OCR PURE — Question 2 4 marks

Exam BoardOCR
ModulePURE
Marks4
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, then solving a simple equation. It requires knowing the standard form (x-a)²+(y-b)²=r² and basic algebraic manipulation, but involves minimal problem-solving beyond the routine procedure. Slightly easier than average due to being a direct single-concept question with clear method.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle
2 The circle \(x ^ { 2 } + y ^ { 2 } - 4 x + k y + 12 = 0\) has radius 1.
Find the two possible values of the constant \(k\).
Question 2:
*Note: Other methods are probably equivalent to the ones shown. Apply whichever method is closest. Allow + or − instead of ± except for final answer.*
\((x-2)^2 + \left(y + \frac{k}{2}\right)^2\) — Allow + or − in both
AnswerMarks Guidance
AnswerMark Guidance
\((x-2)^2 + \left(y + \frac{k}{2}\right)^2 + 12 - 4 - \frac{k^2}{4}\) oe; ignore RHSM1* (3.1a) \((x-2)^2+y^2+ky+12-4=0\) M1; or \((x-2)^2+(y-b)^2=1\) M1; or \(x^2+y^2-4x+ky+13=1\) M1
\(\frac{k^2}{4} - 8 = 1\) or \(-12 + 4 + \frac{k^2}{4} = 1\) oe ft constant termA1 (1.1) \((x-2)^2+y^2+ky+9=1\) M1; or \(4+b^2-1=12\) oe M1; or \(x^2+y^2-4x+ky+4+9=1\) M1
\(k^2 = 36\)M1 (2.1) depM1* \((x-2)^2+(y\pm3)^2=1\) oe A1; or \(b=\pm3\) A1; or \(x^2+y^2-4x\pm6y+4+9=1\) or \((x-2)^2+(y\pm3)^2=1\) A1
\(k = \pm 6\)A1 (1.1) \(k=\pm6\) A1 (all three methods)
[4]
Mark Scheme H230/01 June 2019
## Question 2:

*Note: Other methods are probably equivalent to the ones shown. Apply whichever method is closest. Allow + or − instead of ± except for final answer.*

$(x-2)^2 + \left(y + \frac{k}{2}\right)^2$ — Allow + or − in both

| Answer | Mark | Guidance |
|--------|------|----------|
| $(x-2)^2 + \left(y + \frac{k}{2}\right)^2 + 12 - 4 - \frac{k^2}{4}$ oe; ignore RHS | M1* (3.1a) | $(x-2)^2+y^2+ky+12-4=0$ M1; or $(x-2)^2+(y-b)^2=1$ M1; or $x^2+y^2-4x+ky+13=1$ M1 |
| $\frac{k^2}{4} - 8 = 1$ or $-12 + 4 + \frac{k^2}{4} = 1$ oe ft constant term | A1 (1.1) | $(x-2)^2+y^2+ky+9=1$ M1; or $4+b^2-1=12$ oe M1; or $x^2+y^2-4x+ky+4+9=1$ M1 |
| $k^2 = 36$ | M1 (2.1) depM1* | $(x-2)^2+(y\pm3)^2=1$ oe A1; or $b=\pm3$ A1; or $x^2+y^2-4x\pm6y+4+9=1$ or $(x-2)^2+(y\pm3)^2=1$ A1 |
| $k = \pm 6$ | A1 (1.1) | $k=\pm6$ A1 (all three methods) |
| **[4]** | | |

# Mark Scheme H230/01 June 2019

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2 The circle $x ^ { 2 } + y ^ { 2 } - 4 x + k y + 12 = 0$ has radius 1.\\
Find the two possible values of the constant $k$.

\hfill \mbox{\textit{OCR PURE  Q2 [4]}}
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