CG Coordinate Geometry (2 dimensions only) — revision question pack

9 specification points · notes, questions, answers and worked methods

Checked against AQA 8365 section CG. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

CG1 · Know and use the definition of a gradient

Explanation

The gradient of a straight line measures vertical change per unit horizontal change. For two points $(x_1,y_1)$ and $(x_2,y_2)$ , $m=\frac{y_2-y_1}{x_2-x_1}$ .
Both subtractions must use the same point order; reversing both gives the same result. A positive gradient rises from left to right, a negative gradient falls, and a horizontal line has gradient $0$ .
A vertical line has undefined gradient because its horizontal change is zero.
Gradient may also be read from $y=mx+c$ after an equation is rearranged.
Examiners expect the coordinate substitution and simplification, not an unsupported value read approximately from a diagram.

Gradient is the vertical change divided by the horizontal change between two points.

Worked example

Work out the gradient of the line through $P(-2,5)$ and $Q(4,-7)$ .

1.Use the same point order in numerator and denominator: $m=\frac{-7-5}{4-(-2)}$ .
2.Simplify the changes: $m=\frac{-12}{6}$ .
3.Evaluate the quotient.

Answer: The gradient is $-2$ .

Common mistakes

Don't calculate $\frac{x_2-x_1}{y_2-y_1}$ , giving horizontal change divided by vertical change.
Don't use opposite point orders in the numerator and denominator and reverse the sign.
Don't call a vertical line’s gradient $0$ instead of undefined.

Exam tip

Write the gradient fraction with substituted coordinates before simplifying so the chosen point order is clear.

Tier 1 · Easy

1
Points $A(1,2)$ and $B(5,10)$ lie on a straight line. Work out its gradient.
[2 marks]
2
As $x$ increases by $5$ , a straight line rises by $15$ . Work out the gradient of the line.
[1 mark]

Tier 2 · Standard

1
Work out the gradient of the line $3x-2y=12$ .
[2 marks]
2
A straight line crosses the coordinate axes at $(6,0)$ and $(0,-4)$ . Work out its gradient.
[2 marks]
3
The points $A(-5,8)$ , $B(1,k)$ and $C(7,-4)$ lie on the same straight line. Work out $k$ .
[3 marks]

Tier 3 · Hard

1
The gradient of the line through $A(-3,4)$ and $B(p,2p+1)$ is $\frac{3}{2}$ . Work out $p$ .
[3 marks]
2
The lines $x+y=7$ and $2x-y=5$ meet at $P$ . Work out the gradient of the line joining $P$ to $Q(-1,4)$ .
[3 marks]
3
The gradient of the line through $A(t,t^2+1)$ and $B(t+2,3t+7)$ is $3$ . Given that $t>1$ , work out $t$ .
[4 marks]
4
On a coordinate grid, $1$ cm represents $4$ units horizontally and $7$ units vertically. A straight line rises $6$ cm as it moves $8$ cm to the right. Work out the gradient of the line.
[3 marks]
5
The points $A(2a-1,3a+4)$ and $B(a+5,7-a)$ lie on a vertical straight line. Work out $a$ and explain why the gradient of $AB$ is undefined.
[3 marks]

CG2 · Know the relationship between the gradients of parallel and perpendicular lines

Explanation

Distinct non-vertical parallel lines have equal gradients. Distinct vertical lines are also parallel, although their gradients are undefined.
For two non-vertical perpendicular lines, the gradients satisfy $m_1m_2=-1$ ; equivalently, one is the negative reciprocal of the other. Horizontal and vertical lines form the special perpendicular pair.
To prove that lines are parallel or perpendicular, calculate both gradients exactly and state the comparison: equal gradients establish parallel lines, while a product of $-1$ establishes perpendicular lines.
A diagram alone is not proof.
When an equation is not in $y=mx+c$ form, it should first be rearranged so its gradient can be identified reliably.

Worked example

Line $L$ has equation $3x+2y=8$ . Work out the equation of the line perpendicular to $L$ through $(6,-1)$ in the form $ax+by=c$ .

1.Rearrange $L$ : $y=-\frac32x+4$ , so its gradient is $-\frac32$ .
2.The perpendicular gradient is $\frac23$ because $-\frac32\times\frac23=-1$ .
3.Use $y+1=\frac23(x-6)$ , then multiply by $3$ and rearrange.

Answer: $2x-3y=15$ .

Common mistakes

Don't use $-\frac32$ again and construct a parallel line instead of a perpendicular line.
Don't take the reciprocal $\frac23$ but fail to change the sign when the original gradient is positive.
Don't state that two lines look perpendicular without calculating and comparing their gradients.

Exam tip

In a ‘show that’ proof, display both exact gradients and the equation $m_1m_2=-1$ before giving the conclusion.

Tier 1 · Easy

1
A line has gradient $-3$ . State the gradient of a line perpendicular to it.
[1 mark]
2
Work out the gradient of a line parallel to $5x+2y=7$ .
[2 marks]

Tier 2 · Standard

1
Points $A(-2,1)$ , $B(4,3)$ , $C(1,5)$ and $D(3,-1)$ are given. Show that $AB$ is perpendicular to $CD$ .
[3 marks]
2
The line through $A(1,k)$ and $B(5,11)$ is parallel to $y=2x-3$ . Work out $k$ .
[3 marks]
3
The lines $2x+3y=5$ and $kx-4y=7$ are perpendicular. Work out $k$ .
[3 marks]

Tier 3 · Hard

1
The line $L$ has equation $2y=3x+7$ . A line perpendicular to $L$ passes through the point where $L$ meets the $y$ -axis. Express the new line as $ax+by=c$ using integer coefficients.
[4 marks]
2
Work out the equation of the perpendicular bisector of the segment joining $A(-4,1)$ to $B(2,5)$ . Give your answer in the form $ax+by=c$ .
[4 marks]
3
Points $A(-3,1)$ and $B(6,4)$ are fixed. The point $P(0,p)$ lies on the $y$ -axis and $\angle APB=90^\circ$ . Work out the two possible coordinates of $P$ .
[4 marks]
4
The vertices of quadrilateral $ABCD$ , in order, are $A(-3,1)$ , $B(1,7)$ , $C(4,5)$ and $D(0,-1)$ . Show that $ABCD$ is a rectangle.
[4 marks]
5
The line $2x+ay=7$ is parallel to $y=-\frac{1}{2}x+4$ . The line $bx-3y=5$ is perpendicular to both of them. Work out $a$ and $b$ .
[4 marks]

CG3 · Use Pythagoras' theorem to calculate the distance between two points

Explanation

The horizontal and vertical coordinate differences between two points form the perpendicular sides of a right-angled triangle. Pythagoras’ theorem therefore gives the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .
The same point order should be used for both differences, although squaring makes either consistent order equivalent. Negative differences must be squared with brackets.
If the square root simplifies, an exact integer or surd should be given unless a decimal accuracy is requested.
Distances can establish equal sides, calculate a perimeter or verify a geometric property.
Examiners expect the squared differences to be shown; reading a length from a coordinate grid is insufficient unless the scale and exact endpoints make it explicit.

Worked example

Work out the exact distance between $A(-1,4)$ and $B(5,-3)$ .

1.The coordinate differences are $5-(-1)=6$ and $-3-4=-7$ .
2.Apply Pythagoras: $AB=\sqrt{6^2+(-7)^2}$ .
3.Simplify: $AB=\sqrt{36+49}=\sqrt{85}$ .

Answer: The exact distance is $\sqrt{85}$ units.

Common mistakes

Don't write $-7^2=-49$ instead of using $(-7)^2=49$ .
Don't add the coordinate differences directly instead of adding their squares.
Don't round $\sqrt{85}$ even though an exact distance is requested.

Exam tip

For an exact-distance question, leave the final square root as a simplified surd unless it is a perfect square.

Tier 1 · Easy

1
Work out the distance between $P(1,2)$ and $Q(4,6)$ .
[2 marks]
2
Work out the exact distance between $P(2,-1)$ and $Q(6,1)$ .
[2 marks]

Tier 2 · Standard

1
Work out the exact distance between $R(-2,5)$ and $S(6,-1)$ .
[3 marks]
2
The distance from $P(-1,2)$ to $Q(k,8)$ is $10$ units. Given that $k>-1$ , work out $k$ .
[3 marks]
3
A route consists of the straight segment from $A(-3,4)$ to $B(1,-2)$ followed by the straight segment from $B$ to $C(7,2)$ . Work out the exact length of the route.
[3 marks]

Tier 3 · Hard

1
The vertices of a triangle are $A(-4,1)$ , $B(2,9)$ and $C(8,1)$ . Work out its perimeter and state whether it is isosceles.
[4 marks]
2
The points $A(1,2)$ , $B(7,5)$ and $C(3,-2)$ form a triangle. Use coordinate distances to show that the triangle is right-angled, then work out its area.
[4 marks]
3
The point $P(p,0)$ is equidistant from $A(-3,4)$ and $B(5,2)$ . Work out $p$ and the exact distance $PA$ .
[4 marks]
4
On a map, the points $P(-6,2)$ and $Q(9,10)$ mark two shelters. One coordinate unit represents $40$ metres. A person walks directly from $P$ to $Q$ at $1.6$ metres per second. Work out the travel time in minutes and seconds.
[4 marks]
5
The endpoints of a diagonal of a square are $A(-5,2)$ and $C(7,8)$ . Work out the exact perimeter of the square.
[4 marks]

CG4 · Use ratio to find the coordinates of a point on a line given the coordinates of two other points, including the midpoint

Explanation

The midpoint of endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is found by averaging corresponding coordinates: $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ . More generally, if $AP:PB=m:n$ , point $P$ lies $\frac{m}{m+n}$ of the way from $A$ to $B$ .
The vector method is $P=A+\frac{m}{m+n}(B-A)$ , applied to both coordinates.
Equivalently, the endpoint coordinates receive opposite segment weights.
The resulting point should lie between the endpoints for an internal division ratio.
Examiners expect the ratio to be applied separately and consistently to the $x$ - and $y$ -coordinates, with a midpoint recognised as the special ratio $1:1$ .

A point P dividing the segment AB internally in the ratio AP:PB = m:n.

Worked example

Point $P$ divides the segment from $A(-3,2)$ to $B(12,17)$ in the ratio $AP:PB=2:3$ . Work out $P$ .

1. $B-A=(12-(-3),17-2)=(15,15)$ .
2. $AP$ is $\frac{2}{2+3}=\frac25$ of $AB$ , so $\frac25(15,15)=(6,6)$ .
3.Add this displacement to $A$ : $P=(-3,2)+(6,6)$ .

Answer: $P=(3,8)$ .

Common mistakes

Don't move $\frac35$ of the way from $A$ even though the first segment $AP$ has ratio weight $2$ .
Don't divide coordinates by $2+3$ without first accounting for the displacement from $A$ to $B$ .
Don't use the ratio correctly for $x$ but reverse it for $y$ .

Exam tip

Write the fraction of the displacement from the named starting point before calculating either coordinate.

Tier 1 · Easy

1
Work out the midpoint of the line segment joining $A(-2,5)$ to $B(6,1)$ .
[2 marks]
2
The midpoint of $AB$ is $M(4,-1)$ and $A=(-2,5)$ . Work out the coordinates of $B$ .
[2 marks]

Tier 2 · Standard

1
Point $P$ divides the segment from $A(-4,1)$ to $B(11,16)$ in the ratio $AP:PB=2:3$ . Work out the coordinates of $P$ .
[3 marks]
2
The point $P(1,6)$ lies on the segment joining $A(-5,2)$ to $B(10,12)$ . Work out the ratio $AP:PB$ .
[3 marks]
3
Points $P$ and $Q$ divide the segment from $A(-7,10)$ to $B(8,-5)$ into three equal parts, in the order $A,P,Q,B$ . Work out the coordinates of $P$ and $Q$ .
[3 marks]

Tier 3 · Hard

1
Point $P(3,-4)$ divides $A(-6,5)$ to $B(k,-7)$ in the ratio $AP:PB=3:1$ . Work out $k$ and then the midpoint of $AB$ .
[4 marks]
2
Point $M$ is the midpoint of $AB$ . Point $P(2,5)$ divides $AM$ in the ratio $AP:PM=2:1$ . Given that $A=(-4,1)$ , work out the coordinates of $B$ .
[4 marks]
3
Points $P(0,6)$ and $Q(5,11)$ lie on the segment $AB$ . Given that $AP:PB=1:2$ and $AQ:QB=3:1$ , work out the coordinates of $A$ and $B$ .
[4 marks]
4
Points $P$ and $Q$ lie on the segment from $A(-7,4)$ to $B(13,14)$ . Given that $AP:PB=1:4$ and $AQ:QB=3:2$ , work out the midpoint of $PQ$ .
[4 marks]
5
Point $P$ divides the segment from $A(p,2)$ to $B(8,p)$ in the ratio $AP:PB=1:2$ . The point $P$ lies on $y=2x-5$ . Work out $p$ and the coordinates of $P$ .
[4 marks]

CG5 · The equation of a straight line: y = mx + c and y - y1 = m(x - x1) and other forms, including interpretation of the gradient and y-intercept

Explanation

In the straight-line form $y=mx+c$ , $m$ is the gradient and the line crosses the $y$ -axis at $(0,c)$ . A line of gradient $m$ through $(x_1,y_1)$ can be written in point-gradient form as $y-y_1=m(x-x_1)$ .
When two points are given, calculate the gradient first, then substitute either point.
Equivalent forms such as $ax+by=c$ describe the same line and may be required by the question.
Intercepts are found by setting the other coordinate to zero.
Examiners expect the equation to satisfy the supplied point or points, so substitution provides a useful exact check and exposes sign errors introduced while expanding brackets.

Worked example

Work out the equation of the line through $(-3,7)$ and $(5,-1)$ in the form $y=mx+c$ .

1.Calculate the gradient: $m=\frac{-1-7}{5-(-3)}=\frac{-8}{8}=-1$ .
2.Use $y-7=-1(x+3)$ with the point $(-3,7)$ .
3.Expand and simplify: $y-7=-x-3$ , so $y=-x+4$ .

Answer: $y=-x+4$ .

Common mistakes

Don't use $c=7$ because $7$ is a point’s $y$ -coordinate, even though that point is not on the $y$ -axis.
Don't expand $-(x+3)$ as $-x+3$ and obtain the wrong intercept.
Don't calculate the gradient with inconsistent subtraction orders and change its sign.

Exam tip

After finding a line equation, substitute the second given point as a quick exact verification.

Tier 1 · Easy

1
State the gradient and the $y$ -intercept of the line $y=3x-4$ .
[2 marks]
2
A line has gradient $-2$ and $y$ -intercept $(0,6)$ . Write down its equation.
[1 mark]

Tier 2 · Standard

1
Work out the equation of the line with gradient $4$ that passes through $(2,-1)$ . Give your answer as $y=mx+c$ .
[3 marks]
2
The line $4x+ky=12$ has gradient $-2$ . Work out $k$ and the coordinates of its $y$ -intercept.
[3 marks]
3
A linear function $f$ satisfies $f(-3)=11$ and $f(5)=-1$ . Work out $f(x)$ .
[3 marks]

Tier 3 · Hard

1
A straight line passes through $A(-2,5)$ and $B(4,-7)$ . Work out its equation in the form $ax+by=c$ and its $x$ -intercept.
[4 marks]
2
The lines $x+y=7$ and $2x-y=2$ intersect at $P$ . The straight line $L$ passes through $P$ and through $Q(7,-2)$ . Work out an equation for $L$ in the form $ax+by=c$ .
[4 marks]
3
The line $ax+by=20$ passes through $(2,1)$ . Its positive $x$ -intercept is three times its positive $y$ -intercept. Work out $a$ and $b$ .
[4 marks]
4
A linear conversion sends $x=-4$ to $y=9$ . An increase of $6$ in $x$ produces a decrease of $15$ in $y$ . Work out a formula for $x$ in terms of $y$ .
[4 marks]
5
The midpoint of the two points where a straight line meets the coordinate axes is $M(3,-2)$ . Work out the equation of the line as $ax+by=c$ , using integer coefficients.
[3 marks]

CG6 · Draw a straight line from given information

Explanation

Two distinct points determine a straight line, although a third exact point is a useful check. For $y=mx+c$ , plot the $y$ -intercept $(0,c)$ and use the gradient as rise over run to locate another point.
For $ax+by=c$ , either rearrange to $y=mx+c$ or substitute convenient $x$ -values to build a table.
Intercepts are often efficient, but they must lie within the displayed axes.
Plot points accurately, use a ruler and draw one straight line across the full stated interval.
Examiners award graph marks for a correct straight line in the required range; separate short joins, a freehand curve or an unlabelled line outside the interval can lose accuracy marks.

A straight line drawn through an intercept and a second checked point.

Worked example

Describe the points needed to draw $3x+2y=6$ for $-2\leq x\leq4$ .

1.Rearrange to $y=3-\frac32x$ .
2.At $x=-2$ , $y=6$ ; at $x=0$ , $y=3$ ; at $x=4$ , $y=-3$ .
3.Plot $(-2,6)$ , $(0,3)$ and $(4,-3)$ , then join them with one straight line segment over the stated interval.

Answer: Draw the straight line segment through $(-2,6)$ , $(0,3)$ and $(4,-3)$ .

Common mistakes

Don't use gradient $\frac32$ after losing the negative sign while rearranging the equation.
Don't plot correct points but join them with a freehand curve or separate short segments.
Don't extend the answer beyond the required interval $-2\leq x\leq4$ without marking the requested segment.

Exam tip

Use three exact coordinate pairs, with one serving as a check, and draw a single ruled line over the stated domain.

Tier 1 · Easy

1
On coordinate axes, draw $y=2x+1$ for $-1\leq x\leq2$ .
[2 marks]
2
On coordinate axes, draw the line $x=-2$ for $-4\leq y\leq3$ .
[1 mark]

Tier 2 · Standard

1
On coordinate axes, draw $2x+3y=6$ for $-3\leq x\leq3$ .
[3 marks]
2
A line has gradient $-3$ and passes through $(1,4)$ . Draw the line for $-1\leq x\leq3$ .
[3 marks]
3
On coordinate axes, draw $\frac{x}{4}+\frac{y}{3}=1$ for $-4\leq x\leq8$ .
[3 marks]

Tier 3 · Hard

1
A line passes through $(-2,5)$ and is perpendicular to $y=\frac{1}{2}x-1$ . Draw this line for $-1\leq x\leq3$ .
[4 marks]
2
A straight line has $y$ -intercept $(0,-5)$ and passes through the intersection of $x+y=5$ and $2x-y=4$ . Draw the line for $0\leq x\leq6$ .
[4 marks]
3
Points $A(-4,-1)$ and $B(2,3)$ are fixed. On coordinate axes, draw the set of points equidistant from $A$ and $B$ for $-3\leq x\leq3$ .
[4 marks]
4
A line passes through the midpoint of $A(-5,4)$ and $B(3,-2)$ and is parallel to $3x-2y=7$ . Work out its equation, then draw it for $-3\leq x\leq3$ .
[4 marks]
5
A straight line $L$ passes through $(-2,10)$ . The point on $L$ with $x$ -coordinate $4$ has a $y$ -coordinate equal to half the value of the $y$ -intercept. Work out the equation of $L$ , then draw it for $-2\leq x\leq6$ .
[4 marks]

CG7 · Understand that x^2 + y^2 = r^2 is the equation of a circle with centre (0, 0) and radius r; application of circle geometry facts

Explanation

The equation $x^2+y^2=r^2$ represents the circle with centre $(0,0)$ and positive radius $r$ .
A point lies on the circle exactly when its coordinates satisfy the equation.
Coordinate problems may combine this equation with circle facts: an angle in a semicircle is $90^\circ$ ; the perpendicular from the centre to a chord bisects that chord; a radius is perpendicular to the tangent at its endpoint; and tangents from the same external point have equal lengths.
The radius is $\sqrt{r^2}$ , not $r^2$ .
Examiners expect the relevant circle theorem to be named or applied alongside coordinate calculations rather than inferred only from the appearance of a diagram.

A perpendicular from the centre of a circle bisects a chord.

Worked example

The circle $x^2+y^2=25$ has a vertical chord on the line $x=4$ . Work out the chord length.

1.At the chord endpoints, substitute $x=4$ : $4^2+y^2=25$ .
2.Hence $y^2=9$ , so the endpoints have $y=3$ and $y=-3$ .
3.The chord length is the vertical distance $3-(-3)=6$ .

Answer: The chord length is $6$ units.

Common mistakes

Don't state that the radius of $x^2+y^2=25$ is $25$ instead of $5$ .
Don't keep only $y=3$ and lose the lower chord endpoint $y=-3$ .
Don't call the half-chord length $3$ the full chord length.

Exam tip

When a line meets a circle twice, retain both roots and use the coordinate difference to obtain the full chord length.

Tier 1 · Easy

1
State the centre and radius of the circle $x^2+y^2=49$ .
[2 marks]
2
Show that $(4,-3)$ lies on the circle $x^2+y^2=25$ .
[2 marks]

Tier 2 · Standard

1
Point $P(6,y)$ lies on $x^2+y^2=100$ and $y>0$ . Work out $y$ .
[2 marks]
2
The points $A(-6,8)$ , $B(6,-8)$ and $P(10,0)$ lie on $x^2+y^2=100$ . Explain why $\angle APB=90^\circ$ .
[3 marks]
3
The line $y=5$ meets the circle $x^2+y^2=169$ at $A$ and $B$ . Work out the length $AB$ .
[3 marks]

Tier 3 · Hard

1
On $x^2+y^2=100$ , take $A(6,8)$ and $B(-8,6)$ , with centre $O$ . Show that angle $AOB$ is $90^\circ$ , and work out the area of triangle $AOB$ .
[4 marks]
2
A circle with centre $O$ has equation $x^2+y^2=25$ . From $P(13,0)$ , tangents touch the circle at $A$ and $B$ . Work out the perimeter of quadrilateral $OAPB$ .
[4 marks]
3
The line $y=x+2$ meets the circle $x^2+y^2=34$ at $A$ and $B$ . Work out the exact length $AB$ .
[4 marks]
4
The point $P$ lies on $x^2+y^2=225$ , whose centre is $O=(0,0)$ . The ray from $O$ to $P$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $90^\circ<\theta<180^\circ$ and $\tan\theta=-\frac{4}{3}$ . Work out the coordinates of $P$ .
[4 marks]
5
A square is inscribed in the circle $x^2+y^2=200$ , with the endpoints of each diagonal on the circle. Work out the exact area inside the circle but outside the square.
[4 marks]

CG8 · Understand that (x - a)^2 + (y - b)^2 = r^2 is the equation of a circle with centre (a, b) and radius r

Explanation

The equation $(x-a)^2+(y-b)^2=r^2$ represents a circle with centre $(a,b)$ and positive radius $r$ . The bracket signs are opposite to the centre coordinates: $(x+2)^2$ corresponds to centre coordinate $-2$ .
Given a centre and a point on the circle, calculate $r^2$ as the squared distance between them.
Given diameter endpoints, their midpoint is the centre and one quarter of the squared diameter is $r^2$ .
A candidate point can be checked by substitution.
Examiners expect the final equation in centre-radius form with both brackets, the correct signs and $r^2$ on the right; expanding the equation is unnecessary unless specifically requested.

A circle translated from the origin has centre (a,b) and radius r.

Worked example

A circle has centre $(-2,3)$ and passes through $(4,-5)$ . Work out its equation.

1.The coordinate differences from the centre are $4-(-2)=6$ and $-5-3=-8$ .
2.Calculate $r^2=6^2+(-8)^2=36+64=100$ .
3.Insert centre $(-2,3)$ into centre-radius form.

Answer: $(x+2)^2+(y-3)^2=100$ .

Common mistakes

Don't write $(x-2)^2$ for centre coordinate $-2$ instead of $(x+2)^2$ .
Don't use $r=100$ after calculating the squared radius $r^2=100$ .
Don't subtract the coordinate differences without squaring and obtain a negative radius value.

Exam tip

Read the centre by reversing the signs inside the brackets, then check the equation by substituting the given point.

Tier 1 · Easy

1
State the centre and radius of $(x-3)^2+(y+2)^2=25$ .
[2 marks]
2
Write down the equation of the circle with centre $(-3,5)$ and radius $4$ .
[2 marks]

Tier 2 · Standard

1
A circle has centre $(-4,1)$ and passes through $(2,9)$ . Work out its equation.
[3 marks]
2
The circle $(x-2)^2+(y+1)^2=k$ passes through $(5,3)$ . Work out $k$ and the radius of the circle.
[3 marks]
3
A circle has centre $(a,-2)$ and radius $5$ . It passes through $(4,2)$ and $a<4$ . Work out its equation.
[3 marks]

Tier 3 · Hard

1
The endpoints of a diameter of a circle are $A(-3,5)$ and $B(7,-1)$ . Work out the equation of the circle.
[4 marks]
2
The centre of a circle lies on the $x$ -axis. The circle passes through $A(-2,3)$ and $B(4,3)$ . Work out the equation of the circle.
[4 marks]
3
A circle lies in the first quadrant and touches both coordinate axes. It passes through $(6,3)$ and its radius is less than $10$ . Work out the equation of the circle.
[4 marks]
4
A circle passes through $A(0,0)$ , $B(8,0)$ and $C(4,8)$ . Work out the equation of the circle.
[4 marks]
5
The equation $x^2+y^2+8x-6y=0$ represents a circle. Work out its centre and radius, writing the equation in centre-radius form.
[4 marks]

CG9 · The equation of a tangent at a point on a circle

Explanation

A tangent touches a circle at one point and is perpendicular to the radius at that point. For a non-axis-aligned radius, calculate its gradient from the centre to the point of contact, then take the negative reciprocal for the tangent gradient.
A vertical radius gives a horizontal tangent, while a horizontal radius gives a vertical tangent.
The tangent equation follows from point-gradient form $y-y_1=m(x-x_1)$ using the point of contact.
For a circle centred away from the origin, the radius gradient must start at the actual centre.
Examiners expect the perpendicular-gradient step and the substitution of the contact point; using the radius gradient unchanged produces the equation of the radius, not the tangent.

The tangent at P is perpendicular to the radius through P.

Worked example

Work out the tangent to $(x-1)^2+(y+2)^2=25$ at the point $P(5,1)$ .

1.The centre is $(1,-2)$ , so the radius gradient is $\frac{1-(-2)}{5-1}=\frac34$ .
2.The tangent gradient is the negative reciprocal, $-\frac43$ .
3.Use $y-1=-\frac43(x-5)$ and rearrange.

Answer: $4x+3y=23$ .

Common mistakes

Don't calculate the radius gradient from $(0,0)$ instead of from the centre $(1,-2)$ .
Don't use tangent gradient $\frac34$ and write the radius line again.
Don't find the correct gradient but substitute the circle’s centre rather than the contact point into the tangent equation.

Exam tip

Show centre-to-contact gradient, negative reciprocal, then point-gradient form as three distinct method lines.

Tier 1 · Easy

1
The point $(3,4)$ lies on $x^2+y^2=25$ . Work out the gradient of the tangent there.
[2 marks]
2
Work out the equation of the tangent to $x^2+y^2=49$ at the point $(0,-7)$ .
[2 marks]

Tier 2 · Standard

1
Work out the equation of the tangent to $x^2+y^2=65$ at $(1,8)$ .
[3 marks]
2
Work out the equation of the tangent to $(x+1)^2+(y-2)^2=25$ at $P(2,6)$ .
[3 marks]
3
Show that $3x-4y=25$ is tangent to the circle $x^2+y^2=25$ at $(3,-4)$ .
[3 marks]

Tier 3 · Hard

1
The tangent to $(x-2)^2+(y+3)^2=100$ at $P(8,5)$ meets the $y$ -axis at $Q$ . Work out the coordinates of $Q$ .
[4 marks]
2
The tangent to $x^2+y^2=100$ at $P(6,8)$ meets the positive coordinate axes at $A$ and $B$ . Work out the exact area of triangle $OAB$ , where $O=(0,0)$ .
[4 marks]
3
Work out the equations of the two tangents to $x^2+y^2=25$ that are parallel to $3x+4y=0$ .
[4 marks]
4
The point $P$ lies on $(x-2)^2+(y+1)^2=25$ , has $x$ -coordinate $5$ and lies above the centre. Work out the equation of the tangent to the circle at $P$ .
[3 marks]
5
Tangents to $x^2+y^2=25$ at $A(3,4)$ and $B(-4,3)$ meet at $Q$ . Work out the coordinates of $Q$ .
[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CG1 · Know and use the definition of a gradient

Tier 1 · Easy

Mark scheme for CG1 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$m=2$	2	$m=\frac{10-2}{5-1}=\frac{8}{4}=2$ .
2	$m=3$	1	The gradient is $\frac{\text{vertical change}}{\text{horizontal change}}=\frac{15}{5}=3$ .

Tier 2 · Standard

Mark scheme for CG1 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$m=\frac{3}{2}$	2	Rearrange to make $y$ the subject: $-2y=12-3x$ , so $y=\frac{3}{2}x-6$ . The coefficient of $x$ is the gradient.
2	$m=\frac{2}{3}$	2	Using the two intercepts, $m=\frac{0-(-4)}{6-0}=\frac{4}{6}=\frac{2}{3}$ .
3	$k=2$	3	The gradient of $AC$ is $\frac{-4-8}{7-(-5)}=-1$ . The gradient of $AB$ must also be $-1$ , so $\frac{k-8}{1-(-5)}=-1$ . Hence $k-8=-6$ and $k=2$ .

Tier 3 · Hard

Mark scheme for CG1 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$p=15$	3	Using the two points, $\frac{2p+1-4}{p-(-3)}=\frac{2p-3}{p+3}$ . Set this equal to $\frac{3}{2}$ : $2(2p-3)=3(p+3)$ , so $4p-6=3p+9$ and $p=15$ .
2	$m=-\frac{1}{5}$	3	Adding the equations gives $3x=12$ , so $x=4$ and $y=3$ . Hence $P=(4,3)$ . The gradient of $PQ$ is $\frac{3-4}{4-(-1)}=-\frac{1}{5}$ .
3	$t=3$	4	The gradient is $\frac{3t+7-(t^2+1)}{(t+2)-t}=\frac{-t^2+3t+6}{2}$ . Equating this to $3$ gives $-t^2+3t+6=6$ , so $t(t-3)=0$ . The roots are $0$ and $3$ ; since $t>1$ , $t=3$ .
4	$m=\frac{21}{16}$	3	The horizontal coordinate change is $8\times4=32$ units and the vertical coordinate change is $6\times7=42$ units. Therefore the gradient is $\frac{42}{32}=\frac{21}{16}$ .
5	$a=6$ ; both points have $x=11$ , so the horizontal change is $0$ and the gradient is undefined.	3	A vertical line has equal $x$ -coordinates, so $2a-1=a+5$ and $a=6$ . Then $A=(11,22)$ and $B=(11,1)$ . The gradient fraction has denominator $11-11=0$ , so the gradient is undefined.

CG2 · Know the relationship between the gradients of parallel and perpendicular lines

Tier 1 · Easy

Mark scheme for CG2 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\frac{1}{3}$	1	The perpendicular gradient is the negative reciprocal. The negative reciprocal of $-3$ is $\frac{1}{3}$ , and $-3\times\frac{1}{3}=-1$ .
2	$-\frac{5}{2}$	2	Rearrange to $y=-\frac{5}{2}x+\frac{7}{2}$ . Parallel lines have equal gradients, so the required gradient is $-\frac{5}{2}$ .

Tier 2 · Standard

Mark scheme for CG2 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$AB\perp CD$	3	$m_{AB}=\frac{3-1}{4-(-2)}=\frac{2}{6}=\frac{1}{3}$ and $m_{CD}=\frac{-1-5}{3-1}=\frac{-6}{2}=-3$ . Their product is $\frac{1}{3}\times(-3)=-1$ , so the lines are perpendicular.
2	$k=3$	3	The gradient of $AB$ is $\frac{11-k}{5-1}=\frac{11-k}{4}$ . Parallel lines have equal gradients, so $\frac{11-k}{4}=2$ . Hence $11-k=8$ and $k=3$ .
3	$k=6$	3	The first line has gradient $-\frac{2}{3}$ and the second has gradient $\frac{k}{4}$ . Perpendicular gradients have product $-1$ , so $-\frac{2}{3}\times\frac{k}{4}=-1$ . Hence $2k=12$ and $k=6$ .

Tier 3 · Hard

Mark scheme for CG2 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$4x+6y=21$	4	$L$ is $y=\frac{3}{2}x+\frac{7}{2}$ , so its gradient is $\frac{3}{2}$ and its $y$ -intercept is $(0,\frac{7}{2})$ . The perpendicular gradient is $-\frac{2}{3}$ . Hence $y-\frac{7}{2}=-\frac{2}{3}x$ . Multiplying by $6$ and rearranging gives $4x+6y=21$ .
2	$3x+2y=3$ (accept any non-zero multiple, e.g. $6x+4y=6$ )	4	The midpoint of $AB$ is $(-1,3)$ . The gradient of $AB$ is $\frac{5-1}{2-(-4)}=\frac{2}{3}$ , so the perpendicular gradient is $-\frac{3}{2}$ . Thus $y-3=-\frac{3}{2}(x+1)$ , which rearranges to $3x+2y=3$ .
3	$P=(0,7)$ or $P=(0,-2)$	4	The gradients of $PA$ and $PB$ are $\frac{p-1}{3}$ and $\frac{4-p}{6}$ . Their product is $-1$ , so $(p-1)(4-p)=-18$ . This simplifies to $p^2-5p-14=0$ , so $(p-7)(p+2)=0$ , giving $p=7$ or $p=-2$ . Therefore $P=(0,7)$ or $P=(0,-2)$ .
4	$m_{AB}=m_{CD}=\frac{3}{2}$ and $m_{BC}=m_{DA}=-\frac{2}{3}$ ; opposite sides are parallel and adjacent sides are perpendicular, so $ABCD$ is a rectangle.	4	$m_{AB}=\frac{7-1}{1-(-3)}=\frac{3}{2}$ and $m_{DC}=\frac{5-(-1)}{4-0}=\frac{3}{2}$ , so $AB\parallel DC$ . Also $m_{BC}=\frac{5-7}{4-1}=-\frac{2}{3}$ and $m_{AD}=\frac{-1-1}{0-(-3)}=-\frac{2}{3}$ , so $BC\parallel AD$ . Since $\frac{3}{2}\times(-\frac{2}{3})=-1$ , adjacent sides are perpendicular. Therefore $ABCD$ is a rectangle.
5	$a=4$ and $b=6$	4	The line $2x+ay=7$ cannot be vertical, because it is parallel to a line of gradient $-\frac12$ ; so $a\neq0$ and its gradient is $-\frac{2}{a}$ . Parallel gradients are equal, so $-\frac{2}{a}=-\frac{1}{2}$ and $a=4$ . The gradient of $bx-3y=5$ is $\frac{b}{3}$ . Perpendicular gradients have product $-1$ , so $-\frac{1}{2}\times\frac{b}{3}=-1$ , giving $b=6$ .

CG3 · Use Pythagoras' theorem to calculate the distance between two points

Tier 1 · Easy

Mark scheme for CG3 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$PQ=5$	2	The coordinate differences are $4-1=3$ and $6-2=4$ . Therefore $PQ=\sqrt{3^2+4^2}=\sqrt{25}=5$ .
2	$PQ=2\sqrt{5}$	2	$PQ=\sqrt{(6-2)^2+(1-(-1))^2}=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}$ .

Tier 2 · Standard

Mark scheme for CG3 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$RS=10$	3	$RS=\sqrt{(6-(-2))^2+(-1-5)^2}=\sqrt{8^2+(-6)^2}=\sqrt{100}=10$ .
2	$k=7$	3	$(k-(-1))^2+(8-2)^2=10^2$ , so $(k+1)^2+36=100$ and $(k+1)^2=64$ . Since $k>-1$ , $k+1=8$ , giving $k=7$ .
3	$4\sqrt{13}$ units	3	$AB=\sqrt{(1-(-3))^2+(-2-4)^2}=\sqrt{52}=2\sqrt{13}$ . Also $BC=\sqrt{(7-1)^2+(2-(-2))^2}=\sqrt{52}=2\sqrt{13}$ . The route length is $2\sqrt{13}+2\sqrt{13}=4\sqrt{13}$ units.

Tier 3 · Hard

Mark scheme for CG3 Tier 3 · Hard
Q	Answer	Mark	Comments
1	Perimeter $=32$ The triangle is isosceles.	4	$AB=\sqrt{6^2+8^2}=10$ , $BC=\sqrt{6^2+(-8)^2}=10$ , and $AC=\sqrt{12^2+0^2}=12$ . The perimeter is $10+10+12=32$ . Since $AB=BC$ , the triangle is isosceles.
2	The triangle is right-angled at $A$ . Area $=15$ square units	4	$AB^2=6^2+3^2=45$ , $AC^2=2^2+(-4)^2=20$ and $BC^2=(-4)^2+(-7)^2=65$ . Since $45+20=65$ , the triangle is right-angled at $A$ . Its area is $\frac{1}{2}\sqrt{45}\sqrt{20}=15$ .
3	$p=\frac{1}{4}$ $PA=\frac{5\sqrt{17}}{4}$ units	4	Equating squared distances gives $(p+3)^2+4^2=(p-5)^2+2^2$ . Expanding and simplifying gives $16p=4$ , so $p=\frac{1}{4}$ . Then $PA=\sqrt{(\frac{1}{4}+3)^2+4^2}=\sqrt{\frac{425}{16}}=\frac{5\sqrt{17}}{4}$ units.
4	$7$ minutes $5$ seconds	4	The coordinate changes are $15$ and $8$ , so $PQ=\sqrt{15^2+8^2}=17$ coordinate units. The actual distance is $17\times40=680$ metres. The time is $680\div1.6=425$ seconds, which is $7$ minutes $5$ seconds.
5	$12\sqrt{10}$ units	4	The diagonal length is $AC=\sqrt{(7-(-5))^2+(8-2)^2}=\sqrt{180}=6\sqrt{5}$ . If the side length is $s$ , then the diagonal is $s\sqrt{2}$ , so $s=\frac{6\sqrt{5}}{\sqrt{2}}=3\sqrt{10}$ . The perimeter is $4s=12\sqrt{10}$ units.

CG4 · Use ratio to find the coordinates of a point on a line given the coordinates of two other points, including the midpoint

Tier 1 · Easy

Mark scheme for CG4 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$(2,3)$	2	Average the coordinates: $x=\frac{-2+6}{2}=2$ and $y=\frac{5+1}{2}=3$ . The midpoint is $(2,3)$ .
2	$B=(10,-7)$	2	If $B=(x,y)$ , then $\frac{-2+x}{2}=4$ and $\frac{5+y}{2}=-1$ . These give $x=10$ and $y=-7$ , so $B=(10,-7)$ .

Tier 2 · Standard

Mark scheme for CG4 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$P=(2,7)$	3	The vector from $A$ to $B$ is $(15,15)$ . Since $AP$ is $\frac{2}{5}$ of $AB$ , add $\frac{2}{5}(15,15)=(6,6)$ to $A$ . This gives $P=(-4,1)+(6,6)=(2,7)$ .
2	$AP:PB=2:3$	3	From $A$ to $P$ the displacement is $(6,4)=2(3,2)$ . From $P$ to $B$ it is $(9,6)=3(3,2)$ . The displacements are in the same direction, so $AP:PB=2:3$ .
3	$P=(-2,5)$ and $Q=(3,0)$	3	$\overrightarrow{AB}=(15,-15)$ , so one third of this displacement is $(5,-5)$ . Hence $P=A+(5,-5)=(-2,5)$ and $Q=A+2(5,-5)=(3,0)$ .

Tier 3 · Hard

Mark scheme for CG4 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$k=6$ Midpoint of $AB$ is $(0,-1)$	4	For $AP:PB=3:1$ , $P=\frac{A+3B}{4}$ . Using the $x$ -coordinate, $3=\frac{-6+3k}{4}$ , so $12=-6+3k$ and $k=6$ . Thus $B=(6,-7)$ . The midpoint is $(\frac{-6+6}{2},\frac{5+(-7)}{2})=(0,-1)$ .
2	$B=(14,13)$	4	$\overrightarrow{AP}=(6,4)$ . Since $AP:PM=2:1$ , $\overrightarrow{AM}=\frac{3}{2}(6,4)=(9,6)$ , so $M=(5,7)$ . As $M$ is the midpoint of $AB$ , $B=2M-A=(10,14)-(-4,1)=(14,13)$ .
3	$A=(-4,2)$ and $B=(8,14)$	4	Using the given ratios, $3P=2A+B$ and $4Q=A+3B$ . From the first equation, $B=3P-2A$ . Substitution into the second gives $4Q=9P-5A$ , so $A=\frac{9P-4Q}{5}=(-4,2)$ . Then $B=3P-2A=(8,14)$ .
4	$(1,8)$	4	$\overrightarrow{AB}=(20,10)$ . Point $P$ is one fifth of the way from $A$ to $B$ , so $P=(-7,4)+\frac{1}{5}(20,10)=(-3,6)$ . Point $Q$ is three fifths of the way from $A$ to $B$ , so $Q=(-7,4)+\frac{3}{5}(20,10)=(5,10)$ . The midpoint of $PQ$ is $(\frac{-3+5}{2},\frac{6+10}{2})=(1,8)$ .
5	$p=1$ and $P=(\frac{10}{3},\frac{5}{3})$	4	Since $AP:PB=1:2$ , $P=\frac{2A+B}{3}$ , so $P=(\frac{2p+8}{3},\frac{p+4}{3})$ . Substitution into $y=2x-5$ gives $\frac{p+4}{3}=2(\frac{2p+8}{3})-5=\frac{4p+1}{3}$ . Hence $p+4=4p+1$ , so $p=1$ . Therefore $P=(\frac{10}{3},\frac{5}{3})$ .

CG5 · The equation of a straight line: y = mx + c and y - y1 = m(x - x1) and other forms, including interpretation of the gradient and y-intercept

Tier 1 · Easy

Mark scheme for CG5 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Gradient $=3$ $y$ -intercept $=(0,-4)$	2	Compare $y=3x-4$ with $y=mx+c$ . Therefore $m=3$ and $c=-4$ , so the line crosses the $y$ -axis at $(0,-4)$ .
2	$y=-2x+6$	1	In $y=mx+c$ , the gradient is $m=-2$ and the $y$ -intercept gives $c=6$ . Therefore $y=-2x+6$ .

Tier 2 · Standard

Mark scheme for CG5 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$y=4x-9$	3	Use point-gradient form: $y-(-1)=4(x-2)$ . Hence $y+1=4x-8$ , so $y=4x-9$ .
2	$k=2$ $y$ -intercept $=(0,6)$	3	Rearranging gives $y=-\frac{4}{k}x+\frac{12}{k}$ . Hence $-\frac{4}{k}=-2$ , so $k=2$ . The equation is $2x+y=6$ , which meets the $y$ -axis at $(0,6)$ .
3	$f(x)=-\frac{3}{2}x+\frac{13}{2}$ (or $f(x)=\frac{13-3x}{2}$ )	3	The corresponding points are $(-3,11)$ and $(5,-1)$ , so the gradient is $\frac{-1-11}{5-(-3)}=-\frac{3}{2}$ . Write $f(x)=-\frac{3}{2}x+c$ and use $f(5)=-1$ : $-1=-\frac{15}{2}+c$ , giving $c=\frac{13}{2}$ .

Tier 3 · Hard

Mark scheme for CG5 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$2x+y=1$ $x$ -intercept $=(\frac{1}{2},0)$	4	The gradient is $\frac{-7-5}{4-(-2)}=\frac{-12}{6}=-2$ . Through $A$ , $y-5=-2(x+2)$ , so $y=-2x+1$ and hence $2x+y=1$ . At the $x$ -intercept $y=0$ , giving $2x=1$ and $x=\frac{1}{2}$ .
2	$3x+2y=17$ (accept any non-zero multiple)	4	Adding the two equations gives $3x=9$ , so $P=(3,4)$ . The gradient of $PQ$ is $\frac{-2-4}{7-3}=-\frac{3}{2}$ . Therefore $y-4=-\frac{3}{2}(x-3)$ , which rearranges to $3x+2y=17$ .
3	$a=4$ and $b=12$	4	The intercepts are $\frac{20}{a}$ and $\frac{20}{b}$ . Since the $x$ -intercept is three times the $y$ -intercept, $\frac{20}{a}=3\times\frac{20}{b}$ , so $b=3a$ . Substituting $(2,1)$ into the line gives $2a+b=20$ . Hence $5a=20$ , so $a=4$ and $b=12$ .
4	$x=-\frac{2}{5}(y+1)$ (or $x=-\frac{2}{5}y-\frac{2}{5}$ )	4	The gradient is $\frac{-15}{6}=-\frac{5}{2}$ . Write $y=-\frac{5}{2}x+c$ and substitute $(-4,9)$ : $9=10+c$ , so $c=-1$ . Hence $y=-\frac{5}{2}x-1$ . Rearranging gives $x=-\frac{2}{5}(y+1)$ .
5	$2x-3y=12$ (accept any non-zero multiple)	3	If the intercepts are $A=(u,0)$ and $B=(0,v)$ , their midpoint is $(\frac{u}{2},\frac{v}{2})=(3,-2)$ . Hence $A=(6,0)$ and $B=(0,-4)$ . The gradient is $\frac{0-(-4)}{6-0}=\frac{2}{3}$ , so $y=\frac{2}{3}x-4$ , which rearranges to $2x-3y=12$ .

CG6 · Draw a straight line from given information

Tier 1 · Easy

Mark scheme for CG6 Tier 1 · Easy
Q	Answer	Mark	Comments
1	The straight line segment through $(-1,-1)$ , $(0,1)$ , $(1,3)$ and $(2,5)$	2	Substitute $x=-1,0,1,2$ to obtain $y=-1,1,3,5$ . Plot the points and join them with one straight line segment over the stated interval.
2	The vertical line segment from $(-2,-4)$ to $(-2,3)$	1	Every point has $x$ -coordinate $-2$ . Plot $(-2,-4)$ and $(-2,3)$ and join them with a vertical straight line segment.

Tier 2 · Standard

Mark scheme for CG6 Tier 2 · Standard
Q	Answer	Mark	Comments
1	The straight line segment through $(-3,4)$ , $(0,2)$ and $(3,0)$	3	Rearrange to $y=2-\frac{2}{3}x$ . Substitution gives the convenient points $(-3,4)$ , $(0,2)$ and $(3,0)$ . Plot them and draw one straight line through them across the interval.
2	The line segment $y=-3x+7$ from $(-1,10)$ to $(3,-2)$	3	Use $y-4=-3(x-1)$ to get $y=-3x+7$ . Plot points such as $(-1,10)$ , $(1,4)$ and $(3,-2)$ , then join them with one straight line segment.
3	The straight line segment $y=3-\frac{3}{4}x$ from $(-4,6)$ to $(8,-3)$	3	Rearrange to $y=3-\frac{3}{4}x$ . Convenient points are $(-4,6)$ , $(0,3)$ , $(4,0)$ and $(8,-3)$ . Plot the points and join them with one straight line segment over the stated interval.

Tier 3 · Hard

Mark scheme for CG6 Tier 3 · Hard
Q	Answer	Mark	Comments
1	The line segment $y=-2x+1$ from $(-1,3)$ to $(3,-5)$	4	The given gradient is $\frac{1}{2}$ , so the perpendicular gradient is $-2$ . Through $(-2,5)$ , $y-5=-2(x+2)$ , giving $y=-2x+1$ . Plot points such as $(-1,3)$ , $(0,1)$ and $(3,-5)$ , then draw the straight segment.
2	The line segment $y=\frac{7}{3}x-5$ from $(0,-5)$ to $(6,9)$	4	Adding the simultaneous equations gives $3x=9$ , so their intersection is $(3,2)$ . The gradient from $(0,-5)$ to $(3,2)$ is $\frac{7}{3}$ , giving $y=\frac{7}{3}x-5$ . Plot $(0,-5)$ , $(3,2)$ and $(6,9)$ and draw the segment.
3	The line segment $3x+2y=-1$ from $(-3,4)$ to $(3,-5)$	4	The required set is the perpendicular bisector of $AB$ . Its midpoint is $(-1,1)$ , and the gradient of $AB$ is $\frac{3-(-1)}{2-(-4)}=\frac{2}{3}$ , so the perpendicular gradient is $-\frac{3}{2}$ . Thus $y-1=-\frac{3}{2}(x+1)$ , or $3x+2y=-1$ . Plot $(-3,4)$ , $(-1,1)$ and $(3,-5)$ and draw the segment.
4	The line segment $y=\frac{3}{2}x+\frac{5}{2}$ from $(-3,-2)$ to $(3,7)$	4	The midpoint of $AB$ is $(-1,1)$ . Rearranging $3x-2y=7$ shows that its gradient is $\frac{3}{2}$ , so the required parallel line satisfies $y-1=\frac{3}{2}(x+1)$ . Hence $y=\frac{3}{2}x+\frac{5}{2}$ . Plot $(-3,-2)$ , $(-1,1)$ and $(3,7)$ and draw the line segment.
5	The line segment $y=-x+8$ from $(-2,10)$ to $(6,2)$	4	Write $L$ as $y=mx+c$ . The point $(-2,10)$ gives $-2m+c=10$ . At $x=4$ , the condition gives $4m+c=\frac{c}{2}$ , so $8m+c=0$ . Solving the two equations gives $m=-1$ and $c=8$ . Plot $(-2,10)$ , $(0,8)$ and $(6,2)$ and draw the line segment.

CG7 · Understand that x^2 + y^2 = r^2 is the equation of a circle with centre (0, 0) and radius r; application of circle geometry facts

Tier 1 · Easy

Mark scheme for CG7 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Centre $=(0,0)$ Radius $=7$	2	Compare with $x^2+y^2=r^2$ . Here $r^2=49$ , so the centre is $(0,0)$ and the positive radius is $r=7$ .
2	$4^2+(-3)^2=16+9=25$ , so $(4,-3)$ lies on the circle.	2	Substitute the coordinates into the left-hand side: $4^2+(-3)^2=16+9=25$ . This equals the right-hand side, so the point lies on the circle.

Tier 2 · Standard

Mark scheme for CG7 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$y=8$	2	Substitute $x=6$ : $6^2+y^2=100$ , so $y^2=64$ and $y=\pm8$ . Since $y>0$ , $y=8$ .
2	$AB$ is a diameter, so the angle in the semicircle is $90^\circ$ (a gradient argument showing $AP\perp BP$ also scores full marks).	3	$B=-A$ , so the segment $AB$ passes through the centre $(0,0)$ and is a diameter. Since $P$ lies on the circle, the angle subtended by diameter $AB$ at $P$ is $90^\circ$ .
3	$AB=24$ units	3	At the intersections, $x^2+5^2=169$ , so $x^2=144$ and $x=\pm12$ . Thus the points are $(-12,5)$ and $(12,5)$ , whose separation is $12-(-12)=24$ units.

Tier 3 · Hard

Mark scheme for CG7 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\angle AOB=90^\circ$ Area of triangle $AOB=50$	4	The gradient of $OA$ is $\frac{8}{6}=\frac{4}{3}$ and the gradient of $OB$ is $\frac{6}{-8}=-\frac{3}{4}$ . Their product is $-1$ , so $OA\perp OB$ and $\angle AOB=90^\circ$ . Both are radii of length $10$ , so the area is $\frac{1}{2}\times10\times10=50$ .
2	Perimeter $=34$ units	4	$OA=OB=5$ and $OP=13$ . A radius is perpendicular to a tangent, so $AP=\sqrt{13^2-5^2}=12$ . Tangents from the same external point are equal, so $BP=12$ . The perimeter is $5+12+12+5=34$ .
3	$AB=8\sqrt{2}$ units	4	Substitution gives $x^2+(x+2)^2=34$ , so $x^2+2x-15=0$ and hence $(x+5)(x-3)=0$ . The intersection points are $(-5,-3)$ and $(3,5)$ . Therefore $AB=\sqrt{8^2+8^2}=\sqrt{128}=8\sqrt{2}$ units.
4	$P=(-9,12)$	4	Because $90^\circ<\theta<180^\circ$ , the ray points into the second quadrant, so $x<0$ and $y>0$ . With $\tan\theta=\frac{y}{x}=-\frac{4}{3}$ , write $x=-3k$ and $y=4k$ with $k>0$ (the opposite point $(9,-12)$ would give an angle between $270^\circ$ and $360^\circ$ , outside the stated range). Substitution into the circle gives $9k^2+16k^2=225$ , so $25k^2=225$ and $k=3$ . Therefore $P=(-9,12)$ .
5	$200\pi-400$ square units	4	The circle has $r^2=200$ , so its area is $200\pi$ . Each diagonal of the square is a diameter, so its squared length is $(2r)^2=4r^2=800$ . The area of a square with diagonal $d$ is $\frac{d^2}{2}$ , giving square area $\frac{800}{2}=400$ . The required area is $200\pi-400$ square units.

CG8 · Understand that (x - a)^2 + (y - b)^2 = r^2 is the equation of a circle with centre (a, b) and radius r

Tier 1 · Easy

Mark scheme for CG8 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Centre $=(3,-2)$ Radius $=5$	2	Write $y+2$ as $y-(-2)$ . Comparing with $(x-a)^2+(y-b)^2=r^2$ gives centre $(3,-2)$ and radius $\sqrt{25}=5$ .
2	$(x+3)^2+(y-5)^2=16$	2	Use $(x-a)^2+(y-b)^2=r^2$ with $(a,b)=(-3,5)$ and $r=4$ . This gives $(x+3)^2+(y-5)^2=16$ .

Tier 2 · Standard

Mark scheme for CG8 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$(x+4)^2+(y-1)^2=100$	3	The squared radius is the squared distance from $(-4,1)$ to $(2,9)$ : $r^2=(2+4)^2+(9-1)^2=6^2+8^2=100$ . Insert the centre and $r^2$ into the circle equation.
2	$k=25$ Radius $=5$	3	Substitute $(5,3)$ : $k=(5-2)^2+(3+1)^2=3^2+4^2=25$ . Since $k=r^2$ , the radius is $r=5$ .
3	$(x-1)^2+(y+2)^2=25$	3	Using the point $(4,2)$ gives $(4-a)^2+(2-(-2))^2=25$ . Hence $(4-a)^2=9$ , so $a=1$ or $a=7$ . Since $a<4$ , the centre is $(1,-2)$ and the equation is $(x-1)^2+(y+2)^2=25$ .

Tier 3 · Hard

Mark scheme for CG8 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$(x-2)^2+(y-2)^2=34$	4	The centre is the midpoint: $(\frac{-3+7}{2},\frac{5+(-1)}{2})=(2,2)$ . The diameter has squared length $10^2+(-6)^2=136$ , so $r^2=\frac{136}{4}=34$ . Hence the equation is $(x-2)^2+(y-2)^2=34$ .
2	$(x-1)^2+y^2=18$	4	The centre is equidistant from $A$ and $B$ , so it lies on the perpendicular bisector of $AB$ . The midpoint of the horizontal segment $AB$ has $x$ -coordinate $1$ , and the centre is on the $x$ -axis, so the centre is $(1,0)$ . Then $r^2=(-2-1)^2+(3-0)^2=18$ , giving $(x-1)^2+y^2=18$ .
3	$(x-3)^2+(y-3)^2=9$	4	If the radius is $r$ , touching both axes in the first quadrant makes the centre $(r,r)$ . Substituting $(6,3)$ gives $(6-r)^2+(3-r)^2=r^2$ , so $r^2-18r+45=0$ and hence $(r-3)(r-15)=0$ . The radius is less than $10$ , so $r=3$ . The centre is $(3,3)$ and the equation is $(x-3)^2+(y-3)^2=9$ .
4	$(x-4)^2+(y-3)^2=25$	4	The perpendicular bisector of $AB$ is $x=4$ , so write the centre as $(4,b)$ . Equating the squared distances from the centre to $A$ and $C$ gives $4^2+b^2=(8-b)^2$ . Hence $16+b^2=64-16b+b^2$ , so $b=3$ . The squared radius is $4^2+3^2=25$ , giving $(x-4)^2+(y-3)^2=25$ .
5	$(x+4)^2+(y-3)^2=25$ ; centre $(-4,3)$ and radius $5$	4	Complete both squares: $x^2+8x=(x+4)^2-16$ and $y^2-6y=(y-3)^2-9$ . Therefore $(x+4)^2-16+(y-3)^2-9=0$ , so $(x+4)^2+(y-3)^2=25$ . The centre is $(-4,3)$ and the radius is $5$ .

CG9 · The equation of a tangent at a point on a circle

Tier 1 · Easy

Mark scheme for CG9 Tier 1 · Easy
Q	Answer	Mark	Comments
1	Tangent gradient $=-\frac{3}{4}$	2	The radius from $(0,0)$ to $(3,4)$ has gradient $\frac{4}{3}$ . The tangent is perpendicular, so its gradient is the negative reciprocal, $-\frac{3}{4}$ .
2	$y=-7$	2	The radius from $(0,0)$ to $(0,-7)$ is vertical. The tangent is perpendicular to the radius, so it is the horizontal line through $(0,-7)$ : $y=-7$ .

Tier 2 · Standard

Mark scheme for CG9 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x+8y=65$	3	The radius gradient is $\frac{8}{1}=8$ , so the tangent gradient is $-\frac{1}{8}$ . Thus $y-8=-\frac{1}{8}(x-1)$ . Multiplying by $8$ and rearranging gives $x+8y=65$ .
2	$3x+4y=30$ (or any equivalent form, e.g. $y=-\frac{3}{4}x+\frac{15}{2}$ )	3	The centre is $(-1,2)$ , so the radius gradient is $\frac{6-2}{2-(-1)}=\frac{4}{3}$ . The tangent gradient is $-\frac{3}{4}$ . Hence $y-6=-\frac{3}{4}(x-2)$ , which rearranges to $3x+4y=30$ .
3	$3(3)-4(-4)=25$ and $3^2+(-4)^2=25$ , so $(3,-4)$ lies on both the line and the circle; $\frac{3}{4}\times(-\frac{4}{3})=-1$ , so the line is perpendicular to the radius there and is tangent.	3	The point lies on the line because $3(3)-4(-4)=9+16=25$ , and on the circle because $3^2+(-4)^2=25$ . The radius to the point has gradient $-\frac{4}{3}$ , while $3x-4y=25$ has gradient $\frac{3}{4}$ . Their product is $-1$ , so the line is perpendicular to the radius at the point of contact and is therefore tangent.

Tier 3 · Hard

Mark scheme for CG9 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$Q=(0,11)$	4	The centre is $(2,-3)$ . The radius to $P$ has gradient $\frac{5-(-3)}{8-2}=\frac{8}{6}=\frac{4}{3}$ , so the tangent gradient is $-\frac{3}{4}$ . Its equation is $y-5=-\frac{3}{4}(x-8)$ , or $y=-\frac{3}{4}x+11$ . At the $y$ -axis $x=0$ , so $Q=(0,11)$ .
2	Area $=\frac{625}{6}$ square units	4	The radius gradient is $\frac{8}{6}=\frac{4}{3}$ , so the tangent gradient is $-\frac{3}{4}$ . Its equation is $y-8=-\frac{3}{4}(x-6)$ , or $3x+4y=50$ . Thus the intercepts are $A=(\frac{50}{3},0)$ and $B=(0,\frac{25}{2})$ . The area is $\frac{1}{2}\times\frac{50}{3}\times\frac{25}{2}=\frac{625}{6}$ .
3	$3x+4y=25$ and $3x+4y=-25$ (accept any non-zero multiples or equivalent forms, e.g. $y=-\frac{3}{4}x\pm\frac{25}{4}$ )	4	The tangents have gradient $-\frac{3}{4}$ , so the radii to their contact points have gradient $\frac{4}{3}$ . The radial line $y=\frac{4}{3}x$ meets the circle at $(3,4)$ and $(-3,-4)$ . The tangents through these points are $y-4=-\frac{3}{4}(x-3)$ and $y+4=-\frac{3}{4}(x+3)$ , which simplify to $3x+4y=25$ and $3x+4y=-25$ .
4	$3x+4y=27$ (or $y=-\frac{3}{4}x+\frac{27}{4}$ )	3	Substituting $x=5$ gives $3^2+(y+1)^2=25$ , so $(y+1)^2=16$ . Since $P$ lies above the centre, $y=3$ and $P=(5,3)$ . The radius from $(2,-1)$ to $P$ has gradient $\frac{4}{3}$ , so the tangent gradient is $-\frac{3}{4}$ . Hence $y-3=-\frac{3}{4}(x-5)$ , which rearranges to $3x+4y=27$ .
5	$Q=(-1,7)$	4	At $A$ , the radius gradient is $\frac{4}{3}$ , so the tangent is $3x+4y=25$ . At $B$ , the radius gradient is $-\frac{3}{4}$ , so the tangent is $-4x+3y=25$ . Solving these equations simultaneously gives $x=-1$ and $y=7$ . Therefore $Q=(-1,7)$ .