| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Tangency condition for line and curve |
| Difficulty | Standard +0.8 This question requires setting up a circle equation from geometric constraints, substituting a line equation, and applying the tangent condition (discriminant = 0) to solve a quadratic in r. While the algebraic manipulation is substantial, the conceptual steps are standard A-level techniques, making it moderately challenging but not requiring novel insight. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.03d 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 |
| \(C\): \((x-r)^2+(y-r)^2=r^2\) or \(x^2+y^2-2rx-2ry+r^2=0\) | B1 | Deduces correct circle equation |
| \(y=12-2x\); \(x^2+(12-2x)^2-2rx-2r(12-2x)+r^2=0\) | M1 | Forms equation with \(x^2\), \(x\), \(r^2\), \(xr\) terms using \(y=12-2x\) |
| \(5x^2+(2r-48)x+(r^2-24r+144)=0\) | A1* | Correct algebra to given solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2-4ac=0 \Rightarrow (2r-48)^2-4\times5\times(r^2-24r+144)=0\) | M1 | Attempts \(b^2-4ac=0\) with \(a=5\), \(b=2r-48\), \(c=r^2-24r+144\) |
| \(r^2-18r+36=0\) | A1 | Correct quadratic in \(r\) |
| \((r-9)^2-81+36=0 \Rightarrow r=\ldots\) | dM1 | Correct attempt to solve quadratic; dependent on previous M |
| \(r = 9\pm3\sqrt{5}\) | A1 | Both answers in simplified surd form |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C$: $(x-r)^2+(y-r)^2=r^2$ or $x^2+y^2-2rx-2ry+r^2=0$ | B1 | Deduces correct circle equation |
| $y=12-2x$; $x^2+(12-2x)^2-2rx-2r(12-2x)+r^2=0$ | M1 | Forms equation with $x^2$, $x$, $r^2$, $xr$ terms using $y=12-2x$ |
| $5x^2+(2r-48)x+(r^2-24r+144)=0$ | A1* | Correct algebra to given solution |
## Question 14(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2-4ac=0 \Rightarrow (2r-48)^2-4\times5\times(r^2-24r+144)=0$ | M1 | Attempts $b^2-4ac=0$ with $a=5$, $b=2r-48$, $c=r^2-24r+144$ |
| $r^2-18r+36=0$ | A1 | Correct quadratic in $r$ |
| $(r-9)^2-81+36=0 \Rightarrow r=\ldots$ | dM1 | Correct attempt to solve quadratic; dependent on previous M |
| $r = 9\pm3\sqrt{5}$ | A1 | Both answers in simplified surd form |
---\begin{enumerate}
\item A circle $C$ with radius $r$
\end{enumerate}
\begin{itemize}
\item lies only in the 1st quadrant
\item touches the $x$-axis and touches the $y$-axis
\end{itemize}
The line $l$ has equation $2 x + y = 12$\\
(a) Show that the $x$ coordinates of the points of intersection of $l$ with $C$ satisfy
$$5 x ^ { 2 } + ( 2 r - 48 ) x + \left( r ^ { 2 } - 24 r + 144 \right) = 0$$
Given also that $l$ is a tangent to $C$,\\
(b) find the two possible values of $r$, giving your answers as fully simplified surds.
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q14 [7]}}