| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle from diameter endpoints |
| Difficulty | Easy -1.2 This is a straightforward two-part question requiring only basic coordinate geometry formulas: midpoint formula and the circle equation from center and radius. Both are standard recall exercises with minimal problem-solving, making it easier than average for A-level. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\left(\frac{5+13}{2}, \frac{-1+11}{2}\right) = (9,5)\) | M1, A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Equation of circle: \((x-9)^2 + (y-5)^2 = 52\) | M1, M1, A1 | A1 |
**(a)** $\left(\frac{5+13}{2}, \frac{-1+11}{2}\right) = (9,5)$ | M1, A1 | (2 marks) | M1: For some use of correct formula; can be implied. Use of $\left(\frac{1}{2}(x_A - x_B), \frac{1}{2}(y_A - y_B)\right) \to (4,6)$ is M0A0.
**(b)** $r^2 = (9-5)^2 + (5--1)^2 (= 52)$
Equation of circle: $(x-9)^2 + (y-5)^2 = 52$ | M1, M1, A1 | A1 | (4 marks) | M1: Attempt to find $r$ or $r^2$, given $(9,5)$. $r = AB = \sqrt{208}$ is M0.
2nd M1 for $(x-9)^2 + (y-5)^2 = \text{constant}$, given their $(9,5)$.
A1 for $(x-9)^2 + (y-5)^2 = 52$ only.
**(Subtotal: 6 marks)**The points $A$ and $B$ have coordinates $( 5 , - 1 )$ and $( 13,11 )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the mid-point of $A B$.
Given that $A B$ is a diameter of the circle $C$,
\item find an equation for $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2005 Q2 [6]}}