Circles

1. A circle with centre C has equation x2 + y2 − 2x + 10y − 19 = 0.

i. Find the coordinates of C and the radius of the circle.

[3]

ii. Verify that the point (7, −2) lies on the circumference of the circle.

[1]

iii. Find the equation of the tangent to the circle at the point (7, −2), giving your answer in
the form ax + by + c = 0, where a, b and c are integers.

[5]

2. A circle C has equation x2 + y2 + 8y − 24 = 0.

i. Find the centre and radius of the circle.

[3

ii. The point A (2, 2) lies on the circumference of C. Given that AB is a diameter of the
circle, find the coordinates of B.

[2
3. A circle with centre C has equation (x − 2)2 + (y + 5)2 = 25.

(i) Show that no part of the circle lies above the x-axis.
[3]
(ii) The point P has coordinates (6, k) and lies inside the circle. Find the set of possible values
of k.
[5]
(iii) Prove that the line 2y = x does not meet the circle.
[4]

4. A circle with centre C has equation x2 + y2 − 10x + 4y + 4 = 0.

i. Find the coordinates of C and the radius of the circle.

[3]

ii. Show that the tangent to the circle at the point P (8, 2) has equation 3x + 4y = 32.

[5]

iii. The circle meets the y-axis at Q and the tangent meets the y-axis at R. Find the area of
triangle PQR.

[4]

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5.

The diagram shows the circle with equation x2 + y2 − 8x − 6y − 20 = 0.

i. Find the centre and radius of the circle.

[3]

The circle crosses the positive x-axis at the point A.

ii. Find the equation of the tangent to the circle at A.

[6]

iii. A second tangent to the circle is parallel to the tangent at A. Find the equation
of this second tangent.

[3]

iv. Another circle has centre at the origin O and radius r. This circle lies wholly
inside the first circle. Find the set of possible values of r.

[2]

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6. Points A and B have coordinates (3, 0) and (9, 8) respectively. The line AB is a diameter of a
circle.

(a) Find the coordinates of the centre of the circle. [2]

(b) Find the equation of the tangent to the circle at the point B. [3]

7.

A circle with centre C has equation x2 + y2 + 8x – 4y + 7 = 0, as shown in the diagram. The

circle meets the
x-axis at A and B.
(a) Find

• the coordinates of C,

• the radius of the circle. [3]

(b) Find the coordinates of the points A and B. [2]

The chord DE passes through the point and is perpendicular to OC, where O is the
origin.

(c) Find the coordinates of the points D and E. [7]

(d) Hence find the area of the quadrilateral BEAD. [2]

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8. In this question you must show detailed reasoning.

A circle touches the lines and y = 2x at (6, 3) and (3, 6) respectively.

Find the equation of the circle. [7]

9. The circle x2 + y2 − 8x + 2y = 0 passes through the origin O. Line OA is a diameter to this

circle.

Find the equation of the line OA, giving your answer in the form ax + by = 0, where a
(i) [5]
and b are integers.

The tangent to the circle at point A meets the x-axis at the point B. Find the area of
(ii) [6]
triangle OAB.

10. A circle with equation x2 + y2 + 6x − 4y = k has a radius of 4.

(a) Find the coordinates of the centre of the circle. [2]

(b) Find the value of the constant k. [2]

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11. The equation of a circle is x2 + y2 + 6x – 2y – 10 = 0.

(a) Find the centre and radius of the circle. [3]

(b) Find the coordinates of any points where the line y = 2x – 3 meets the circle x2 + y2 + [4]
6x – 2y – 10 = 0.

(c) State what can be deduced from the answer to part (b) about the line y = 2x – 3 and
the circle
x2 + y2 + 6x – 2y – 10 = 0. [1]

12. A circle with centre C has equation x2 +y2 +8x − 2y − 7 = 0.

Find

(a) the coordinates of C, [2]

(b) the radius of the circle. [1]

13. In this question you must show detailed reasoning.

The lines and are tangents to a circle at (2, 1) and (–2, 1) respectively.
Find the equation of the circle in the form x2 +y2 + ax + by + c = 0, where a, b and c are
[6]
constants.

14. A line has equation y = 2x and a circle has equation x2 + y2 + 2x − 16y + 56 = 0.

(a) [3]
Show that the line does not meet the circle.

(b) (i) Find the equation of the line through the centre of the circle that is perpendicular [4]
to the line y = 2x.

(ii) Hence find the shortest distance between the line y = 2x and the circle, giving
your answer in an exact form.
[4]

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15.

The diagram shows a circle with centre (a, −a) that passes through the origin.
(a) Write down an equation for the circle in terms of a. [2]

(b) Given that the point (1, −5) lies on the circle, find the exact area of the circle. [3]

END OF QUESTION paper

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Mark scheme Circles

Question Answer/Indicative content Marks Part marks and guidance

1 i Centre (1, −5) B1 Correct centre

(x − 1)2 + (y + 5)2 − 19 − 1 − 25 = 0
i M1 Correct method to find r2 r2 = (±5)2 + (±1)2 + 19 for the M mark
(x − 1)2 + (y + 5)2 = 45

Correct radius. Do not allow if wrong centre used in calculation of

radius.

Examiner's Comments

i A1 This standard piece of bookwork was generally done very well, with
around three-quarters of candidates scoring all three marks. Only
occasionally was the centre seen as (2, –10). The most common
cause of errors was again dealing with negative numbers, particularly
when squaring to find the radius, or not subtracting appropriately
after completing the square.

Substitution of coordinates into equation of circle in any form or use

of Pythagoras' theorem to calculate the distance of (7, −2) from C

No follow through for this part as AG. Must be consistent –

72 + (−2)2 − 14 − 20 − 19 Examiner's Comments
ii B1 do not allow finding the distance as if no / wrong
=0
radius found in 9(i).
This was managed well by most candidates, with substitution of the
point into the original equation generally a more successful approach
than using Pythagoras’ theorem.

uses with their C

iii M1 Follow through from 9(i) until final mark.
gradient of radius =
(3/4 correct)

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iii A1√ If (−1,5) is used for C, then expect
Follow through from their C allow unsimplified single fraction e.g.

Follow through from their gradient, even if M0 scored. Allow

iii gradient of tangent = −2 B1√

Gradient of radius =

correct equation of straight line through (7, −2), any non-zero

iii y + 2 = −2 (x − 7) M1
numerical gradient Gradient of tangent =

oe 3 term equation in correct form i.e. k(2x + y − 12) = 0 where k is

an integer cao

Examiner's Comments Alternative markscheme for implicit differentiation:

A large number of candidates secured full marks on this question and

almost all managed to secure partial credit. Some candidates M1 Attempt at implicit diff as evidenced by
term
simplified to . The incorrect simplification of to was
iii 2x + y − 12 = 0 A1

also common. The majority remembered to find the negative A1

reciprocal of their gradient and then substituted this correctly to find

an equation of a straight line. Some candidates still miss the detail of A1 Substitution of (7, −2) to obtain gradient of tangent =

the question and do not give the correct answer in the required form, −2

needlessly losing the final mark. Another not uncommon error was to Then M1 A1 as main scheme

attempt to differentiate implicitly when candidates clearly had either

not yet met this technique or did not understand the process;
successful solutions using this approach were extremely rare.

Total 9

2 i Centre (0, −4) B1

i x2 + (y + 4)2 − 16 − 24 = 0 M1 (y ± 4)2 − 42 seen (or implied by correct answer) Or attempt at r2 = f2 + g2 − c

Do not allow A mark from (y − 4)2

i A1
Examiner's Comments

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Over two-thirds of candidates secured all three marks interpreting the

given equation of a circle correctly. Marks were lost mainly due to
sign errors, both in attempting to find the centre and in attempts to
complete the square.

ii (−2, −10) B1FT FT through centre given in (i) i.e. (their 2x − 2, their 2y − 2)

FT through centre given in (i)

Examiner's Comments

Apply same scheme if equation of diameter found and

Candidates who drew a diagram usually recognised that the
ii B1FT attempt to solve simultaneously; no marks until a correct
coordinates of B could be found by simple addition or subtraction
value of x/y found.
and were then usually successful in scoring both marks, especially as
there was a follow-through from part (i). Those who tried to apply
standard techniques involving Pythagoras’ theorem and resulting
quadratics were very rarely successful in finding either value.

2) If the candidate attempts to solve by using the formula

a. If the formula is quoted incorrectly then M0. c.

ii
b. If the formula is quoted correctly then one sign slip is
permitted. Substituting the wrong numerical value for a or b
or c scores M0

Total 5

3 i y coordinate of the centre is −5 B1 Correct y value Alt

i Radius = 5 B1 Correct radius Shows only meets x axis at one point B1

Correct explanation based on the above — allow clear diagram www

Centre is five units below x axis and radius is five, so just touches the Examiner's Comments Correct y value for the centre B1
i B1
x-axis Correct explanation B1 www
The simplest, and most common, approach to show that the circle
did not go above the x axis was to identify the centre and radius from

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the equation and state/show on a diagram that the circle just touched
the axis at a single point. The majority of candidates showed clear
solutions to this effect. Some, however, tested just a single point
(usually y = 1) and showed this was not a point on the circle, which
was of course insufficient.

CP2 = (6 − 2)2 + (k + 5)2

Attempt to find CP or CP2 Alternative
ii M1
Puts x = 6 to into equation of circle M1
CP2 < 25 ⇒ 16 + k2 + 10k + 25 < 25

Correct three term quadratic equation*, could be in terms

ii k2 + 10k + 16 < 0 A1 Correct three term quadratic expression*
of y A1

ii (k + 2)(k + 8) < 0 A1 k = −2 and k = −8 found k = −2 and k = −8 found (allow y) A1

ii M1 Chooses “inside region” for their roots of their quadratic Then as main scheme
−8 < k < −2

Must be strict inequalities for the A mark

* Or (k + 5)2 < 9

Examiner's Comments
* Or (k + 5)2 = 9
SC
Most candidates took the correct approach to this, substituting x = 6
Trial and improvement
ii A1 and then solving the quadratic and finding the values of k that
B2 if final answer correct
corresponded to the points on the circumference. A large number of
(B1 if inequalities are not strict)
candidates then stopped and failed to identify the correct range of
Can only get 5/5 if fully explained
values being any value between these. Those who carried on were
usually correct, but it was fairly common to not give the answer as the
strict inequalities required for the point to be inside the circle. There
were some neat alternative solutions using Pythagoras’ theorem to
find the values of k from a good sketch.

(2y − 2)2 + (y + 5)2 = 25

iii 5y2 + 2 y + 4 = 0 M1* Attempts to eliminate x or y from equation of circle If y eliminated: 5x2 + 4x + 16 = 0
b2 −4ac = 4 − 4 × 5 × 4

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= −76
< 0, so line and circle do not meet

b2 −4ac = 16 − 4 × 5 × 16
iii A1 Correct three term quadratic obtained
= −304

Correct method to establish quadratic has no roots e.g. considers

iii M1dep*
value of b2 − 4ac, tries to find roots from quadratic formula
No marks for purely graphical attempts

Correct clear conclusion www AG

Examiner's Comments

There were a large number of fully correct solutions to the request to

prove that the line and circle do not meet. Most performed the easier
iii A1
substitution for x, but (2y)2 = 2y2 was a fairly common error. Many
were able to use the discriminant, or the quadratic formula, to explain
their reasoning clearly. Only a few claimed that the line and circle did
not meet because the quadratic could not be factorised. Some of the
weaker candidates again resorted to testing a single point
(sometimes the centre) or drawing a poor diagram.

Total 12

4 i C = (5, −2) B1 Correct centre

i (x − 5)2 + (y + 2)2 − 25 = 0 M1 (x ± 5)2 − 52 and (y ± 2)2 − 22 seen (or implied by correct answer) Or attempt at r2 = f2 + g2 − c

Correct radius – do not allow A mark from (x + 5)2 and / or (y − 2)2

Examiner's Comments
i Radius = 5 A1
Apart from the usual sign error, most candidates were able to identify
the centre and calculate the radius of the circle with little apparently
difficulty.

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ii M1 Attempt to find gradient of radius (3/4 correct) See also alternative methods on next page

ii A1

ii B1ft

Equation of straight line through P, using their perpendicular gradient

ii M1 Do not allow use of gradient of radius instead of tangent
(not from rearrangement)

Rearrange to required form www AG

Substitute for x/y or attempt to get an equation in 1 variable only M1 Ignore order of terms
k(x − 16x + 64) = 0 or k(y − 4y + 4) = 0 A1
2 2 M*1 Attempt at implicit differentiation as evidenced by
4y + 3x = 32 Correct method to solve quadratic — see appendix 1 M1
x = 8, y = 2 found A1 term

States one root implies tangent B1

Attempts to rearrange equation of line to find

Examiner's Comments
ii A1
and
Unsuccessful attempts at implicit differentiation notwithstanding, A1 Substitution of (8, 2) to obtain

compares with gradient of radius M1 most candidates were able to present a clear accurate solution to this Then as main scheme OR
Multiply gradients to get −1 B1 part of the question. The expected approach of finding the gradient of Attempts to rearrange equation of line to find
Check (8, 2) lies on line B1 the radius, its negative reciprocal and then the equation of the line
through (8, 2) was performed very well. Some candidates merely M1dep
rearranged the given equation to find its gradient and re-substituted; Check (8, 2) lies on line B1
this gained no credit.

For the M mark, allow splitting into two triangles

iii Q = (0, −2) B1 Q found correctly

iii R = (0, 8) B1 R found correctly

Attempt to find area of triangle with their Q, R and height 8 i.e. If using PQ as base then expect to see

iii M1

www

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Examiner's Comments

Most candidates were able to find both points on the y-axis and the
best solutions to this included a sketch diagram to aid candidates on
their way. Some chose to find the lengths of all the sides of the
iii 40 A1
triangle and multiply together sides that were not perpendicular before
halving. Although full marks to this part were comparatively rare, it
was noticeable that some lower attaining candidates who did use a
good sketch were able to outscore many of the higher attaining
students on this particular part.

Total 12

5 i Centre of circle (4, 3) B1 Correct centre

i (x − 4)2 − 16 + (y − 3)2 − 9 − 20 = 0 M1 (x ± 4)2 − 42 and (y ± 3)2 − 32 seen (or implied by correct answer) Or r2 = 42 + 32 + 20 soi

i r2 = 45 or better www ISW after

Examiner's Comments
i r= A1
This proved to be a very successfully answered question,
with around nine in ten candidates securing all three marks.

Alterative for finding gradient: M1 Attempt at implicit

At A, y = 0 so x2 − 8x − 20 = 0 Valid method to find A e.g. put y = 0 and attempt to solve quadratic
ii M1
(x − 10)(x + 2) = 0 (allow slips) or Pythagoras' theorem
differentiation as evidenced by term

ii A = (10, 0) A1 Correct answer found

Attempts to find gradient of radius (3 out of 4 terms correct for their

ii M1
centre, their A)
Gradient of radius =

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A1
ii Gradient of tangent = 2 B1

and substitution of (10, 0) to obtain 2.

ii y − 0 = 2(x − 10) M1 Equation of line through their A, any non-zero gradient

Examiner's Comments

Just over half of candidates obtained full marks in this part,

with errors appearing at all stages. Some put x rather than
ii y = 2x − 20 A1 Correct answer in any three-term form y equal to 0 when trying to find A and the alternative
method of using Pythagoras’ theorem often led to slips.
There were a significant number of problems finding the

gradient and errors such as

were commonly seen.

iii A' = (−2, 6) B1 Finds the opposite end of the diameter

iii y − 6 = 2(x + 2) M1 Line through their A' parallel to their line in (ii) Not through centre of circle

Examiner's Comments

Many candidates did not realise that the point required for
iii y = 2x + 10 A1 Correct answer in any three-term form
the parallel line was the opposite end of the diameter. Most
did use the same gradient as in (ii), but some used the
negative reciprocal. An interesting method sometimes seen
was consideration of translation of the original line.

Attempts to find the distance from O to their centre and subtract from
iv OC = =5 M1 ISW incorrect simplification
their radius

iv (0 <) r < A1 Correct inequality, condone ≤ Examiner's Comments

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This proved very demanding, with many candidates unable
to start; those who drew a diagram were generally more
successful but less than a quarter of candidates secured
both marks. Even amongst those who found the maximum

length of the radius to be it was quite rare

to see the correct inequality.

Total 14

M1
(AO1.1a) Correct working for
either coordinate May
6 a be implied by x = 6 or y
A1 (AO1.1) =4
(6,4)
[2]

M1 (AO1.1)

Gradient of radius through B is

M1 (AO1.1)

b Gradient of tangent is
A1 FT their gradient
(AO2.2a)

So equation of tangent is oe [3]

Total 5

Centre of circle is (–4, 2) B1 (AO1.1)

Correct centre
M1 (AO1.1)
7 a (x + 4)2 –16 + (y –2)2 –4 + 7 = 0
(x ± 4)2 –16 +
A1 (AO1.1) OR r2 = 42 + 22 – 7
(y ± 2)2 – 4 seen
[3]

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r = 3.61 or better
www

M1
y = 0 ⇒ x2 + 8 x + 7 = 0 (AO1.1a) Substitute y =0
and attempt to
b solve
A1 (AO1.1)
A (–7, 0) and B (–1, 0)
[2] BC

M1
(AO3.1a)
Identify gradient of
line OC
Hence mDE = 2
A1FT
(AO1.2)
Use of m1m2 = –1
with their mOC

M1 (AO1.1)

M1 Form equation of
c
(x + 4)2 + (2x + 4)2 = 13
(AO3.1a) line DE

Substitute to get
M1 (AO1.1) quadratic in one
5x2 + 24x + 19 = 0 ⇒x =… variable

Expand and
A1 (AO1.1)
attempt to solve
their 3-term
A1
quadratic
(AO3.2a)

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[7]
BC

M1
(AO1.1a)

d
A1 (AO1.1)

[2]

Total 14

DR
B1(AO3.1a)

or M1(AO1.1a)
Grad of rad = –2

Attempt equation
8 of either radius
y – 3 = –2(x – 6)
M1(AO1.2)

y = –2x + 15 or M1(AO2.1)
or attempt
A1(AO1.1) equation of other
Equation of line from O to centre is y = x radius
M1(AO1.1)

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A1(AO1.1)
Solve their or equns of both
[7]
equation of radius radii
x = –2x + 15 with y = x

C is (5, 5)

r2 = (5 – 3)2 + (5 – 6)2 (= 5)

ISW

(x – 5)2 + (y – 5)2 = 5

Total 7

(x − 4)2 −16 + (y +1)2 − 1 = 0 M1

Correct e.g. (x ± 4)2 and
(x − 4)2 + (y +1)2 = 17 method to (y ± 1)2 seen (or implied
find centre of by correct answer)
circle
A1
M can be implied by
Centre = (4, −1)
correct centre.

B1 Correct Note: Centre (− 4, 1)

centre soi. leads to
9 i
“correct” answer.
M1A0B0M1A0 Max 2/5
Gradient of
OA correct
(could use
M1
OC or CA) [A
= (8, −2) is
not required
for this part,

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but may be
used]
A1
x + 4y = 0 Attempts Alternative for first three
equation of marks:
straight line M1 Attempt at implicit
through O or differentiation as
A or centre of evidenced by
the circle with
their
calculated
term
gradient.
A1
www Correct
equation in
required form
i.e. and substitutes O
k(x + 4y) = 0
for integer k, to obtain
allow
[5] 0 = 4y + x B1 Find correct
etc. negative reciprocal

Examiner’s Comments

There were many full accurate solutions to this question, although the
testing of the identification of the centre of a circle but without that
specific request proved taxing to some. There were both sign errors
and division errors, with (4, −2) being frequently seen. Slips also
followed in finding the gradient of the line. Many candidates chose the
origin as the point to use to find the equation, although the use of the
centre of the circle or even A itself were not uncommon. Some
candidates failed to give the final answer in the required form.

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Must be seen
/ used in (ii);
B1ft

ft their centre
ft their
B1ft
gradient in (i)

M1
Attempts
A = (8, − 2) equation of
perpendicular If centre used here, max
line through B1B1, 2/6.
m' = 4
their A. (Not
M1 (4, −1).)
y + 2 = 4(x − 8) Equation of line/B may
Attempt to not be seen explicitly.
find x value
ii of point B
from their
M1
When equation of Must have used a valid
perpendicular method to find B.
line

Attempt to
find area of
Area
OAB e.g.

A1
their OB
× their Look out for
“correct” answer from
wrong coordinates − A0.
[6] 2, or their
OA ×

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their AB, or
split into two
triangles

Accept 8.5 or
equivalent
fractions but
not
unsimplified
surds. www

Examiner’s Comments

Candidates who made errors in the first part of the question were still
able to score 4 out of 5 in this part as follow through and method
marks were allowed from wrong centres. Just over a quarter of
candidates were able to secure all 5 marks, but a significant minority
were unable to access this part, either through not realising A was the
opposite end of the diameter to O or not realising that the
perpendicular gradient was needed to find the coordinates of B.
Several found B through the use of a sketch rather than finding the
equation of the perpendicular line. In general candidates who used a
sketch were more likely to access, and succeed in, this question than
those who did not.

Total 11

M1(AO1.1a)
Attempt to
(x ± 3) + (y ± 2) …
2 2
complete the
square
10 a A1(AO1.1)

State correct
(−3,2) Ignore constant
centre www
[2] term(s)

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M1(AO1.1a)
13 + k = 16 Attempt to link 9,
A1(AO1.1) 4, 16 and k
b

k=3 Obtain k = 3
[2]

Total 4

Allow x = –3, y = 1

B1
Correct centre of
(AO 1.1) Allow for (x ± 3)2
circle
±9 + (y ± 1)2 ± 1
M1 seen (x ± 3)2 +
Attempt to
centre is (–3, 1) (AO 1.1a) (y ± 1)2 – 10 = 0 is
complete the
M0 as no evidence
square twice
of subtracting the
(x + 3)2 – 9 + (y – 1)2 – 1 – 10 = 0 constant terms to
(x + 3)2 + (y – 1)2 = 20
complete the
squares
11 a
Or attempt to use
r2 = g2 + f 2 – c

From correct
A1 working only,
(AO 1.1) including correct
Correct radius
or factorisation
Allow r = 4.47, or
better

[3]
Examiner’s Comments

Solutions to this question were nearly always correct, with most

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candidates choosing to write the equation in factorised form. There
were a few sign errors when stating the centre of the circle, and also a
few errors when subtracting the constant term when completing the
square each time.

Substitute the
Either substitute
linear equation into
for y, or an attempt
the quadratic
at x
equation
M1 Either use the
(AO 3.1a) given expanded
x2 + (2x – 3)2 + 6x – 2(2x – 3) – 10 = 0
OR
equation or their
(x + 3)2 + (2x – 4)2 = 20
attempt at a
factorised equation
Correct three term
Must be three
quadratic
terms, but not
A1
necessarily on
(AO 1.1)
same side of
b x2 – 2x + 1 = 0 equation
BC, or from any
A0 if additional
valid method
incorrect x value
A1
A0 if additional
(AO 1.1) Allow x = 1, y = –1
x=1 points also given
A1
(AO 2.1)
(1, –1)
Examiner’s Comments
[4]
All candidates attempted to solve the equations simultaneously, either
using the expanded equation of the circle or the factorised equation.
As this question did not specify ‘detailed reasoning’, it was expected
that candidates would solve the ensuing quadratic on their calculator
but instead most still showed the factorisation.

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Allow just mention
Correct deduction of ‘tangent’
Strict follow- Allow other correct
B1ft through on their statements such
(AO 2.2a) number of roots as the line and the
from (b) circle only touch
c The line is a tangent to the circle at (1, –1)
once

Examiner’s Comments

[1] Part (b) shows one point of intersection so it was expected that
candidates would put this information into context and conclude that
the line was a tangent to the circle. If an error had happened in part
(b) resulting in other than one point of intersection then candidates
could still get this mark for a correct deduction from their answer.

Total 8

e.g. (x ± 4)2 and (y

Correct method to
± 1)2 seen (or
find centre of circle
implied)

M1(AO
1.1)E
(x + 4)2 −16 + (y − 1)2 − 1 − 7 = 0

12 a (x + 4)2 + (y − 1)2 = 24
A1(AO
1.1)E
C(−4,1)

[2]

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 Circles

Examiner’s Comments
This proved to be a good start for nearly all candidates with the vast
majority correctly completing the square (twice) to find the coordinates
of the centre of the circle. When errors occurred these were nearly
always down to sign errors inside the two brackets.

oe e.g.

B1(AO
1.1)E
b

[1]

Examiner’s Comments
Nearly all candidates stated the radius of the circle correctly in either
part (a) or part (b).

Total 3

DR
If no wking seen, Alt method using
no marks proportion: Centre
y – 1= –2(x – 2) or y = –2x + c & M1
or y – 1= 2(x–(−2)) is on y-axis, not (0,
sub (2, 1) (AO3.1a)
1) (may be implied)
or solve y = –2x + M1
5 & y = 2x+5
13

or
A1

y = –2x + 5 c=5
(AO1.1)
c=1+2×2
y = 2x + 5 or c = 5 c = 5 A1
A1

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 Circles
(AO3.2a)
Centre is (0, 5)
M1
stated or implied Centre is (0, 5) A1
(AO1.1a)

or r2 = 22 + 42 or ft
their centre

= √20

M1 = 20
(AO1.2)

x2 + (y – 5)2 = 20 oe
or a = 0, b = –10,
c=5
ft their centre and
A1
(AO1.1)
rad2 (≠ 0), however
x2 + y2 – 10y + 5 = 0 found
[6]
cao

Examiner’s Comments

A clear initial sketch proved to be extremely useful for the majority of

candidates that successfully answered this question. Some failed to
consider one or both normals and therefore were unable to make any
significant progress. Some found the centre incorrectly, for example
by finding the midpoint of the line joining (2, 1) and (–2, 1). Others
treated (0, 0) as the centre of the circle. However, despite many false
starts, many used their incorrect centre and radius in the equation of a
circle, and gained at least one mark. However, some candidates
found a centre (p, q) and radius r, and then wrote x2 + y2 + px + qy +
r.

A great many candidates started from the equation x2 + y2 + ax + by +

c = 0, given in the question, and attempted to find a, b and c by

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 Circles
substituting coordinates and various other devices. Not surprisingly,
these generally failed to gain any marks. (Although one candidate did
actually succeed by this method, taking several pages to do so, and
gained full credit.)

Total 6

M1 (AO
2.1) Substitute y = 2x
x + 4x + 2x − 32x + 56 = 0
2 2
into equation of
circle

and rearrange to
5x2 − 30x + 56 = 0
three term
14 a quadratic
M1 (AO
2.4)
30 − 4 × 5 × 56 = −220
2
Consider
A1 (AO
discriminant
2.2a)
b2 − 4ac < 0 hence no real roots so the circle and line do not intersect Conclude with no
real roots
[3]

Centre of circle is (−1, 8) B1 (AO 1.1) Seen or used

B1 (AO

Gradient of perpendicular is −0.5 2.2a) For gradient of

perpendicular
M1 (AO
1.1)
y − 8 = −0.5(x + 1) Attempt equation
b (i)
of line through their
circle centre with
A1 (AO 1.1)
gradient of −0.5

x + 2y = 15 Obtain correct
[4] equation Allow any 3 term

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 Circles
equivalent
M1 (AO
5x = 15
3.1b)

x = 3, y = 6
Attempt to solve
distance from centre to line is M1 (AO simultaneously
1.1) with y = 2x

Use Pythagoras to
(x + 1)2 + (y − 8)2 = 32
find distance
(ii)
between centre of
M1 (AO
circle and point of
1.1a)
intersection

Attempt to find
radius of circle Seen at any point
in solution – allow
hence shortest distance between line
back credit to part
and circle is A1 (AO (a) if the radius is
3.2a) found at the same
time as the centre
of circle
[4] Obtain
Allow any exact
equiv

Total 11

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 Circles
Correct LHS
(x − a )2 + ( y + a )2 = Κ B1 (AO 1.1) (accept if
expanded: x2 + y2
− 2ax + 2ay + 2a2)
Κ = 2a2
15 a B1 (AO 1.1) Correct RHS
Allow full marks for
any equivalent
form, e.g. x2 + y2 −
[2]
2ax + 2ay = 0

M1 (AO
1.1a)

(1 − a)2 + (−5 + a)2 = 2a2

Substitute (1, −5)
into their circle
M1 (AO equation
1.1)
b
Solve for a and
substitute into πr2
A1 (AO
with their r2
2.2a)

[3]

Total 5

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