| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Intersection existence or conditions |
| Difficulty | Standard +0.3 This is a standard circle-line intersection problem requiring substitution of y=mx into the circle equation, forming a quadratic in x, and applying the discriminant condition b²-4ac≥0. The algebra is straightforward and the technique is commonly practiced. Part (b) requires basic geometric interpretation. Slightly above average due to the algebraic manipulation and discriminant application, but still a routine A-level question. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| DR \(x^2 + (mx)^2 - 6x - 2mx + 5 = 0\), \((1+m^2)x^2 - (6+2m)x + 5 = 0\) (I) | M1 | Substitute \(y = mx\) into the other equation, in original or rearranged form even if incorrectly rearranged |
| \((6+2m)^2 - 20(1+m^2) \geq 0\), \(\Delta = -16m^2 + 24m + 16 \geq 0\) | M1, M1 | Attempt find \(\Delta\), ft their equation; attempt rearrange \(\Delta\) as a quadratic expression in \(m\) |
| Roots of \(-16m^2 + 24m + 16 = 0\) are \(m = 2\) and \(m = -\frac{1}{2}\) | A1 | or critical values are \(2\) and \(-\frac{1}{2}\) cao |
| Range for real solutions is \(-\frac{1}{2} \leq m \leq 2\) | A1 | cao. Not "\(<\)" |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m=2 \Rightarrow x^2 + 4x^2 - 6x - 4x + 5 = 0\ (\Rightarrow 5x^2 - 10x + 5 = 0)\) | M1 | Substitute \(m=2\) into their (I), or substitute \(y=2x\) into \(x^2+y^2-6x-2y+5=0\) |
| \(x=1\), & repeated root or only one root oe or \(x=1, x=1\). NB May be implied by next line. | A1 | |
| Line is a tangent | A1 | or "Only one intersection point" oe dep M1 only |
| Alt method 1: \(m=2\) gives \(\Delta = -16\times4 + 24\times2 + 16 = 0\), hence repeated root, NB may be implied by next line; Line is a tangent | M1, A1, A1 | Substitute \(m=2\) into their \(\Delta\); or "Only one intersection point" oe |
| Alt method 2: Attempt draw circle centre \((3,1)\) and line through \(O\); Approx correct diagram showing line touching circle; State "Tangent" or "Only one intersection point" oe | M1, A1, A1 | NB Question allows for diagrammatic solution. Dep M1A1 |
| [3] |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **DR** $x^2 + (mx)^2 - 6x - 2mx + 5 = 0$, $(1+m^2)x^2 - (6+2m)x + 5 = 0$ (I) | M1 | Substitute $y = mx$ into the other equation, in original or rearranged form even if incorrectly rearranged |
| $(6+2m)^2 - 20(1+m^2) \geq 0$, $\Delta = -16m^2 + 24m + 16 \geq 0$ | M1, M1 | Attempt find $\Delta$, ft their equation; attempt rearrange $\Delta$ as a quadratic expression in $m$ |
| Roots of $-16m^2 + 24m + 16 = 0$ are $m = 2$ and $m = -\frac{1}{2}$ | A1 | or critical values are $2$ and $-\frac{1}{2}$ cao |
| Range for real solutions is $-\frac{1}{2} \leq m \leq 2$ | A1 | cao. Not "$<$" |
| **[5]** | | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m=2 \Rightarrow x^2 + 4x^2 - 6x - 4x + 5 = 0\ (\Rightarrow 5x^2 - 10x + 5 = 0)$ | M1 | Substitute $m=2$ into their (I), or substitute $y=2x$ into $x^2+y^2-6x-2y+5=0$ |
| $x=1$, & repeated root or only one root oe or $x=1, x=1$. NB May be implied by next line. | A1 | |
| Line is a tangent | A1 | or "Only one intersection point" oe dep M1 only |
| **Alt method 1:** $m=2$ gives $\Delta = -16\times4 + 24\times2 + 16 = 0$, hence repeated root, NB may be implied by next line; Line is a tangent | M1, A1, A1 | Substitute $m=2$ into their $\Delta$; or "Only one intersection point" oe |
| **Alt method 2:** Attempt draw circle centre $(3,1)$ and line through $O$; Approx correct diagram showing line touching circle; State "Tangent" or "Only one intersection point" oe | M1, A1, A1 | NB Question allows for diagrammatic solution. Dep M1A1 |
| **[3]** | | |
---7
\begin{enumerate}[label=(\alph*)]
\item In this question you must show detailed reasoning.
Find the range of values of the constant $m$ for which the simultaneous equations $y = m x$ and $x ^ { 2 } + y ^ { 2 } - 6 x - 2 y + 5 = 0$ have real solutions.
\item Give a geometrical interpretation of the solution in the case where $m = 2$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q7 [8]}}