Edexcel C2 2006 January — Question 3 7 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Year2006
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeCircle from diameter endpoints
DifficultyEasy -1.2 This is a straightforward C2 circle question requiring only standard formulas: distance formula for diameter length, midpoint formula for center, and circle equation from center and radius. All three parts are direct applications of memorized techniques with no problem-solving or geometric insight required.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle

3. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{84b2d36b-c112-4d35-84a1-bc2b707f162d-04_675_792_287_568}
\end{figure} In Figure \(1 , A ( 4,0 )\) and \(B ( 3,5 )\) are the end points of a diameter of the circle \(C\). Find
  1. the exact length of \(A B\),
  2. the coordinates of the midpoint \(P\) of \(A B\),
  3. an equation for the circle \(C\).

AnswerMarks Guidance
(a) \((AB)^2 = (4-3)^2 + (5)^2\) \([=26]\)M1
\(AB = \sqrt{26}\)A1 (2 marks)
(b) \(p = \left(\frac{4+3}{2}, \frac{5}{2}\right)\)M1
\(= \left(\frac{7}{2}, \frac{5}{2}\right)\)A1 (2 marks)
(c) \((x-x_p)^2 + (y-y_p)^2 = \left(\frac{AB}{2}\right)^2\)LHS M1; RHS M1
\((x-3.5)^2 + (y-2.5)^2 = 6.5\)oe A1 c.a.o (3 marks)
Guidance:
- (a) M1 for an expression for \(AB\) or \(AB^2\). N.B. \((x_1+x_2)^2 + \ldots\) is M0
- (b) M1 for a full method for \(x_p\)
- (c) 1st M1 for using their \(x_p\) and \(y_p\) in LHS; 2nd M1 for using their \(AB\) in RHS
Note: \(x^2 + y^2 - 7x - 5y + 12 = 0\) scores 3/3 for part (c).
Condone use of calculator approximations that lead to correct answer given.
Total: 7 marks
(a) $(AB)^2 = (4-3)^2 + (5)^2$ $[=26]$ | M1
$AB = \sqrt{26}$ | A1 | (2 marks)

(b) $p = \left(\frac{4+3}{2}, \frac{5}{2}\right)$ | M1
$= \left(\frac{7}{2}, \frac{5}{2}\right)$ | A1 | (2 marks)

(c) $(x-x_p)^2 + (y-y_p)^2 = \left(\frac{AB}{2}\right)^2$ | LHS M1; RHS M1
$(x-3.5)^2 + (y-2.5)^2 = 6.5$ | oe A1 c.a.o | (3 marks)

**Guidance:**
- (a) M1 for an expression for $AB$ or $AB^2$. N.B. $(x_1+x_2)^2 + \ldots$ is M0
- (b) M1 for a full method for $x_p$
- (c) 1st M1 for using their $x_p$ and $y_p$ in LHS; 2nd M1 for using their $AB$ in RHS

**Note:** $x^2 + y^2 - 7x - 5y + 12 = 0$ scores 3/3 for part (c).

Condone use of calculator approximations that lead to correct answer given.

**Total: 7 marks**

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3.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
  \includegraphics[alt={},max width=\textwidth]{84b2d36b-c112-4d35-84a1-bc2b707f162d-04_675_792_287_568}
\end{center}
\end{figure}

In Figure $1 , A ( 4,0 )$ and $B ( 3,5 )$ are the end points of a diameter of the circle $C$.

Find
\begin{enumerate}[label=(\alph*)]
\item the exact length of $A B$,
\item the coordinates of the midpoint $P$ of $A B$,
\item an equation for the circle $C$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C2 2006 Q3 [7]}}