OCR C1 2013 June — Question 6 5 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2013
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeFind centre and radius from equation
DifficultyModerate -0.8 This is a straightforward C1 circles question requiring completion of the square to find centre/radius (standard technique) and using the diameter property that the centre is the midpoint of AB (simple coordinate geometry). Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires correct algebraic manipulation.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle

A circle \(C\) has equation \(x^2 + y^2 + 8y - 24 = 0\).
  1. Find the centre and radius of the circle. [3]
  2. The point \(A(2, 2)\) lies on the circumference of \(C\). Given that \(AB\) is a diameter of the circle, find the coordinates of \(B\). [2]

(i)
Answer: Centre \((0, -4)\); \(x^2 + (y + 4)^2 - 16 - 24 = 0\); Radius \(= \sqrt{40}\)
AnswerMarks Guidance
Marks: B1M1 A1 [3]
Guidance:\((y \pm 4)^2 - 4^2\) seen (or implied by correct answer) Do not allow A mark from \((y - 4)^2\)
Additional notes: Or attempt at \(r^2 = f^2 + g^2 - c\); A0 for \(\pm\sqrt{40}\)
(ii)
Answer: \((-2, -10)\)
AnswerMarks
Marks: B1FTB1FT [2]
Guidance: FT through centre given in (i)FT through centre given in (i)
Additional notes: i.e. (their \(2x - 2\), their \(2y - 2\)); Apply same scheme if equation of diameter found and attempt to solve simultaneously; no marks until a correct value of \(x/y\) found.
## (i)
**Answer:** Centre $(0, -4)$; $x^2 + (y + 4)^2 - 16 - 24 = 0$; Radius $= \sqrt{40}$

**Marks:** B1 | M1 | A1 [3]

**Guidance:** | $(y \pm 4)^2 - 4^2$ seen (or implied by correct answer) | Do not allow A mark from $(y - 4)^2$

**Additional notes:** Or attempt at $r^2 = f^2 + g^2 - c$; A0 for $\pm\sqrt{40}$

## (ii)
**Answer:** $(-2, -10)$

**Marks:** B1FT | B1FT [2]

**Guidance:** FT through centre given in (i) | FT through centre given in (i)

**Additional notes:** i.e. (their $2x - 2$, their $2y - 2$); Apply same scheme if equation of diameter found and attempt to solve simultaneously; no marks until a correct value of $x/y$ found.

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A circle $C$ has equation $x^2 + y^2 + 8y - 24 = 0$.
\begin{enumerate}[label=(\roman*)]
\item Find the centre and radius of the circle. [3]
\item The point $A(2, 2)$ lies on the circumference of $C$. Given that $AB$ is a diameter of the circle, find the coordinates of $B$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR C1 2013 Q6 [5]}}