Graph transformations and the equation of a circle – GCSE Maths exam questions with mark scheme

1Grade 6–7HigherNon-calculator4 marks

The point $A(2, 5)$ lies on the curve with equation $y = f(x)$.

Write down the coordinates of the image of $A$ on the curve with equation

(a) $y = f(x) + 3$ [1]

(b) $y = f(x - 4)$ [1]

(d) $y = f(-x)$ [1]

(a) $(2, 8)$B1
(b) $(6, 5)$B1
(c) $(2, -5)$B1
(d) $(-2, 5)$B1
Answer: (a) $(2, 8)$ (b) $(6, 5)$ (c) $(2, -5)$ (d) $(-2, 5)$
Examiner tip: Changes INSIDE the bracket act on $x$ the 'opposite' way: $f(x - 4)$ moves the graph 4 to the RIGHT.

2Grade 6–7HigherNon-calculator3 marks

(a) Describe fully the single transformation that maps the graph of $y = x^2$ onto the graph of $y = (x + 3)^2 - 2$ [2]

(b) Describe fully the single transformation that maps the graph of $y = x^2$ onto the graph of $y = -x^2$ [1]

(a) translation (do not accept 'move', 'shift' or 'transformation'; B0 B0 if more than one transformation is given)B1
(a) by the vector $\begin{pmatrix} -3 \\ -2 \end{pmatrix}$ (accept a correct description in words, e.g. 3 units left and 2 units down; do not accept coordinates $(-3, -2)$ as the vector)B1
(b) reflection in the $x$-axis (accept $y = 0$; allow rotation of $180°$ about the origin, or enlargement scale factor $-1$ centre the origin, since $x^2$ is even; a reflection with no mirror line, or a rotation with no centre, scores B0; B0 if more than one transformation is given)B1
Answer: (a) translation by $\begin{pmatrix} -3 \\ -2 \end{pmatrix}$ (b) reflection in the $x$-axis
Examiner tip: 'Describe fully' needs the type of transformation AND its details – a translation needs a vector, a reflection needs a mirror line.

3Grade 6–7HigherNon-calculator2 marks

The diagram shows the graph of $y = f(x)$ (solid line) and two other graphs, $G$ (dashed) and $H$ (dotted).

(a) Write down the equation of graph $G$ in terms of $f$. [1]

(b) Write down the equation of graph $H$ in terms of $f$. [1]

(a) $y = f(x - 6)$ (accept $f(x - 6)$ without '$y =$'; do not accept $f(x + 6)$)B1
(b) $y = -f(x)$ (accept $-f(x)$ without '$y =$'; do not accept $f(-x)$)B1
Answer: (a) $y = f(x - 6)$ (b) $y = -f(x)$
Examiner tip: Track one clear point (here the peak at $(-4, 4)$, which moves to $(2, 4)$ on $G$): a move 6 to the RIGHT is $f(x - 6)$, not $f(x + 6)$.

4Grade 6–7HigherNon-calculator3 marks

A circle has equation $x^2 + y^2 = 36$

(a) Write down the radius of the circle. [1]

(b) Is the point $(4, 5)$ inside, on or outside the circle? You must show your working. [2]

(a) $6$B1
(b) for $4^2 + 5^2 = 41$ or $\sqrt{41}$ or $6.4\ldots$ (distance from the origin)M1
(b) outside, since $41 > 36$ (or $6.4 > 6$)A1
Answer: (a) 6 (b) outside
Examiner tip: The radius is $\sqrt{36} = 6$, not 36 – the equation is $x^2 + y^2 = r^2$.

5Grade 8–9HigherNon-calculator4 marks

The point $P(3, -4)$ lies on the circle with equation $x^2 + y^2 = 25$

Find the equation of the tangent to the circle at $P$. Give your answer in the form $ax + by = c$, where $a$, $b$ and $c$ are integers. [4]

for the gradient of the radius $OP$: $\dfrac{-4}{3}$M1
for the gradient of the tangent: $\dfrac{3}{4}$ (negative reciprocal of their radius gradient)M1
for substituting $(3, -4)$ with their tangent gradient, e.g. $y + 4 = \tfrac34(x - 3)$M1
$3x - 4y = 25$ (oe with integer coefficients)A1
Answer: $3x - 4y = 25$
Examiner tip: A tangent is perpendicular to the RADIUS at the point of contact – find the radius gradient from the centre $(0, 0)$ first.

6Grade 6–7HigherNon-calculator3 marks

The graph of $y = \sin x°$ for $0 \le x \le 360$ has a maximum point at $(90, 1)$.

Write down the coordinates of the maximum point of each of these graphs, for $0 \le x \le 360$.

(a) $y = \sin x° + 2$ [1]

(b) $y = \sin(x - 60)°$ [1]

(a) $(90, 3)$B1
(b) $(150, 1)$B1
(c) $(270, 1)$B1
Answer: (a) $(90, 3)$ (b) $(150, 1)$ (c) $(270, 1)$
Examiner tip: For $y = -\sin x$ the old MINIMUM $(270, -1)$ becomes the new maximum – it is not $(90, -1)$.

7Grade 8–9HigherNon-calculator5 marks

The point $P(2, 4)$ lies on the circle $x^2 + y^2 = 20$

The tangent to the circle at $P$ crosses the $x$-axis at $A$ and the $y$-axis at $B$.

Work out the area of triangle $OAB$, where $O$ is the origin. [5]

for the gradient of $OP$: $\tfrac{4}{2} = 2$M1
for the tangent gradient $-\tfrac12$ used with $(2, 4)$, e.g. $4 = -\tfrac12 \times 2 + c$M1
$y = -\tfrac12 x + 5$ (oe)A1
for finding both intercepts from their tangent: $A(10, 0)$ and $B(0, 5)$M1
25 square unitsA1
Answer: 25 square units
Examiner tip: Always use the perpendicular gradient for a tangent – using the radius gradient (2) gives a line through the centre, not a tangent.

8Grade 6–7HigherNon-calculator3 marks

The diagram shows a circle with centre the origin $O$.

The circle passes through the point $A(-3, 1)$.

(a) Find the equation of the circle. [2]

(b) $AB$ is a diameter of the circle. Write down the coordinates of $B$. [1]

(a) for using Pythagoras or substituting $A$: $(-3)^2 + 1^2$ or $r^2 = 10$ or $r = \sqrt{10}$M1
(a) $x^2 + y^2 = 10$ (accept $x^2 + y^2 = (\sqrt{10})^2$; do not accept $x^2 + y^2 = \sqrt{10}$)A1
(b) $(3, -1)$B1
Answer: (a) $x^2 + y^2 = 10$ (b) $(3, -1)$
Examiner tip: The number on the right is $r^2$, not $r$: writing $x^2 + y^2 = \sqrt{10}$ loses the accuracy mark.

9Grade 6–7HigherNon-calculator4 marks

The graph of $y = f(x)$ is drawn on the grid. It is made of straight-line segments joining the points $(-4, 1)$, $(-2, -2)$, $(1, -2)$ and $(3, 2)$.

(a) On the grid, sketch the graph of $y = f(x) + 2$ [2]

(b) On the grid, sketch the graph of $y = f(-x)$ [2]

For each graph, give the coordinates of its four vertices.

(a) for a vertical translation of the correct shape (any non-zero amount), or 3 of the 4 vertices $(-4, 3)$, $(-2, 0)$, $(1, 0)$, $(3, 4)$ correctB1
(a) fully correct graph through $(-4, 3)$, $(-2, 0)$, $(1, 0)$ and $(3, 4)$B1
(b) for a reflection of the correct shape in either axis, or 3 of the 4 vertices $(4, 1)$, $(2, -2)$, $(-1, -2)$, $(-3, 2)$ correctB1
(b) fully correct graph through $(4, 1)$, $(2, -2)$, $(-1, -2)$ and $(-3, 2)$ (reflection in the $y$-axis)B1
Answer: (a) vertices $(-4, 3)$, $(-2, 0)$, $(1, 0)$, $(3, 4)$ (b) vertices $(4, 1)$, $(2, -2)$, $(-1, -2)$, $(-3, 2)$
Examiner tip: $f(-x)$ changes the sign of every $x$-coordinate (reflection in the $y$-axis); students who flip it upside down have drawn $-f(x)$ and get at most 1 mark.

10Grade 6–7HigherNon-calculator5 marks

$f(x) = x^2 - 6x + 5$

The graph of $y = f(x)$ has a turning point at $(3, -4)$ and crosses the $x$-axis at $(1, 0)$ and $(5, 0)$.

(a) Write down the coordinates of the turning point of the graph of $y = f(x + 5)$ [1]

(b) Write down the coordinates of the turning point of the graph of $y = -f(x)$ [1]

(d) The graph of $y = f(x - a)$ passes through the origin. Find the two possible values of $a$. [2]

(a) $(-2, -4)$B1
(b) $(3, 4)$B1
(c) $k = 4$B1
(d) for recognising that a root ($x = 1$ or $x = 5$) must move to $x = 0$, e.g. translation of 1 or 5 to the left, or $f(-a) = 0$, or one correct valueM1
(d) $a = -1$ and $a = -5$ (both required, no extras)A1
Answer: (a) $(-2, -4)$ (b) $(3, 4)$ (c) $4$ (d) $a = -1$ or $a = -5$
Examiner tip: $f(x - a)$ moves the graph $a$ to the right, so to slide the roots at 1 and 5 left onto the origin $a$ must be negative; giving $a = 1$ and $a = 5$ is the classic sign slip.

11Grade 6–7HigherNon-calculator3 marks

(a) Describe fully the single transformation that maps the graph of $y = \cos x°$ onto the graph of $y = \cos(x + 40)°$ [2]

(b) Sam says:

“The graph of $y = \cos(-x)°$ is a reflection of the graph of $y = \cos x°$ in the $y$-axis, so it is a different graph.”

Explain why Sam is wrong. [1]

(a) translation (do not accept ‘move’, ‘shift’ or ‘transformation’ alone)B1
(a) by vector $\begin{pmatrix} -40 \\ 0 \end{pmatrix}$ (accept ‘40 to the left’ or ‘−40 in the $x$-direction’); B0 B0 if more than one transformation is givenB1
(b) explains that $y = \cos x°$ is symmetrical about the $y$-axis (accept $\cos(-x) = \cos x$, or e.g. both equal 0.5 at $x = \pm 60$), so the reflection maps the graph onto itself / the graphs are the sameC1
Answer: (a) translation by $\begin{pmatrix} -40 \\ 0 \end{pmatrix}$ (b) the cosine graph is symmetrical in the $y$-axis, so reflecting it in the $y$-axis gives the same graph
Examiner tip: ‘Describe fully’ needs the word translation AND the vector; $+40$ inside the bracket means 40 to the LEFT.

12Grade 6–7HigherNon-calculator5 marks

A circle has equation $x^2 + y^2 = 45$

(a) Write down the radius of the circle. Give your answer as a surd in its simplest form. [1]

(b) The point $(p, 2p)$ lies on the circle. Find the possible values of $p$. [3]

(c) $A$ is the point $(6, 3)$ and $B$ is the point $(-6, -3)$. Both points lie on the circle. Explain why $AB$ is a diameter of the circle. [1]

(a) $3\sqrt{5}$ (B0 for $\sqrt{45}$ alone)B1
(b) for substituting: $p^2 + (2p)^2 = 45$M1
(b) for $5p^2 = 45$ or $p^2 = 9$M1
(b) $p = 3$ and $p = -3$ (accept $\pm 3$; A0 if only $3$)A1
(c) the midpoint of $AB$ is $(0, 0)$, the centre of the circle (accept: $A$, $O$ and $B$ lie on one straight line $y = \tfrac{1}{2}x$, so the chord passes through the centre)C1
Answer: (a) $3\sqrt{5}$ (b) $p = 3$ or $p = -3$ (c) midpoint of $AB$ is $(0, 0)$, the centre, so the chord $AB$ passes through the centre
Examiner tip: Write $(2p)^2 = 4p^2$, not $2p^2$, and remember the negative root: both $(3, 6)$ and $(-3, -6)$ are on the circle.

13Grade 8–9HigherNon-calculator5 marks

The points $P(-1, 7)$ and $R(7, -1)$ lie on the circle with equation $x^2 + y^2 = 50$

The tangent to the circle at $P$ and the tangent to the circle at $R$ meet at the point $T$.

Find the coordinates of $T$. [5]

for the gradient of a radius: gradient $OP = -7$ or gradient $OR = -\tfrac{1}{7}$M1
for using the negative reciprocal with the correct point, e.g. $y - 7 = \tfrac{1}{7}(x + 1)$ or $y + 1 = 7(x - 7)$M1
one correct tangent: $x - 7y = -50$ or $y = 7x - 50$ (oe)A1
for a correct method to solve their two tangent equations simultaneously (second tangent may come from symmetry about $y = x$, i.e. $T$ lies on $y = x$)M1
$T\left(\tfrac{25}{3}, \tfrac{25}{3}\right)$ (accept $(8\tfrac13, 8\tfrac13)$ or $(8.33, 8.33)$ or better)A1
Answer: $T\left(\tfrac{25}{3}, \tfrac{25}{3}\right)$
Examiner tip: The tangent is perpendicular to the RADIUS at that point: use gradient $OP = -7$, so the tangent gradient is $\tfrac17$ (not $-7$, and not $7$).

14Grade 8–9HigherNon-calculator4 marks

$f(x) = x^3 - 4x$

(a) Show that reflecting the graph of $y = f(x)$ in the $y$-axis gives the same graph as reflecting it in the $x$-axis. [3]

(b) The graph of $y = f(x) + k$ passes through the point $(1, 0)$. Find the value of $k$. [1]

(a) for writing the reflection in the $y$-axis as $f(-x) = (-x)^3 - 4(-x)$M1
(a) $f(-x) = -x^3 + 4x$A1
(a) for showing $-f(x) = -(x^3 - 4x) = -x^3 + 4x$ and concluding $f(-x) = -f(x)$, so the two reflections give the same graph (accept equivalent: both reflected graphs have equation $y = -x^3 + 4x$)C1
(b) $k = 3$B1
Answer: (a) $f(-x) = -x^3 + 4x = -f(x)$ (b) $k = 3$
Examiner tip: Use brackets: $(-x)^3 = -x^3$ and $-4(-x) = +4x$; checking with one or two numbers is not a proof and will not earn the communication mark.

15Grade 6–7HigherCalculator4 marks

(a) On the grid, draw the graph of $x^2 + y^2 = 9$ [2]

(b) Use your graph to find estimates for the solutions of the simultaneous equations

$$x^2 + y^2 = 9 \qquad y = x + 1$$

[2]

(a) for a circle with centre $(0, 0)$ (may be freehand), or a curve through at least 3 of $(3, 0)$, $(0, 3)$, $(-3, 0)$, $(0, -3)$B1
(a) fully correct circle, centre the origin, radius 3B1
(b) for drawing the line $y = x + 1$ (at least from $x = -3$ to $x = 2$) or reading one correct intersectionM1
(b) $x = 1.6, y = 2.6$ and $x = -2.6, y = -1.6$ (each value $\pm 0.2$; must be paired correctly)A1
Answer: (a) circle centre $O$, radius 3 (b) $x \approx 1.6, y \approx 2.6$ and $x \approx -2.6, y \approx -1.6$
Examiner tip: Give the solutions as matched pairs of $x$ and $y$ read from where the line crosses the circle; a list of four numbers without pairing loses the final mark.

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