| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Moderate -0.8 This is a straightforward C2 circle question requiring only direct application of standard formulas: distance formula for diameter length, midpoint formula, and the circle equation formula using center and radius. All three parts are routine calculations with no problem-solving or geometric insight required. |
| Spec | 1.03a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt \(f(3)\) or \(f(-3)\) | M1 | Numbers substituted into expression; or assume \(a=-9\) and attempt \(f(3)\) or \(f(-3)\) |
| \(f(3)=54-45+3a+18=0 \Rightarrow 3a=-27 \Rightarrow a=-9\) | A1* cso | For applying \(f(3)\) correctly, setting result equal to 0; do not accept \(x=-9\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = (x-3)(2x^2+x-6)\) | M1 A1 | 1st M1: dividing by \((x-3)\) leading to 3TQ beginning with correct term \(2x^2\) |
| \(= (x-3)(2x-3)(x+2)\) | M1 A1 | 2nd M1: valid attempt to factorise quadratic; 2nd A1 cao needs all three factors. NB: \((x-3)(x-\frac{3}{2})(x+2)\) is M1A1M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3^y=3 \Rightarrow y=1\) or \(g(1)=0\) | B1 | \(y=1\) seen as solution; may be spotted as answer, no working needed |
| \(3^y=1.5 \Rightarrow \log(3^y)=\log 1.5\) or \(y=\log_3 1.5\) | M1 | Attempt to take logs to solve \(3^y=\alpha\) or \(3^{ky}=\alpha\), but not \(6^y=\alpha\), where \(\alpha>0\) and \(\alpha\neq 3\) and was a root of \(f(x)=0\) |
| \(y = \text{awrt } 0.37\) | A1 | If a third answer included (and not rejected) such as \(\ln(-2)\), lose final A mark |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt $f(3)$ or $f(-3)$ | M1 | Numbers substituted into expression; or assume $a=-9$ and attempt $f(3)$ or $f(-3)$ |
| $f(3)=54-45+3a+18=0 \Rightarrow 3a=-27 \Rightarrow a=-9$ | A1* cso | For applying $f(3)$ correctly, setting result equal to 0; do not accept $x=-9$ |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = (x-3)(2x^2+x-6)$ | M1 A1 | 1st M1: dividing by $(x-3)$ leading to 3TQ beginning with correct term $2x^2$ |
| $= (x-3)(2x-3)(x+2)$ | M1 A1 | 2nd M1: valid attempt to factorise quadratic; 2nd A1 cao needs all three factors. NB: $(x-3)(x-\frac{3}{2})(x+2)$ is M1A1M0A0 |
**Way 2:** Uses trial/factor theorem to obtain $x=-2$ **or** $x=\frac{3}{2}$ (M1, A1), both roots (M1), correct factorisation (A1). Way 3: No working, three factors $(x-3)(2x-3)(x+2)$ otherwise need working (M1A1M1A1).
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $3^y=3 \Rightarrow y=1$ or $g(1)=0$ | B1 | $y=1$ seen as solution; may be spotted as answer, no working needed |
| $3^y=1.5 \Rightarrow \log(3^y)=\log 1.5$ or $y=\log_3 1.5$ | M1 | Attempt to take logs to solve $3^y=\alpha$ or $3^{ky}=\alpha$, but not $6^y=\alpha$, where $\alpha>0$ and $\alpha\neq 3$ and was a root of $f(x)=0$ |
| $y = \text{awrt } 0.37$ | A1 | If a third answer included (and not rejected) such as $\ln(-2)$, lose final A mark |3.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{c1a3d21d-38fe-4619-9e99-5c4788cdb891-019_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 Q3}}