| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard circle equation and line-circle intersection techniques. Part (i) is direct substitution into the circle formula, part (ii) requires substituting the line equation into the circle equation and solving a quadratic (which factors nicely), and part (iii) uses the distance formula with given coordinates. All steps are routine C1 procedures with no conceptual challenges, making it slightly easier than average. |
| 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 | Mark | Notes |
| \((x+1)^2 + (y-6)^2 = (2\sqrt{5})^2\) | M1 | |
| \((x+1)^2 + (y-6)^2 = 20\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| Sub. \(y = 3x-1\) into eqn of \(C\): \((x+1)^2 + [(3x-1)-6]^2 = 20\) | M1 | |
| \((x+1)^2 + (3x-7)^2 = 20\) | ||
| \(x^2 - 4x + 3 = 0\) | A1 | |
| \((x-1)(x-3) = 0\) | M1 | |
| \(x = 1, 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(x=1 \Rightarrow y=2 \therefore (1,2)\), \(\quad x=3 \Rightarrow y=8 \therefore (3,8)\) | B1 | |
| \(AB = \sqrt{(3-1)^2+(8-2)^2} = \sqrt{4+36} = \sqrt{40} = \sqrt{4\times10} = 2\sqrt{10}\) | M1 A1 | (9) |
# Question 7:
## Part (i):
| Answer | Mark | Notes |
|--------|------|-------|
| $(x+1)^2 + (y-6)^2 = (2\sqrt{5})^2$ | M1 | |
| $(x+1)^2 + (y-6)^2 = 20$ | A1 | |
## Part (ii):
| Answer | Mark | Notes |
|--------|------|-------|
| Sub. $y = 3x-1$ into eqn of $C$: $(x+1)^2 + [(3x-1)-6]^2 = 20$ | M1 | |
| $(x+1)^2 + (3x-7)^2 = 20$ | | |
| $x^2 - 4x + 3 = 0$ | A1 | |
| $(x-1)(x-3) = 0$ | M1 | |
| $x = 1, 3$ | A1 | |
## Part (iii):
| Answer | Mark | Notes |
|--------|------|-------|
| $x=1 \Rightarrow y=2 \therefore (1,2)$, $\quad x=3 \Rightarrow y=8 \therefore (3,8)$ | B1 | |
| $AB = \sqrt{(3-1)^2+(8-2)^2} = \sqrt{4+36} = \sqrt{40} = \sqrt{4\times10} = 2\sqrt{10}$ | M1 A1 | **(9)** |
---7. The circle $C$ has centre $( - 1,6 )$ and radius $2 \sqrt { 5 }$.\\
(i) Find an equation for $C$.
The line $y = 3 x - 1$ intersects $C$ at the points $A$ and $B$.\\
(ii) Find the $x$-coordinates of $A$ and $B$.\\
(iii) Show that $A B = 2 \sqrt { 10 }$.\\
\hfill \mbox{\textit{OCR C1 Q7 [9]}}