| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| 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 multi-part question testing standard circle techniques: substituting a point to find a parameter, completing the square to find centre and radius, and finding a tangent equation using the perpendicular gradient method. All steps are routine procedures covered in Core Mathematics with no novel problem-solving required, making it slightly easier than average but not trivial due to the multiple parts and algebraic manipulation involved. |
| 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 |
|---|---|---|
| Working | Mark | Guidance |
| \((1)^2 + (16)^2 + 4(1) + 16p + 123 = 0 \Rightarrow p = ...\) | M1 | Substitute \(x=1, y=16\) into circle equation and solve for \(p\) |
| \(16p = -384 \Rightarrow p = -24\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Centre is \((-2, 12)\) | B1ft | Follow through on their \(p\); coordinates of form \((-2, -\frac{p}{2})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(r^2 = ``2"^2 + ``12"^2 - 123\) or \(r^2 = (1-``-2")^2 + (16-``12")^2\) | M1 | Must have \(r>0\) and \(r^2 \neq 123\) |
| \(r = 5\) | A1cao | Can only be awarded from correct \(p=-24\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(m_N = \frac{16 - ``12"}{1 - ``-2"} \left(= \frac{4}{3}\right)\) | M1 | Attempt gradient between centre \(C\) and \((1,16)\) |
| \(m_r = ``-\frac{3}{4}"\) | M1 | Use \(m_r = -\frac{1}{m_N}\) |
| \(y - 16 = ``-\frac{3}{4}"(x-1)\) | M1 | Line through \((1,16)\) with changed gradient; both brackets correct |
| \(3x + 4y - 67 = 0\) | A1 | Accept \(\pm A(3x+4y-67=0)\) where \(A \in \mathbb{N}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2x + 2y\frac{dy}{dx} + 4 - 24\frac{dy}{dx} = 0\) | M1 | Implicit differentiation to form \(...x + ...y\frac{dy}{dx} + ... + ...\frac{dy}{dx} = 0\) |
| \(``2"\times1 + ``2"\times16\frac{dy}{dx} + ``4" - ``24"\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = ... \left(= -\frac{3}{4}\right)\) | M1 | Substitute \((1,16)\) and solve for \(\frac{dy}{dx}\) |
| \(y - 16 = ``-\frac{3}{4}"(x-1)\) | M1 | Line through \((1,16)\) using their \(\frac{dy}{dx}\) |
| \(3x + 4y - 67 = 0\) | A1 |
## Question 10:
### Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $(1)^2 + (16)^2 + 4(1) + 16p + 123 = 0 \Rightarrow p = ...$ | M1 | Substitute $x=1, y=16$ into circle equation and solve for $p$ |
| $16p = -384 \Rightarrow p = -24$ | A1 | |
### Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Centre is $(-2, 12)$ | B1ft | Follow through on their $p$; coordinates of form $(-2, -\frac{p}{2})$ |
### Part (a)(iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $r^2 = ``2"^2 + ``12"^2 - 123$ or $r^2 = (1-``-2")^2 + (16-``12")^2$ | M1 | Must have $r>0$ and $r^2 \neq 123$ |
| $r = 5$ | A1cao | Can only be awarded from correct $p=-24$ |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $m_N = \frac{16 - ``12"}{1 - ``-2"} \left(= \frac{4}{3}\right)$ | M1 | Attempt gradient between centre $C$ and $(1,16)$ |
| $m_r = ``-\frac{3}{4}"$ | M1 | Use $m_r = -\frac{1}{m_N}$ |
| $y - 16 = ``-\frac{3}{4}"(x-1)$ | M1 | Line through $(1,16)$ with changed gradient; both brackets correct |
| $3x + 4y - 67 = 0$ | A1 | Accept $\pm A(3x+4y-67=0)$ where $A \in \mathbb{N}$ |
### Part (b) Alternative:
| Working | Mark | Guidance |
|---------|------|----------|
| $2x + 2y\frac{dy}{dx} + 4 - 24\frac{dy}{dx} = 0$ | M1 | Implicit differentiation to form $...x + ...y\frac{dy}{dx} + ... + ...\frac{dy}{dx} = 0$ |
| $``2"\times1 + ``2"\times16\frac{dy}{dx} + ``4" - ``24"\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = ... \left(= -\frac{3}{4}\right)$ | M1 | Substitute $(1,16)$ and solve for $\frac{dy}{dx}$ |
| $y - 16 = ``-\frac{3}{4}"(x-1)$ | M1 | Line through $(1,16)$ using their $\frac{dy}{dx}$ |
| $3x + 4y - 67 = 0$ | A1 | |
---\begin{enumerate}
\item The circle $C$ has equation
\end{enumerate}
$$x ^ { 2 } + y ^ { 2 } + 4 x + p y + 123 = 0$$
where $p$ is a constant.
Given that the point $( 1,16 )$ lies on $C$,\\
(a) find\\
(i) the value of $p$,\\
(ii) the coordinates of the centre of $C$,\\
(iii) the radius of $C$.\\
(b) Find an equation of the tangent to $C$ at the point ( 1,16 ), giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.
\includegraphics[max width=\textwidth, alt={}, center]{de511cb3-35c7-4225-b459-a136b6304b78-31_33_19_2668_1896}\\
\hfill \mbox{\textit{Edexcel C12 2019 Q10 [9]}}