| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 8 |
| 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 standard circle question requiring completing the square to find centre and radius, then using the perpendicular gradient property to find the tangent equation. All techniques are routine for C1/C2 level with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-step nature 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 circle1.03f Circle properties: angles, chords, tangents1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((x \pm 5)^2 + (y \pm 3)^2 = \ldots\) | M1 | For \((x\pm5)^2 + (y\pm3)^2 = \ldots\) or Centre \(= (\pm5, \pm3)\) |
| Centre \(= (-5, 3)\) | A1 | Answer without working is M1A1. Allow written as \(x=-5, y=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\ldots = (\pm5)^2 + (\pm3)^2 - 9\) | M1 | |
| Radius \(= 5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Gradient \(OP = \frac{7-3}{-2--5} = \frac{4}{3}\) | M1 | Attempts gradient of radius from \((-5,3)\) to \(P(-2,7)\). Allow one slip in sign or order |
| Gradient of tangent \(= -\frac{3}{4}\) | M1 | Uses perpendicular gradient rule |
| Equation of tangent \(y - 7 = -\frac{3}{4}(x+2)\) | dM1 | Uses correct perpendicular gradient with \(P(-2,7)\). If \(y=mx+c\), must proceed to \(c=\ldots\) |
| \(3x + 4y - 22 = 0\) | A1 | Or multiple thereof |
## Question 9:
**Part (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(x \pm 5)^2 + (y \pm 3)^2 = \ldots$ | M1 | For $(x\pm5)^2 + (y\pm3)^2 = \ldots$ or Centre $= (\pm5, \pm3)$ |
| Centre $= (-5, 3)$ | A1 | Answer without working is M1A1. Allow written as $x=-5, y=3$ |
**Part (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\ldots = (\pm5)^2 + (\pm3)^2 - 9$ | M1 | |
| Radius $= 5$ | A1 | |
**Part (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Gradient $OP = \frac{7-3}{-2--5} = \frac{4}{3}$ | M1 | Attempts gradient of radius from $(-5,3)$ to $P(-2,7)$. Allow one slip in sign or order |
| Gradient of tangent $= -\frac{3}{4}$ | M1 | Uses perpendicular gradient rule |
| Equation of tangent $y - 7 = -\frac{3}{4}(x+2)$ | dM1 | Uses correct perpendicular gradient with $P(-2,7)$. If $y=mx+c$, must proceed to $c=\ldots$ |
| $3x + 4y - 22 = 0$ | A1 | Or multiple thereof |
---9. The circle $C$ has equation
$$x ^ { 2 } + y ^ { 2 } + 10 x - 6 y + 9 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of $C$.
\item Find the radius of $C$.
The point $P ( - 2,7 )$ lies on $C$.
\item Find an equation of the tangent to $C$ at the point $P$.
Write your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q9 [8]}}