| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.3 This is a straightforward multi-part circle question requiring standard techniques: completing the square to find centre and radius, substituting a point to find a coordinate, using the distance formula, and finding a line equation from two points. All methods are routine C1 procedures with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-3)^2 - 9 + y^2 - k = 0\) | B1 | \((x-3)^2\) soi |
| \((x-3)^2 + y^2 = 9 + k\), Centre \((3, 0)\) | B1 | Correct centre |
| \(9 + k = 4^2\), \(k = 7\) | M1, A1 (4) | Correct value for \(k\) (may be embedded) |
| Answer | Marks |
|---|---|
| Centre \((-g, -f)\) | M1 |
| Centre \((3, 0)\) | A1 |
| \(4 = \sqrt{f^2 + g^2 - (-k)}\) | M1 |
| \(k = 7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((3-3)^2 + y^2 = 16\), \(y^2 = 16\), \(y = 4\) | M1, A1 | Attempt to substitute \(x = 3\); \(y = 4\) (do not allow \(\pm 4\)) |
| Length of \(AB = \sqrt{(-1-3)^2 + (0-4)^2}\) | M1 | Correct method using Pythagoras |
| \(= \sqrt{32}\) | A1ft | \(\sqrt{32}\) or \(\sqrt{16 + a^2}\) |
| \(= 4\sqrt{2}\) | A1 (5) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(AB = 1\) or \(\frac{a}{4}\) | B1ft | |
| \(y - 0 = m(x+1)\) or \(y - 4 = m(x-3)\) | M1 | Attempts equation of straight line through their A or B with their gradient |
| \(y = x + 1\) | A1 (3) | Correct equation in any form with simplified constants |
# Question 9:
**(i)**
$(x-3)^2 - 9 + y^2 - k = 0$ | B1 | $(x-3)^2$ soi
$(x-3)^2 + y^2 = 9 + k$, Centre $(3, 0)$ | B1 | Correct centre
$9 + k = 4^2$, $k = 7$ | M1, A1 (4) | Correct value for $k$ (may be embedded)
**Alternative (expanded form):**
Centre $(-g, -f)$ | M1 |
Centre $(3, 0)$ | A1 |
$4 = \sqrt{f^2 + g^2 - (-k)}$ | M1 |
$k = 7$ | A1 |
**(ii)**
$(3-3)^2 + y^2 = 16$, $y^2 = 16$, $y = 4$ | M1, A1 | Attempt to substitute $x = 3$; $y = 4$ (do not allow $\pm 4$)
Length of $AB = \sqrt{(-1-3)^2 + (0-4)^2}$ | M1 | Correct method using Pythagoras
$= \sqrt{32}$ | A1ft | $\sqrt{32}$ or $\sqrt{16 + a^2}$
$= 4\sqrt{2}$ | A1 (5) | cao
**(iii)**
Gradient of $AB = 1$ or $\frac{a}{4}$ | B1ft |
$y - 0 = m(x+1)$ or $y - 4 = m(x-3)$ | M1 | Attempts equation of straight line through their A or B with their gradient
$y = x + 1$ | A1 (3) | Correct equation in any form with simplified constants
---9 The circle with equation $x ^ { 2 } + y ^ { 2 } - 6 x - k = 0$ has radius 4 .\\
(i) Find the centre of the circle and the value of k .
The points $\mathrm { A } ( 3 , \mathrm { a } )$ and $\mathrm { B } ( - 1,0 )$ lie on the circumference of the circle, with $\mathrm { a } > 0$.\\
(ii) Calculate the length of $A B$, giving your answer in simplified surd form.\\
(iii) Find an equation for the line $A B$.
\hfill \mbox{\textit{OCR C1 2007 Q9 [12]}}