| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Circle through three points using perpendicular bisectors |
| Difficulty | Moderate -0.3 This is a multi-part coordinate geometry question covering standard C1 topics: line equations, area of triangle, perpendicular bisector, and circle equations. While it has multiple parts (12 marks total), each component uses routine techniques—gradient formula, y-intercept, midpoint, perpendicular gradient, and circle properties. Part (iv) requires combining the perpendicular bisector with x=3 to find the centre, which adds mild problem-solving, but overall this is slightly easier than average due to straightforward application of standard methods with clear scaffolding. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| 4 (i) | 1−3 |
| Answer | Marks |
|---|---|
| isw | M1 |
| Answer | Marks |
|---|---|
| A1 | or use of y = their gradient x + c |
| Answer | Marks |
|---|---|
| 4 (ii) | when y = 0, x = 8; when x = 0, |
| Answer | Marks |
|---|---|
| [Area =] ½ × 8/3 × 8 o.e. cao isw | M1 |
| M1 | allow y = 8/3 used without |
| Answer | Marks |
|---|---|
| 4 (iii) | grad perp = −1/grad AB stated, or |
| Answer | Marks |
|---|---|
| showing midpt of AB is (2, 2) | M1 |
| Answer | Marks |
|---|---|
| M1 | or showing 3 × −1/3 = −1 |
Question 4:
--- 4 (i) ---
4 (i) | 1−3
grad AB = [= −1/3]
5−(−1)
y − 3 = their grad (x – (−1)) or
y − 1 = their grad (x − 5)
y = −1/3x + 8/3 or 3y = −x + 8 o.e
isw | M1
M1
A1 | or use of y = their gradient x + c
with coords of A or B
y−3 x−(−1)
or M2 for = o.e.
1−3 5−(−1)
o.e. eg x + 3y − 8 = 0 or 6y = 16 −
2x
allow B3 for correct eqn www
--- 4 (ii) ---
4 (ii) | when y = 0, x = 8; when x = 0,
y = 8/3 or ft their (i)
[Area =] ½ × 8/3 × 8 o.e. cao isw | M1
M1 | allow y = 8/3 used without
explanation if already seen in eqn in
(i)
NB answer 32/3 given;
allow 4 × 8/3 if first M1 earned;
or
M1 for
8 1(8−x)dx=1 ( 8x−1 x2) 8
3 3 2
0 0
and M1 dep for 1(64−32[−0])
3
--- 4 (iii) ---
4 (iii) | grad perp = −1/grad AB stated, or
used after their grad AB stated in
this part
midpoint [of AB] = (2, 2)
y − 2 = their grad perp (x − 2) or ft
their midpoint
alt method working back from
ans:
grad perp = −1/grad AB and
showing/stating same as given
line
finding intn of their
y = −1/3x − 8/3and y = 3x − 4 is
(2, 2)
showing midpt of AB is (2, 2) | M1
M1
M1
or
M1
M1
M1 | or showing 3 × −1/3 = −1
if (i) is wrong, allow the first M1
here ft, provided the answer is
correct ft
must state ‘midpoint’ or show
working
for M3 this must be correct, starting
from grad AB = −1/3, and also
needs correct completion to given
ans y = 3x − 4
mark one method or the other, to
benefit of candidate, not a mixture
eg stating − 1/3 × 3 = −1
or showing that (2, 2) is on y = 3x −
4, having found (2, 2) first
[for both methods: for M3 must be
fully correct]\includegraphics{figure_1}
Fig. 11 shows the line through the points A $(-1, 3)$ and B $(5, 1)$.
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the line through A and B. [3]
\item Show that the area of the triangle bounded by the axes and the line through A and B is $\frac{32}{3}$ square units. [2]
\item Show that the equation of the perpendicular bisector of AB is $y = 3x - 4$. [3]
\item A circle passing through A and B has its centre on the line $x = 3$. Find the centre of the circle and hence find the radius and equation of the circle. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 Q4 [12]}}