OCR PURE — Question 4 3 marks

Exam BoardOCR
ModulePURE
Marks3
PaperDownload PDF ↗
TopicCircles
TypeFind centre and radius from equation
DifficultyModerate -0.5 This is a straightforward application of completing the square to find the centre and radius of a circle. Students need to rearrange to standard form $(x-a)^2 + (y-b)^2 = r^2$ and use the given radius to solve for k. It's slightly easier than average as it's a direct single-method question with no problem-solving required, though it does require careful algebraic manipulation.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle
4 The circle \(x ^ { 2 } + y ^ { 2 } - 6 x + 4 y + k = 0\) has radius 5.
Determine the value of \(k\).
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(x^2 + y^2 - 6x + 4y + k = 0 \Rightarrow (x-3)^2 - 9 + (y+2)^2 - 4 + k = 0\)M1* Attempt to complete the square for both \(x\) and \(y\) terms. Must have \((x \pm 3)^2 + (y \pm 2)^2 + \ldots\)
\(r^2 = 9 + 4 - k = 5^2\)M1dep* Setting up an equation for \(k\) correctly using either 5 or \(5^2\). e.g. \(\sqrt{13-k} = 5\) or \(13 - k = 25\)
\(k = -12\)A1 cao
[3] 
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + y^2 - 6x + 4y + k = 0 \Rightarrow (x-3)^2 - 9 + (y+2)^2 - 4 + k = 0$ | M1* | Attempt to complete the square for both $x$ and $y$ terms. Must have $(x \pm 3)^2 + (y \pm 2)^2 + \ldots$ |
| $r^2 = 9 + 4 - k = 5^2$ | M1dep* | Setting up an equation for $k$ correctly using either 5 or $5^2$. e.g. $\sqrt{13-k} = 5$ or $13 - k = 25$ |
| $k = -12$ | A1 | cao |
| **[3]** | | |

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4 The circle $x ^ { 2 } + y ^ { 2 } - 6 x + 4 y + k = 0$ has radius 5.\\
Determine the value of $k$.

\hfill \mbox{\textit{OCR PURE  Q4 [3]}}
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