Edexcel C2 2007 January — Question 3 6 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2007
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeCircle from diameter endpoints
DifficultyModerate -0.8 This is a straightforward application of the circle-from-diameter formula requiring only finding the midpoint (center) and distance formula (radius), then substituting into the standard circle equation. It's a routine C2 exercise with clear steps and no problem-solving insight needed, making it easier than average.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^2
3. The line joining the points \(( - 1,4 )\) and \(( 3,6 )\) is a diameter of the circle \(C\). Find an equation for \(C\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Centre \(\left(\frac{-1+3}{2}, \frac{6+4}{2}\right) = (1, 5)\)M1, A1 Some use of correct midpoint formula in \(x\) or \(y\); \((5,1)\) gains M1 A0
\(r = \frac{\sqrt{(3-(-1))^2 + (6-4)^2}}{2}\) or \(r^2 = (1-(-1))^2 + (5-4)^2\) or \(r^2 = (3-1)^2 + (6-5)^2\)M1 Correct method to find \(r\) or \(r^2\); attempting \(r = \sqrt{\frac{(\text{diameter})^2}{2}}\) is M0
\((x-1)^2 + (y-5)^2 = 5\)M1, A1, A1 (6) LHS \((x-1)^2+(y-5)^2\); RHS \(= 5\); or correct equivalent e.g. \(x^2+y^2-2x-10y+21=0\) 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre $\left(\frac{-1+3}{2}, \frac{6+4}{2}\right) = (1, 5)$ | M1, A1 | Some use of correct midpoint formula in $x$ or $y$; $(5,1)$ gains M1 A0 |
| $r = \frac{\sqrt{(3-(-1))^2 + (6-4)^2}}{2}$ or $r^2 = (1-(-1))^2 + (5-4)^2$ or $r^2 = (3-1)^2 + (6-5)^2$ | M1 | Correct method to find $r$ or $r^2$; attempting $r = \sqrt{\frac{(\text{diameter})^2}{2}}$ is M0 |
| $(x-1)^2 + (y-5)^2 = 5$ | M1, A1, A1 **(6)** | LHS $(x-1)^2+(y-5)^2$; RHS $= 5$; or correct equivalent e.g. $x^2+y^2-2x-10y+21=0$ |

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3. The line joining the points $( - 1,4 )$ and $( 3,6 )$ is a diameter of the circle $C$. Find an equation for $C$.\\

\hfill \mbox{\textit{Edexcel C2 2007 Q3 [6]}}
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