| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle touching axes |
| Difficulty | Standard +0.3 This is a straightforward circle problem requiring completion of the square to find the center, using the tangency condition (distance from center to x-axis equals radius) to find k, then finding intersection points with axes and calculating triangle area. All steps are standard techniques with no novel insight 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 circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation is \((x-3)^2 + (y+4)^2 - 9 - 16 + k = 0\) | M1 | Attempts to complete the square; may be implied by centre \((\pm3, \pm4)\) |
| Radius must be \(4 \Rightarrow 25 - k = 16\) | M1 | Uses valid method to find \(k\) |
| \(\Rightarrow k = 9\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=0 \Rightarrow x^2 - 6x + k = 0\) has one solution | M1 | Substitutes \(y=0\) into circle equation |
| \(\Rightarrow 6^2 - 4\times1\times k = 0\) | M1 | Sets discriminant to 0 |
| \(\Rightarrow k = 9\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C\) intersects \(x\)-axis at \((3,0)\) | B1 | Coordinates where circle touches \(x\)-axis |
| \(y^2 + 8y + 9 = 0 \Rightarrow y = \ldots\) or \(y = -4 \pm \sqrt{\text{"16"} - \text{"9"}}\); or base of triangle is \(2\sqrt{\text{"16"}-\text{"9"}}\) | M1 | Attempts \(y\)-coordinates of intersections with \(y\)-axis, or distance between them |
| Area \(RST = \frac{1}{2} \times 2\sqrt{7} \times 3\) | M1 | Uses correct method for area of triangle \(RST\) |
| \(= 3\sqrt{7}\) | A1 |
## Question 6:
### Part (a) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation is $(x-3)^2 + (y+4)^2 - 9 - 16 + k = 0$ | M1 | Attempts to complete the square; may be implied by centre $(\pm3, \pm4)$ |
| Radius must be $4 \Rightarrow 25 - k = 16$ | M1 | Uses valid method to find $k$ |
| $\Rightarrow k = 9$ | A1 | |
### Part (a) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=0 \Rightarrow x^2 - 6x + k = 0$ has one solution | M1 | Substitutes $y=0$ into circle equation |
| $\Rightarrow 6^2 - 4\times1\times k = 0$ | M1 | Sets discriminant to 0 |
| $\Rightarrow k = 9$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C$ intersects $x$-axis at $(3,0)$ | B1 | Coordinates where circle touches $x$-axis |
| $y^2 + 8y + 9 = 0 \Rightarrow y = \ldots$ or $y = -4 \pm \sqrt{\text{"16"} - \text{"9"}}$; or base of triangle is $2\sqrt{\text{"16"}-\text{"9"}}$ | M1 | Attempts $y$-coordinates of intersections with $y$-axis, or distance between them |
| Area $RST = \frac{1}{2} \times 2\sqrt{7} \times 3$ | M1 | Uses correct method for area of triangle $RST$ |
| $= 3\sqrt{7}$ | A1 | |
---\begin{enumerate}
\item A circle has equation
\end{enumerate}
$$x ^ { 2 } - 6 x + y ^ { 2 } + 8 y + k = 0$$
where $k$ is a positive constant.
Given that the $x$-axis is a tangent to this circle,\\
(a) find the value of $k$.
The circle meets the coordinate axes at the points $R , S$ and $T$.\\
(b) Find the exact area of the triangle $R S T$.\\
\includegraphics[max width=\textwidth, alt={}, center]{515f245f-9c5b-4263-ab2c-0a4f96f3bff0-21_2647_1840_118_111}\\
\hfill \mbox{\textit{Edexcel P2 2021 Q6 [7]}}