OCR MEI C1 — Question 6 13 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircles
TypeCircle through three points using right angle in semicircle
DifficultyModerate -0.3 This is a multi-part coordinate geometry question requiring standard techniques: gradient calculations to prove perpendicularity, area formula for right-angled triangle, circle theorem (angle in semicircle), midpoint/distance for circle equation, and diameter properties. While it has 13 marks and multiple parts, each step uses routine C1 methods with no novel insight required. The perpendicularity proof and circle justification are slightly above pure recall, but overall this is a straightforward structured question slightly easier than average.
Spec1.03b Straight lines: parallel and perpendicular relationships1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents
The points A \((-1, 6)\), B \((1, 0)\) and C \((13, 4)\) are joined by straight lines.
  1. Prove that the lines AB and BC are perpendicular. [3]
  2. Find the area of triangle ABC. [3]
  3. A circle passes through the points A, B and C. Justify the statement that AC is a diameter of this circle. Find the equation of this circle. [6]
  4. Find the coordinates of the point on this circle that is furthest from B. [1]
Question 6:
AnswerMarks
606
(ii) rad AB = oe [= −3] isw
11
04
grad BC = oe [= 1/3] isw
113
product of grads = −1 [so lines perp]
AnswerMarks
stated or shown numericallyM1
M1
AnswerMarks
M1for full marks, it should be clear that
grads are independently obtained
or ‘one grad is neg. reciprocal of other’
or
M1 for length of one side (or square of
it)
M1 for length of other two sides (or
their squares) found independently
M1 for showing or stating that Pythag
AnswerMarks
holds [so triangle rt angled]eg grads of 3 and 1/3 without earlier working earn
M1M0
for M3, must be fully correct, with gradients evaluated
at least to 6/2 and 4/12 stage
AB2 = 62 + 22 = 40, BC2 = 42 + 122 = 160, AC2 = 142
+ 2 = 200
AnswerMarks
6(iiii))A 40 or BC = 160
½ × 40 × 160 oe or ft their AB, BC
AnswerMarks
40M1
M1
AnswerMarks
A1or M1 for one of area under AC (=70),
under AB (=6) under BC (=24) (accept
unsimplified) and M1 for their trap. −
AnswerMarks
two trianglesallow M1 for (11)2 (60)2 or for
(131)2 (40)2
or for rectangle  3 triangles method,
1 1 1
[614 (2)(6) (4)(12) (2)(14)
2 2 2
=84  6  24  14]
M1 for two of the 4 areas correct and M1 for the
subtraction
AnswerMarks
6(iiiiii) le subtended by diameter =
90° soi
mid point M of AC = (6, 5)
rad of circle = 1 142 22 1 200
2 2
oe or equiv using r2
(x − a)2 + (y − b)2 = r2 seen or
(x – their 6)2 + (y – their 5)2 = k used,
with k > 0
AnswerMarks
(x − 6)2 + (y − 5)2 = 50 caoB1
B2
M1
M1
AnswerMarks
A1or angle at centre = twice angle at
circumf = 2 × 90 = 180 soi
or showing BM = AM or CM, where M
is midpt of AC; or showing that BM =
½ AC
allow if seen in circle equation ; M1 for
correct working seen for both coords
accept unsimplified; or eg r2 = 72 + 12
or 52 + 52; may be implied by correct
equation for circle or by correct method
for AM, BM or CM ft their M
AnswerMarks
or x2  y2 12x10y110condone ‘AB and BC are perpendicular’ or ‘ABC is
right angled triangle’ provided no spurious extra
reasoning
allow M1 bod intent for AC = 200 followed by r =
100
must be simplified (no surds)
AnswerMarks Guidance
6(iivv))(11, 10) 1 
Question 6:
6 | 06
(ii) rad AB = oe [= −3] isw
11
04
grad BC = oe [= 1/3] isw
113
product of grads = −1 [so lines perp]
stated or shown numerically | M1
M1
M1 | for full marks, it should be clear that
grads are independently obtained
or ‘one grad is neg. reciprocal of other’
or
M1 for length of one side (or square of
it)
M1 for length of other two sides (or
their squares) found independently
M1 for showing or stating that Pythag
holds [so triangle rt angled] | eg grads of 3 and 1/3 without earlier working earn
M1M0
for M3, must be fully correct, with gradients evaluated
at least to 6/2 and 4/12 stage
AB2 = 62 + 22 = 40, BC2 = 42 + 122 = 160, AC2 = 142
+ 2 = 200
6 | (iiii))A 40 or BC = 160
½ × 40 × 160 oe or ft their AB, BC
40 | M1
M1
A1 | or M1 for one of area under AC (=70),
under AB (=6) under BC (=24) (accept
unsimplified) and M1 for their trap. −
two triangles | allow M1 for (11)2 (60)2 or for
(131)2 (40)2
or for rectangle  3 triangles method,
1 1 1
[614 (2)(6) (4)(12) (2)(14)
2 2 2
=84  6  24  14]
M1 for two of the 4 areas correct and M1 for the
subtraction
6 | (iiiiii) le subtended by diameter =
90° soi
mid point M of AC = (6, 5)
rad of circle = 1 142 22 1 200
2 2
oe or equiv using r2
(x − a)2 + (y − b)2 = r2 seen or
(x – their 6)2 + (y – their 5)2 = k used,
with k > 0
(x − 6)2 + (y − 5)2 = 50 cao | B1
B2
M1
M1
A1 | or angle at centre = twice angle at
circumf = 2 × 90 = 180 soi
or showing BM = AM or CM, where M
is midpt of AC; or showing that BM =
½ AC
allow if seen in circle equation ; M1 for
correct working seen for both coords
accept unsimplified; or eg r2 = 72 + 12
or 52 + 52; may be implied by correct
equation for circle or by correct method
for AM, BM or CM ft their M
or x2  y2 12x10y110 | condone ‘AB and BC are perpendicular’ or ‘ABC is
right angled triangle’ provided no spurious extra
reasoning
allow M1 bod intent for AC = 200 followed by r =
100
must be simplified (no surds)
6 | (iivv))(11, 10) | 1
The points A $(-1, 6)$, B $(1, 0)$ and C $(13, 4)$ are joined by straight lines.

\begin{enumerate}[label=(\roman*)]
\item Prove that the lines AB and BC are perpendicular. [3]

\item Find the area of triangle ABC. [3]

\item A circle passes through the points A, B and C. Justify the statement that AC is a diameter of this circle. Find the equation of this circle. [6]

\item Find the coordinates of the point on this circle that is furthest from B. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C1  Q6 [13]}}
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Q1 5 Q2 12 Q3 10 Q4 12 Q5 14 Q6 13