Substitutes $P(5,4)$ into an expression of the form $ax + by\frac{dy}{dx} + c + d\frac{dy}{dx} = 0$ AND finds the value of $\frac{dy}{dx} = (7)$. The values of $a, b, c$ and $d$ must be non-zero. | M1 |
Uses $m = \frac{dy}{dx}\bigg|_{x=5}$ with $P(5,4)$ to find equation of tangent | M1 |
$y = 7x - 31$ | A1 |
---2. A circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } + 4 x - 10 y - 21 = 0$$
Find
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item the coordinates of the centre of $C$,
\item the exact value of the radius of $C$.
The point $P ( 5,4 )$ lies on $C$.
\end{enumerate}\item Find the equation of the tangent to $C$ at $P$, writing your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q2 [7]}}