| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Chord length calculation |
| Difficulty | Moderate -0.3 This is a straightforward circle question requiring completion of the square to find centre and radius, then using Pythagoras to find a chord length. All steps are standard C1 techniques with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part nature and algebraic manipulation needed. |
| 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 |
|---|---|---|
| (i) \((x-6)^2 - 36 + (y+4)^2 - 16 + 16 = 0\) | M1 | |
| \(\therefore\) centre \((6, -4)\) | A1 | |
| (ii) \((x-6)^2 + (y+4)^2 = 36\) | M1 | |
| \(\therefore\) radius \(= 6\) | A1 | |
| (iii) Sketch of circle with centre at \((6, -4)\) and radius 6 | B2 | |
| (iv) \(y = 0 \therefore (x-6)^2 + 16 = 36\) | M1 | |
| \(x = 6 \pm \sqrt{20} = 6 \pm 2\sqrt{5}\) | A1 | |
| \(AB = 6 + 2\sqrt{5} - (6 - 2\sqrt{5}) = 4\sqrt{5}\) | M1 A1 | (10) |
**(i)** $(x-6)^2 - 36 + (y+4)^2 - 16 + 16 = 0$ | M1 |
$\therefore$ centre $(6, -4)$ | A1 |
**(ii)** $(x-6)^2 + (y+4)^2 = 36$ | M1 |
$\therefore$ radius $= 6$ | A1 |
**(iii)** Sketch of circle with centre at $(6, -4)$ and radius 6 | B2 |
**(iv)** $y = 0 \therefore (x-6)^2 + 16 = 36$ | M1 |
$x = 6 \pm \sqrt{20} = 6 \pm 2\sqrt{5}$ | A1 |
$AB = 6 + 2\sqrt{5} - (6 - 2\sqrt{5}) = 4\sqrt{5}$ | M1 A1 | (10)
---9. The circle $C$ has the equation
$$x ^ { 2 } + y ^ { 2 } - 12 x + 8 y + 16 = 0$$
(i) Find the coordinates of the centre of $C$.\\
(ii) Find the radius of $C$.\\
(iii) Sketch $C$.
Given that $C$ crosses the $x$-axis at the points $A$ and $B$,\\
(iv) find the length $A B$, giving your answer in the form $k \sqrt { 5 }$.\\
\hfill \mbox{\textit{OCR C1 Q9 [10]}}