| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Line-circle intersection points |
| Difficulty | Moderate -0.3 This is a straightforward C1 circle question requiring completing the square to find centre/radius, substituting a point into the equation, and solving simultaneous equations for line-circle intersection. All techniques are standard and routine, making it slightly easier than average, though the multi-part structure and algebraic manipulation keep it from being trivial. |
| 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 | Marks | Guidance |
| Centre \((-1, 2)\) | B1 | Correct centre |
| \((x+1)^2 - 1 + (y-2)^2 - 4 - 8 = 0\) | M1 | Attempt at completing the square |
| \((x+1)^2 + (y-2)^2 = 13\) | ||
| Radius \(\sqrt{13}\) | A1 [3] | Correct radius |
| *Alternative:* Centre \((-g,-f)\) is \((-1,2)\) | B1 | |
| \(g^2+f^2-c\) | M1 | |
| Radius \(= \sqrt{13}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2)^2 + (k-2)^2 = 13\) | M1 | Attempt to substitute \(x=-3\) into circle equation |
| \((k-2)^2 = 9\) | M1 | Correct method to solve quadratic |
| \(k - 2 = \pm 3\) | ||
| \(k = -1\) | A1 [3] | \(k=-1\) (negative value chosen) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER \(y = 6 - x\) | M1 | Attempt to solve equations simultaneously |
| \((x+1)^2 + (6-x-2)^2 = 13\) | M1 | Substitute into their circle equation for \(x/y\) or attempt to get an equation in 1 variable only |
| \((x+1)^2 + (4-x)^2 = 13\) | ||
| \(x^2+2x+1+16-8x+x^2 = 13\) | A1 | Obtain correct 3 term quadratic |
| \(2x^2 - 6x + 4 = 0\) | M1 | Correct method to solve quadratic of form \(ax^2+bx+c=0\) (\(b\neq0\)) |
| \(2(x-1)(x-2) = 0\) | ||
| \(x = 1, 2\) | A1 | Both \(x\) values correct |
| \(\therefore y = 5, 4\) | A1 [6] | Both \(y\) values correct. or one correct pair www B1; second correct pair B1 |
| OR \(x = 6-y\) substitution leads to \(y=4,5\); \(x=2,1\) | T&I M1 A1 one correct \(x\) (or \(y\)) value; A1 correct associated coordinate |
## Question 10:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre $(-1, 2)$ | B1 | Correct centre |
| $(x+1)^2 - 1 + (y-2)^2 - 4 - 8 = 0$ | M1 | Attempt at completing the square |
| $(x+1)^2 + (y-2)^2 = 13$ | | |
| Radius $\sqrt{13}$ | A1 [3] | Correct radius |
| *Alternative:* Centre $(-g,-f)$ is $(-1,2)$ | B1 | |
| $g^2+f^2-c$ | M1 | |
| Radius $= \sqrt{13}$ | A1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2)^2 + (k-2)^2 = 13$ | M1 | Attempt to substitute $x=-3$ into circle equation |
| $(k-2)^2 = 9$ | M1 | Correct method to solve quadratic |
| $k - 2 = \pm 3$ | | |
| $k = -1$ | A1 [3] | $k=-1$ (negative value chosen) |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER** $y = 6 - x$ | M1 | Attempt to solve equations simultaneously |
| $(x+1)^2 + (6-x-2)^2 = 13$ | M1 | Substitute into their circle equation for $x/y$ or attempt to get an equation in 1 variable only |
| $(x+1)^2 + (4-x)^2 = 13$ | | |
| $x^2+2x+1+16-8x+x^2 = 13$ | A1 | Obtain correct 3 term quadratic |
| $2x^2 - 6x + 4 = 0$ | M1 | Correct method to solve quadratic of form $ax^2+bx+c=0$ ($b\neq0$) |
| $2(x-1)(x-2) = 0$ | | |
| $x = 1, 2$ | A1 | Both $x$ values correct |
| $\therefore y = 5, 4$ | A1 [6] | Both $y$ values correct. **or** one correct pair www B1; second correct pair B1 |
| **OR** $x = 6-y$ substitution leads to $y=4,5$; $x=2,1$ | | **T&I** M1 A1 one correct $x$ (or $y$) value; A1 correct associated coordinate |10 A circle has equation $x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 8 = 0$.\\
(i) Find the centre and radius of the circle.\\
(ii) The circle passes through the point $( - 3 , k )$, where $k < 0$. Find the value of $k$.\\
(iii) Find the coordinates of the points where the circle meets the line with equation $x + y = 6$.
\hfill \mbox{\textit{OCR C1 2007 Q10 [12]}}