| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Topic | Circles |
| Type | Find centre and radius from equation |
| Difficulty | Easy -1.2 This is a straightforward application of completing the square to convert a circle equation to standard form. It requires only routine algebraic manipulation with no problem-solving insight, making it easier than average but not trivial since students must remember the technique and execute it carefully. |
| 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 |
|---|---|---|
| \((x+4)^2 - 16 + (y-1)^2 - 1 - 7 = 0\) | M1 1.1 | Correct method to find centre; e.g. \((x \pm 4)^2\) and \((y \pm 1)^2\) seen or implied |
| \((x+4)^2 + (y-1)^2 = 24\) | — | — |
| \(C(-4, 1)\) | A1 1.1 [2] | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Radius} = \sqrt{24}\) | B1 1.1 [1] | oe e.g. \(2\sqrt{6}\) |
## Question 1(i):
$(x+4)^2 - 16 + (y-1)^2 - 1 - 7 = 0$ | M1 1.1 | Correct method to find centre; e.g. $(x \pm 4)^2$ and $(y \pm 1)^2$ seen or implied
$(x+4)^2 + (y-1)^2 = 24$ | — | —
$C(-4, 1)$ | A1 1.1 [2] | —
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## Question 1(ii):
$\text{Radius} = \sqrt{24}$ | B1 1.1 [1] | oe e.g. $2\sqrt{6}$
---1 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } + 8 x - 2 y - 7 = 0$.\\
Find\\
(i) the coordinates of $C$,\\
(ii) the radius of the circle.
\hfill \mbox{\textit{OCR H240/03 2018 Q1 [3]}}