3 Coordinate geometry in the (x, y) plane — revision question pack

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

Checked against Edexcel 9MA0 section 3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

3.1 · Understand and use the equation of a straight line, including the forms y − y1 = m(x − x1) and ax + by + c = 0; gradient conditions for two straight lines to be parallel or perpendicular; use straight line models in a variety of contexts.

Explanation

A non-vertical straight line has constant gradient $m$ and may be written as $y-y_1=m(x-x_1)$ through a known point or rearranged into $ax+by+c=0$ .
Parallel lines have equal gradients, while non-vertical perpendicular lines have gradients whose product is $-1$ .
For two known points, first calculate $m=(y_2-y_1)/(x_2-x_1)$ , then substitute one point into a line equation and check the other point.
In a linear model, interpret the gradient and intercept in context and respect practical restrictions.
A common error is to round a limiting input upwards when this breaks the stated constraint.

Worked example

The line $L$ has equation $2x-5y+7=0$ . Find an equation of the line through $(4,-1)$ that is perpendicular to $L$ . Write your equation as $ax+by+c=0$ with integer coefficients.

1.Rearrange $L$ to $y=\dfrac{2}{5}x+\dfrac{7}{5}$ , so its gradient is $\dfrac{2}{5}$ .
2.A perpendicular line has gradient $-\dfrac{5}{2}$ .
3.Hence $y+1=-\dfrac{5}{2}(x-4)$ , which rearranges to $5x+2y-18=0$ .

Answer: $5x+2y-18=0$

Common mistakes

Don't treat the constant $c$ in $ax+by+c=0$ as the $y$ -intercept without first rearranging the equation.
Don't use the negative reciprocal rule on the coefficients without first finding the line's gradient.

Exam tip

When asked for a perpendicular line, extract the original gradient, use its negative reciprocal, then substitute the given point.

Tier 1 · Easy

1.
Find the equation of the line with gradient $-3$ that passes through $(2,4)$ . Write it as $y=mx+c$ .
(2)
(Total for Question 1 is 2 marks)
2.
The points $A(-2,7)$ and $B(4,-5)$ lie on a straight line. Find an equation of $AB$ in the form $ax+by+c=0$ . Use integer coefficients with greatest common divisor $1$ , and choose $a>0$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The lines $y=2x+1$ and $y=-x+7$ meet at $P$ . Find the coordinates of $P$ and an equation of the line through $P$ that is parallel to $3x+y=4$ . Write your equation as $ax+by+c=0$ . Use integer coefficients with greatest common divisor $1$ , and choose $a>0$ .
(4)
(Total for Question 1 is 4 marks)
2.
The lines $L_1$ and $L_2$ have equations $y=(k-1)x$ and $2x+(k+3)y=34$ respectively, where $k$ is a constant. Given that $L_1$ and $L_2$ are perpendicular, find $k$ and the coordinates of their point of intersection.
(4)
(Total for Question 2 is 4 marks)
3.
The perpendicular bisector of the segment joining $A(-3,2)$ to $B(5,6)$ is the line $L$ . Find an equation of $L$ in the form $ax+by+c=0$ . Use integer coefficients with greatest common divisor $1$ , and choose $a>0$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A delivery company models the charge $C$ pounds for a journey of $d$ kilometres by a straight line. Journeys of $8$ km and $14$ km cost $\pounds 19.60$ and $\pounds 29.80$ respectively. Find the model for $C$ in terms of $d$ , interpret its gradient, and find the greatest whole number of kilometres that can be travelled for at most $\pounds 45$ .
(5)
(Total for Question 1 is 5 marks)
2.
The points $A(2,-1)$ and $B(p,7)$ lie on a line that is perpendicular to $3x-2y+5=0$ . Find $p$ and the equation of $AB$ . The line $AB$ meets the positive coordinate axes at $P$ and $Q$ . Find the exact area of triangle $OPQ$ , where $O$ is the origin.
(6)
(Total for Question 2 is 6 marks)
3.
The point $P(7,-2)$ does not lie on the line $L$ with equation $3x-4y+12=0$ . The point $Q$ on $L$ is closest to $P$ . Using straight-line methods, find the exact coordinates of $Q$ and the distance $PQ$ . The reflection of $P$ in $L$ is the point $R$ . Find the exact coordinates of $R$ .
(6)
(Total for Question 3 is 6 marks)
4.
The line $L_k$ has equation $(k+1)x+(2-k)y=3k+4$ , where $k$ is a real constant. Show that every line in this family passes through a fixed point $P$ . Find $P$ . Find the value of $k$ for which $L_k$ is parallel to the line $2x-3y=7$ . For this value of $k$ , find the exact area of the triangle formed by $L_k$ and the coordinate axes.
(6)
(Total for Question 4 is 6 marks)
5.
The points $A(-3,1)$ , $B(5,3)$ and $C(k,9)$ are consecutive vertices of a parallelogram $ABCD$ . The midpoint of its diagonals lies on the line $2x-y=1$ . Using straight-line methods, find $k$ and the coordinates of $D$ . Find an equation of the diagonal $BD$ and the exact perpendicular distance from $A$ to $BD$ .
(6)
(Total for Question 5 is 6 marks)

3.2 · Understand and use the coordinate geometry of the circle, including the equation (x − a)² + (y − b)² = r²; complete the square for centre and radius; use circle properties (semicircle angle, chord bisector, tangent-radius).

Explanation

A circle with centre $(a,b)$ and radius $r$ has equation $(x-a)^2+(y-b)^2=r^2$ .
Complete the square separately in $x$ and $y$ to reveal the centre and radius of a circle given in expanded form.
A radius is perpendicular to the tangent at its endpoint; perpendicular chord bisectors meet at the circumcentre, and an angle subtended by a diameter at the circumference is $90^\circ$ .
When testing tangency, require exactly one intersection or set the perpendicular distance from the centre to the line equal to the radius.
A common error is to read the centre as $(a,b)$ from $(x+a)^2+(y+b)^2$ instead of $(-a,-b)$ .

A point

(x,y)

lies on the circle when its distance from centre

(a,b)

r

Worked example

The point $P(5,3)$ lies on the circle with centre $(2,-1)$ . Determine the tangent at $P$ . Write your equation as $ax+by+c=0$ with integer coefficients.

1.The gradient of the radius from $(2,-1)$ to $(5,3)$ is $4/3$ , so the tangent gradient is $-3/4$ .
2.Hence $y-3=-\dfrac{3}{4}(x-5)$ .
3.Multiplying by $4$ and rearranging gives $3x+4y-27=0$ .

Answer: $3x+4y-27=0$

Common mistakes

Don't complete the square but treat $r^2$ as the radius instead of taking its positive square root.
Don't find the radius gradient but forget that the tangent gradient is its negative reciprocal.

Exam tip

For a tangent equation, join the centre to the contact point first, then use perpendicular gradients.

Tier 1 · Easy

1.
Find the centre and radius of the circle $x^2+y^2-6x+8y-11=0$ .
(3)
(Total for Question 1 is 3 marks)
2.
A circle has centre $(-3,4)$ and passes through $(1,1)$ . Find its equation.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The points $A(-2,1)$ and $B(6,5)$ are the endpoints of a diameter of a circle. Find the equation of the circle.
(4)
(Total for Question 1 is 4 marks)
2.
For the circle $C$ with equation $x^2+y^2+4x-6y-12=0$ , determine both $x$ -axis intercepts. At the intercept whose $x$ -coordinate is positive, find the tangent and write it as $ax+by+c=0$ .
(5)
(Total for Question 2 is 5 marks)
3.
The points $A(-4,0)$ and $B(6,0)$ are the endpoints of a diameter of a circle. The point $C=(p,4)$ lies on the circle. Using the angle in a semicircle, find all possible values of $p$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The points $A(-1,2)$ , $B(5,2)$ and $C(1,6)$ lie on a circle. Use perpendicular bisectors to find the equation of this circumcircle.
(5)
(Total for Question 1 is 5 marks)
2.
The circle $C$ has centre $(2,-1)$ and radius $5$ . The point $P(15,-1)$ lies outside $C$ . Find the equations of both tangents from $P$ to $C$ and the exact coordinates of both points of contact.
(7)
(Total for Question 2 is 7 marks)
3.
The circles $C_1$ and $C_2$ have equations $x^2+y^2=25$ and $(x-6)^2+y^2=25$ respectively. Find an equation of their common chord and the exact coordinates of its endpoints. Hence find the equation of the circle whose diameter is this common chord.
(6)
(Total for Question 3 is 6 marks)
4.
A circle has its centre in the first quadrant and is tangent to both coordinate axes. It passes through the point $(6,8)$ . Find the equations of all possible circles, giving the radii in exact form.
(5)
(Total for Question 4 is 5 marks)
5.
The line $y=2x+c$ cuts the circle $(x+1)^2+(y-3)^2=49$ in a chord of length $6$ . Find all possible values of $c$ . For each value, find the exact coordinates of the midpoint of the chord.
(6)
(Total for Question 5 is 6 marks)

3.3 · Understand and use the parametric equations of curves and conversion between Cartesian and parametric forms.

Explanation

Parametric equations express both coordinates in terms of a third variable: $x=f(t)$ and $y=g(t)$ .
To obtain a Cartesian equation, eliminate the parameter by making $t$ the subject of one equation and substituting into the other, or by using a suitable identity.
To parametrise a Cartesian curve, choose expressions that satisfy it identically, such as $x=a\cos t$ and $y=b\sin t$ for $x^2/a^2+y^2/b^2=1$ .
Record any parameter interval because it determines which part of the Cartesian curve is traced.
A common error is to eliminate $t$ but lose a domain restriction.

Worked example

The curve $x=2t+1$ , $y=t^2-3$ is defined for $t\ge0$ . Find a Cartesian equation and state the corresponding restriction on $x$ .

1.Rearrange $x=2t+1$ to $t=(x-1)/2$ .
2.Therefore $y=((x-1)/2)^2-3=(x-1)^2/4-3$ .
3.Since $t\ge0$ , $x=2t+1\ge1$ .

Answer: $y=\dfrac{(x-1)^2}{4}-3$ , with $x\ge1$

Common mistakes

Don't square an equation to eliminate the parameter and retain Cartesian points not generated by any allowed parameter.
Don't eliminate the parameter correctly but lose the restriction inherited from its stated domain.

Exam tip

After converting to Cartesian form, translate the parameter interval into the corresponding restriction on the Cartesian variable.

Tier 1 · Easy

1.
For $x=2t-1$ and $y=t+4$ , eliminate $t$ to obtain a Cartesian equation.
(2)
(Total for Question 1 is 2 marks)
2.
The curve has parametric equations $x=5\cos t$ and $y=2\sin t$ , where $0\leq t<2\pi$ . Eliminate $t$ to find a Cartesian equation of the curve.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The curve $y=(x-1)^2+2$ is parameterised so that $x=2t+1$ . Express $y$ in terms of $t$ and find the parameter values where the curve meets the line $y=6$ .
(4)
(Total for Question 1 is 4 marks)
2.
A curve $C$ is defined by $x=t^2+2$ and $y=3t-1$ for $t\geq0$ . Obtain a Cartesian equation with its restriction. Hence determine exactly where $C$ meets $x=6$ .
(4)
(Total for Question 2 is 4 marks)
3.
The curve $C$ has parametric equations $x=3e^t$ and $y=2e^{-t}$ , where $t$ is real. Find a Cartesian equation for $C$ in the form $xy=k$ and state the restrictions on $x$ and $y$ . Find the exact point on $C$ for which $t=\ln 4$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
The curve $x=t+1$ , $y=t^2-2t$ meets the line $y=3x-6$ at two points. Find the exact coordinates of both points.
(5)
(Total for Question 1 is 5 marks)
2.
The Cartesian curve $C$ has equation $y^2=4x+8$ . By setting $y=2t$ , obtain parametric equations for $C$ . The line $y=x+1$ meets $C$ at two points. Find the corresponding exact values of $t$ and the exact coordinates of both points.
(5)
(Total for Question 2 is 5 marks)
3.
For $-2\leq t\leq2$ , a point $P$ has coordinates $x=t^2-1$ and $y=t^3-t$ . Show that its path satisfies $y^2=x^2(x+1)$ . State the range of $x$ and identify all parameter values that produce the self-intersection $(0,0)$ .
(5)
(Total for Question 3 is 5 marks)
4.
A curve is defined for real $t$ by $x=\dfrac{1-t^2}{1+t^2}$ and $y=\dfrac{2t}{1+t^2}$ . Show that every point generated lies on the circle $x^2+y^2=1$ . Identify the point on this circle that is not generated by any real value of $t$ . Find all values of $t$ for which the generated point lies on $y=x$ , and give the corresponding exact coordinates.
(6)
(Total for Question 4 is 6 marks)
5.
For $t>0$ , a point $P$ is defined by $x=t+\dfrac1t$ and $y=t-\dfrac1t$ . Obtain a Cartesian equation for the path of $P$ and state the restriction on $x$ . Prove that the least possible value of $x$ is $2$ and give the point where it occurs. Find also the exact point where the path meets the line $x+y=6$ .
(5)
(Total for Question 5 is 5 marks)

3.4 · Use parametric equations in modelling in a variety of contexts.

Explanation

In a parametric model, the parameter often represents time and the pair $(x(t),y(t))$ gives the object's position at that time.
Translate a contextual event into equations: meeting a boundary fixes one coordinate, while a collision requires both coordinates of two objects to agree at the same time.
Eliminating the parameter can reveal the path, but the parameter range still controls the physically modelled section of that path.
Check units, permitted times and whether a calculated event occurs within the model's domain.
A common error is to find where two paths cross without checking that the objects arrive there simultaneously.

Worked example

An arch is modelled by $x=6t$ , $y=12t-3t^2$ for $0\le t\le4$ , with $x$ and $y$ in metres. Find a Cartesian equation for the arch and its horizontal span.

1.Since $x=6t$ , $t=x/6$ .
2.Substitute into $y$ : $y=12(x/6)-3(x/6)^2=2x-x^2/12$ .
3.The parameter interval gives $0\le x\le24$ .
4.The arch meets ground level at its two endpoints, so its horizontal span is $24-0=24$ m.

Answer: $y=2x-\dfrac{x^2}{12}$ for $0\le x\le24$ ; Span $24$ m

Common mistakes

Don't solve for a geometric intersection but assign the two objects different parameter values at the same time.
Don't treat the parameter as a physical coordinate rather than eliminating it and interpreting the stated interval.

Exam tip

In a parametric model, connect the parameter endpoints to the physical endpoints before reporting dimensions.

Tier 1 · Easy

1.
A particle's position after $t$ seconds is modelled by $x=12t$ , $y=20t-5t^2$ , where distances are in metres. Find its position when $t=1.5$ .
(2)
(Total for Question 1 is 2 marks)
2.
A survey drone is modelled as having position $(x,y)=(2+6t,15-3t)$ at time $t$ seconds, where distances are in metres and $0\leq t\leq5$ . Find the time and position at which it reaches the line $y=6$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A robot moves with position $x=2t+1$ , $y=t^2$ for $0\leq t\leq4$ , where distances are in metres and $t$ is in seconds. Find the time at which it crosses the line $y=3x-5$ and give the exact coordinates of the crossing point.
(4)
(Total for Question 1 is 4 marks)
2.
An inspection vehicle moves with position $(x,y)=(4t,t^2-6t+10)$ for $0\leq t\leq5$ . Coordinates are measured in metres and $t$ denotes seconds. Find both crossings of the marker line $y=2$ , including their times and positions, and the elapsed time between them.
(4)
(Total for Question 2 is 4 marks)
3.
A survey vessel has position $(x,y)=(3+4t,1+3t)$ at time $t$ hours, where distances are in kilometres and $0\leq t\leq5$ . A beacon is at $(14,10)$ . Find the exact time at which the vessel is closest to the beacon and its minimum distance from the beacon.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Two particles move for $t\ge0$ seconds. Particle $P$ has position $(3t+2,t^2+2)$ and particle $Q$ has position $(12-2t,t+4)$ . Determine whether they collide and, if they do, find the time and position of the collision.
(4)
(Total for Question 1 is 4 marks)
2.
For $0\leq t\leq7$ , a drone's position is modelled by $x=2t+1$ and $y=t^2-6t+13$ , with distances in kilometres and $t$ in hours. A signal corridor is the set of points on or below the line $y=x$ . Find when the drone enters and leaves the corridor, the two boundary positions, and the total time spent in the corridor.
(5)
(Total for Question 2 is 5 marks)
3.
Warehouse robot $P$ follows the route $x=2s+1$ , $y=s+4$ , where $s$ is the number of minutes after $P$ starts. Robot $Q$ follows the route $x=11-u$ , $y=2u-1$ , where $u$ is the number of minutes after $Q$ starts. Find the intersection of the two routes and the corresponding values of $s$ and $u$ . Hence determine how much earlier or later $Q$ must start for the robots to meet there.
(5)
(Total for Question 3 is 5 marks)
4.
A tracking beacon moves with position $(x,y)=(8\cos t,6\sin t)$ at time $t$ hours, where $0\leq t\leq2\pi$ . The beacon is inside a monitored region when $x\geq0$ and $y\geq3$ . Find the complete interval of times for which the beacon is inside the region. Give the exact entry and exit positions and the fraction of the modelled time spent inside the region.
(6)
(Total for Question 4 is 6 marks)
5.
During a six-hour survey, a craft follows $(x,y)=(-6+2t,-8+3t)$ , with $t$ measured in hours and coordinates in kilometres ( $0\leq t\leq6$ ). It can communicate with a relay at the origin while it is at most $5$ km from the relay. Find the exact times at which communication starts and ends, and hence the total communication time. Eliminate $t$ to find a Cartesian equation of the craft's route, stating the restriction on $x$ .
(7)
(Total for Question 5 is 7 marks)

Answer key

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

3.1 · Understand and use the equation of a straight line, including the forms y − y1 = m(x − x1) and ax + by + c = 0; gradient conditions for two straight lines to be parallel or perpendicular; use straight line models in a variety of contexts.

Tier 1 · Easy

Mark scheme for 3.1 Tier 1 · Easy
Question	Scheme	Marks
1	$y=-3x+10$	2
(2 marks)		2
Notes
Use point-gradient form: $y-4=-3(x-2)$ . Expanding gives $y-4=-3x+6$ , so $y=-3x+10$ .
2	$2x+y-3=0$	3
(3 marks)		3
Notes
The gradient of $AB$ is $(-5-7)/(4-(-2))=-12/6=-2$ . Using point $A$ , $y-7=-2(x+2)$ , so $y=-2x+3$ . Rearranging gives $2x+y-3=0$ .

Tier 2 · Standard

Mark scheme for 3.1 Tier 2 · Standard
Question	Scheme	Marks
1	$P=(2,5)$ $3x+y-11=0$	4
(4 marks)		4
Notes
At $P$ , $2x+1=-x+7$ , so $3x=6$ and $x=2$ ; then $y=5$ . The line $3x+y=4$ has gradient $-3$ , so a parallel line through $(2,5)$ is $y-5=-3(x-2)$ . Rearranging gives $3x+y-11=0$ .
2	$k=5$ $(1,4)$	4
(4 marks)		4
Notes
If $k=-3$ , $L_2$ is vertical but $L_1$ is not horizontal, so the lines are not perpendicular. Otherwise their gradients are $k-1$ and $-2/(k+3)$ . Perpendicularity gives $(k-1)[-2/(k+3)]=-1$ , so $2(k-1)=k+3$ and $k=5$ . The lines are then $y=4x$ and $2x+8y=34$ . Substitution gives $34x=34$ , so $x=1$ and $y=4$ .
3	$2x+y-6=0$	4
(4 marks)		4
Notes
The midpoint of $AB$ is $(1,4)$ and the gradient of $AB$ is $(6-2)/[5-(-3)]=1/2$ . Therefore the perpendicular bisector has gradient $-2$ . Its equation is $y-4=-2(x-1)$ , which rearranges to $2x+y-6=0$ .

Tier 3 · Hard

Mark scheme for 3.1 Tier 3 · Hard
Question	Scheme	Marks
1	$C=1.7d+6$ The model charges $\pounds 1.70$ per kilometre. $22$ km	5
(5 marks)		5
Notes
The gradient is $(29.80-19.60)/(14-8)=10.20/6=1.70$ , representing the charge per kilometre. Write $C=1.7d+c$ and use $(8,19.60)$ : $19.60=1.7(8)+c$ , so $c=6$ . For the budget, solve $1.7d+6\le45$ , giving $d\le390/17\approx22.94$ . The greatest permitted whole number is therefore $22$ km; $23$ km would cost $\pounds 45.10$ .
2	$p=-10$ $AB: 2x+3y-1=0$ Area $=\dfrac{1}{12}$ square units	6
(6 marks)		6
Notes
The line $3x-2y+5=0$ has gradient $3/2$ , so $AB$ has gradient $-2/3$ . Also the gradient from $A$ to $B$ is $8/(p-2)$ , hence $8/(p-2)=-2/3$ and $p=-10$ . Through $A$ , $y+1=-\dfrac{2}{3}(x-2)$ , which gives $2x+3y-1=0$ . Its positive intercepts are $P=(1/2,0)$ and $Q=(0,1/3)$ , so the area is $\tfrac12(1/2)(1/3)=1/12$ square units.
3	$Q=\left(\dfrac{52}{25},\dfrac{114}{25}\right)$ $PQ=\dfrac{41}{5}$ $R=\left(-\dfrac{71}{25},\dfrac{278}{25}\right)$	6
(6 marks)		6
Notes
The gradient of $L$ is $3/4$ , so the perpendicular through $P$ has equation $y+2=-\dfrac43(x-7)$ , or $4x+3y-22=0$ . Solving this with $3x-4y+12=0$ gives $Q=(52/25,114/25)$ . Hence $PQ=\sqrt{(123/25)^2+(164/25)^2}=41/5$ . Since reflection in $L$ makes $Q$ the midpoint of $PR$ , $R=2Q-P=(-71/25,278/25)$ .
4	$P=\left(\dfrac{10}{3},\dfrac{1}{3}\right)$ $k=-7$ Area $=\dfrac{289}{108}$ square units	6
(6 marks)		6
Notes
Rearrange the family as $k(x-y-3)+(x+2y-4)=0$ . A point on every member must satisfy $x-y-3=0$ and $x+2y-4=0$ , giving $P=(10/3,1/3)$ . Parallel lines have proportional $x$ - and $y$ -coefficients, so $(k+1)/(2-k)=2/(-3)$ . Hence $-3(k+1)=2(2-k)$ and $k=-7$ . The selected line is $6x-9y=17$ , with intercepts $(17/6,0)$ and $(0,-17/9)$ . Its triangle with the coordinate axes therefore has area $\tfrac12(17/6)(17/9)=289/108$ square units.
5	$k=9$ $D=(1,7)$ $BD: x+y-8=0$ Distance $=5\sqrt2$	6
(6 marks)		6
Notes
The diagonals of a parallelogram bisect each other, so their midpoint is the midpoint of $AC$ , namely $M=((k-3)/2,5)$ . Since $M$ lies on $2x-y=1$ , $k-3-5=1$ , giving $k=9$ . Also $A+C=B+D$ , so $D=A+C-B=(1,7)$ . The gradient of $BD$ is $(7-3)/(1-5)=-1$ , hence $BD$ has equation $y-3=-(x-5)$ , or $x+y-8=0$ . The perpendicular distance from $A(-3,1)$ to this line is $\|-3+1-8\|/\sqrt2=10/\sqrt2=5\sqrt2$ .

3.2 · Understand and use the coordinate geometry of the circle, including the equation (x − a)² + (y − b)² = r²; complete the square for centre and radius; use circle properties (semicircle angle, chord bisector, tangent-radius).

Tier 1 · Easy

Mark scheme for 3.2 Tier 1 · Easy
Question	Scheme	Marks
1	Centre $(3,-4)$ Radius $6$	3
(3 marks)		3
Notes
Complete both squares: $x^2-6x=(x-3)^2-9$ and $y^2+8y=(y+4)^2-16$ . The equation becomes $(x-3)^2+(y+4)^2=36$ , so the centre is $(3,-4)$ and the radius is $6$ .
2	$(x+3)^2+(y-4)^2=25$	2
(2 marks)		2
Notes
The squared radius is $(1+3)^2+(1-4)^2=16+9=25$ . Therefore the circle is $(x+3)^2+(y-4)^2=25$ .

Tier 2 · Standard

Mark scheme for 3.2 Tier 2 · Standard
Question	Scheme	Marks
1	$(x-2)^2+(y-3)^2=20$	4
(4 marks)		4
Notes
The centre is the midpoint of $AB$ , namely $((-2+6)/2,(1+5)/2)=(2,3)$ . The squared radius is the squared distance from the centre to $A$ : $(-2-2)^2+(1-3)^2=16+4=20$ . Hence the circle is $(x-2)^2+(y-3)^2=20$ .
2	$(-6,0)$ and $(2,0)$ $4x-3y-8=0$	5
(5 marks)		5
Notes
On the $x$ -axis, $y=0$ , so $x^2+4x-12=0=(x+6)(x-2)$ , giving $(-6,0)$ and $(2,0)$ . Completing the square gives $(x+2)^2+(y-3)^2=25$ , so the centre is $(-2,3)$ . The radius to $(2,0)$ has gradient $-3/4$ , hence the tangent gradient is $4/3$ . Thus $y=\dfrac43(x-2)$ , or $4x-3y-8=0$ .
3	$p=-2$ or $p=4$	4
(4 marks)		4
Notes
The angle $ACB$ is $90^\circ$ , so $AC$ and $BC$ are perpendicular. The cases $p=-4$ and $p=6$ give one vertical side but the other side is not horizontal, so neither works. Otherwise the gradients are $4/(p+4)$ and $-4/(6-p)$ . Hence $-16/[(p+4)(6-p)]=-1$ , so $(p+4)(6-p)=16$ . This simplifies to $p^2-2p-8=0=(p+2)(p-4)$ , giving $p=-2$ or $p=4$ .

Tier 3 · Hard

Mark scheme for 3.2 Tier 3 · Hard
Question	Scheme	Marks
1	$(x-2)^2+(y-3)^2=10$	5
(5 marks)		5
Notes
The midpoint of horizontal chord $AB$ is $(2,2)$ , so its perpendicular bisector is $x=2$ . The midpoint of $AC$ is $(0,4)$ and the gradient of $AC$ is $2$ , so its perpendicular bisector is $y-4=-\dfrac{1}{2}x$ . At $x=2$ this gives $y=3$ , so the centre is $(2,3)$ . The squared radius is $(-1-2)^2+(2-3)^2=10$ , hence the circumcircle is $(x-2)^2+(y-3)^2=10$ .
2	$5x-12y-87=0$ , touching at $\left(\dfrac{51}{13},-\dfrac{73}{13}\right)$ $5x+12y-63=0$ , touching at $\left(\dfrac{51}{13},\dfrac{47}{13}\right)$	7
(7 marks)		7
Notes
Let a tangent through $P$ have equation $y+1=m(x-15)$ . Its perpendicular distance from the centre $(2,-1)$ must be $5$ , so $\|13m\|/\sqrt{m^2+1}=5$ . Hence $169m^2=25m^2+25$ , giving $m=\pm5/12$ . The two tangent equations are therefore $5x-12y-87=0$ and $5x+12y-63=0$ . The point of contact is the foot of the perpendicular from the centre. Solving each tangent together with its perpendicular through $(2,-1)$ gives respectively $\left(51/13,-73/13\right)$ and $\left(51/13,47/13\right)$ .
3	$x=3$ $(3,4)$ and $(3,-4)$ $(x-3)^2+y^2=16$	6
(6 marks)		6
Notes
Subtracting the two circle equations gives $x^2-(x-6)^2=0$ , hence $12x-36=0$ and the common chord is $x=3$ . Substitution into $x^2+y^2=25$ gives $9+y^2=25$ , so its endpoints are $(3,4)$ and $(3,-4)$ . Their midpoint is $(3,0)$ and half their separation is $4$ , so the circle with this chord as diameter is $(x-3)^2+y^2=16$ .
4	$(x-r)^2+(y-r)^2=r^2$ , where $r=14-4\sqrt6$ or $r=14+4\sqrt6$	5
(5 marks)		5
Notes
Tangency to both positive coordinate axes means that the centre is $(r,r)$ and the radius is $r>0$ . Since $(6,8)$ lies on the circle, $(6-r)^2+(8-r)^2=r^2$ . This simplifies to $r^2-28r+100=0$ , so $r=[28\pm\sqrt{384}]/2=14\pm4\sqrt6$ . Both values are positive, so both give valid circles. Substituting either exact value for $r$ in $(x-r)^2+(y-r)^2=r^2$ gives the two equations.
5	$c=5+10\sqrt2$ , midpoint $\left(-1-4\sqrt2,3+2\sqrt2\right)$ $c=5-10\sqrt2$ , midpoint $\left(-1+4\sqrt2,3-2\sqrt2\right)$	6
(6 marks)		6
Notes
The circle has centre $(-1,3)$ and radius $7$ . If $d$ is the perpendicular distance from the centre to the chord, then $d^2+3^2=7^2$ , so $d=2\sqrt{10}$ . Writing the line as $2x-y+c=0$ , its distance from the centre is $\|c-5\|/\sqrt5$ . Thus $\|c-5\|/\sqrt5=2\sqrt{10}$ , giving $c=5\pm10\sqrt2$ . The chord midpoint is the foot of the perpendicular from $(-1,3)$ to the line. If $q=c-5$ , this foot is $(-1-2q/5,3+q/5)$ , which gives the two stated midpoints.

3.3 · Understand and use the parametric equations of curves and conversion between Cartesian and parametric forms.

Tier 1 · Easy

Mark scheme for 3.3 Tier 1 · Easy
Question	Scheme	Marks
1	$y=\dfrac{x}{2}+\dfrac{9}{2}$	2
(2 marks)		2
Notes
From $x=2t-1$ , $t=(x+1)/2$ . Substitute into $y=t+4$ to get $y=(x+1)/2+4=x/2+9/2$ .
2	$\dfrac{x^2}{25}+\dfrac{y^2}{4}=1$	2
(2 marks)		2
Notes
From the parametric equations, $\cos t=x/5$ and $\sin t=y/2$ . Using $\cos^2t+\sin^2t=1$ gives $x^2/25+y^2/4=1$ .

Tier 2 · Standard

Mark scheme for 3.3 Tier 2 · Standard
Question	Scheme	Marks
1	$y=4t^2+2$ $t=-1$ or $t=1$	4
(4 marks)		4
Notes
Substitute $x=2t+1$ into the Cartesian equation: $y=((2t+1)-1)^2+2=4t^2+2$ . At an intersection with $y=6$ , $4t^2+2=6$ , so $t^2=1$ and therefore $t=-1$ or $t=1$ .
2	$x=\dfrac{(y+1)^2}{9}+2$ , with $y\geq-1$ $(6,5)$	4
(4 marks)		4
Notes
From $y=3t-1$ , $t=(y+1)/3$ . Substitution into $x=t^2+2$ gives $x=(y+1)^2/9+2$ . Since $t\geq0$ , the restriction is $y\geq-1$ . When $x=6$ , $t^2+2=6$ , so the permitted value is $t=2$ , giving $y=5$ .
3	$xy=6$ , with $x>0$ and $y>0$ $(12,\dfrac12)$	4
(4 marks)		4
Notes
Multiplying the parametric equations gives $xy=(3e^t)(2e^{-t})=6$ . Since exponentials are positive for every real $t$ , the traced branch has $x>0$ and $y>0$ . When $t=\ln4$ , $e^t=4$ and $e^{-t}=1/4$ , so the point is $(12,1/2)$ .

Tier 3 · Hard

Mark scheme for 3.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\left(\dfrac{7+\sqrt{13}}{2},\dfrac{9+3\sqrt{13}}{2}\right)$ $\left(\dfrac{7-\sqrt{13}}{2},\dfrac{9-3\sqrt{13}}{2}\right)$	5
(5 marks)		5
Notes
At an intersection, $t^2-2t=3(t+1)-6$ , so $t^2-5t+3=0$ . Hence $t=(5\pm\sqrt{13})/2$ . Using $x=t+1$ gives $x=(7\pm\sqrt{13})/2$ , and then $y=3x-6=(9\pm3\sqrt{13})/2$ , with matching signs.
2	$x=t^2-2$ , $y=2t$ $t=1+\sqrt2$ gives $(1+2\sqrt2,2+2\sqrt2)$ $t=1-\sqrt2$ gives $(1-2\sqrt2,2-2\sqrt2)$	5
(5 marks)		5
Notes
Putting $y=2t$ into $y^2=4x+8$ gives $4t^2=4x+8$ , so $x=t^2-2$ . At an intersection with $y=x+1$ , $2t=t^2-1$ , hence $t^2-2t-1=0$ and $t=1\pm\sqrt2$ . Since $x=y-1=2t-1$ , matching the signs gives the two stated points.
3	$y^2=x^2(x+1)$ $-1\leq x\leq3$ $t=-1$ and $t=1$	5
(5 marks)		5
Notes
Since $y=t(t^2-1)=tx$ , squaring gives $y^2=t^2x^2$ . Also $t^2=x+1$ , so $y^2=x^2(x+1)$ . Over $-2\leq t\leq2$ , $0\leq t^2\leq4$ , hence $-1\leq x\leq3$ . The point $(0,0)$ requires $x=t^2-1=0$ , so $t=-1$ or $t=1$ ; both values also give $y=0$ .
4	$x^2+y^2=1$ $(-1,0)$ is not generated $t=-1+\sqrt2$ gives $\left(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2}\right)$ $t=-1-\sqrt2$ gives $\left(-\dfrac{\sqrt2}{2},-\dfrac{\sqrt2}{2}\right)$	6
(6 marks)		6
Notes
Substitution gives $x^2+y^2=[(1-t^2)^2+4t^2]/(1+t^2)^2=(1+t^2)^2/(1+t^2)^2=1$ . The point $(-1,0)$ would require $(1-t^2)/(1+t^2)=-1$ , which leads to $1=-1$ , so no real $t$ generates it. On $y=x$ , $2t=1-t^2$ , hence $t^2+2t-1=0$ and $t=-1\pm\sqrt2$ . The line and circle give $2x^2=1$ ; matching the sign obtained from each parameter gives the two stated points.
5	$x^2-y^2=4$ , with $x\geq2$ Least $x=2$ at $(2,0)$ Intersection $\left(\dfrac{10}{3},\dfrac{8}{3}\right)$	5
(5 marks)		5
Notes
Since $x+y=2t$ and $x-y=2/t$ , multiplication gives $x^2-y^2=4$ . For $t>0$ , $(t-1)^2\geq0$ gives $t+1/t\geq2$ , so $x\geq2$ , with equality only at $t=1$ . This gives the point $(2,0)$ . At the intersection with $x+y=6$ , the relation $x+y=2t$ gives $t=3$ . Therefore $x=3+1/3=10/3$ and $y=3-1/3=8/3$ .

3.4 · Use parametric equations in modelling in a variety of contexts.

Tier 1 · Easy

Mark scheme for 3.4 Tier 1 · Easy
Question	Scheme	Marks
1	$(18,18.75)$ metres	2
(2 marks)		2
Notes
Substitute $t=1.5$ : $x=12(1.5)=18$ and $y=20(1.5)-5(1.5)^2=30-11.25=18.75$ .
2	$t=3$ seconds $(20,6)$ metres	3
(3 marks)		3
Notes
Set $15-3t=6$ , giving $t=3$ , which lies in the modelled interval. Then $x=2+6(3)=20$ , so the position is $(20,6)$ metres.

Tier 2 · Standard

Mark scheme for 3.4 Tier 2 · Standard
Question	Scheme	Marks
1	$t=3-\sqrt7$ seconds Crossing point $\left(7-2\sqrt7,\,16-6\sqrt7\right)$	4
(4 marks)		4
Notes
At a crossing, $t^2=3(2t+1)-5$ , so $t^2-6t+2=0$ and $t=3\pm\sqrt7$ . Only $3-\sqrt7$ lies in the modelled interval $0\leq t\leq4$ . Substitution gives $x=2(3-\sqrt7)+1=7-2\sqrt7$ and $y=(3-\sqrt7)^2=16-6\sqrt7$ .
2	$t=2$ and $t=4$ seconds $(8,2)$ and $(16,2)$ $2$ seconds	4
(4 marks)		4
Notes
At the marker line, $t^2-6t+10=2$ , so $t^2-6t+8=0=(t-2)(t-4)$ . Both values lie in the modelled interval. Since $x=4t$ , the positions are $(8,2)$ and $(16,2)$ , separated by $4-2=2$ seconds.
3	$t=\dfrac{71}{25}$ hours Minimum distance $=\dfrac35$ km	5
(5 marks)		5
Notes
The squared distance from the beacon is $D^2=(4t-11)^2+(3t-9)^2=25t^2-142t+202$ . Completing the square gives $D^2=25(t-71/25)^2+9/25$ . Since $71/25$ lies in the modelled interval, the minimum occurs then and is $D=\sqrt{9/25}=3/5$ km.

Tier 3 · Hard

Mark scheme for 3.4 Tier 3 · Hard
Question	Scheme	Marks
1	They collide at $t=2$ seconds at $(8,6)$ .	4
(4 marks)		4
Notes
Equal $x$ -coordinates require $3t+2=12-2t$ , so $5t=10$ and $t=2$ . Check the other coordinate at this same time: for $P$ , $y=2^2+2=6$ ; for $Q$ , $y=2+4=6$ . Both positions are therefore $(8,6)$ at $t=2$ , so a collision occurs. The linear $x$ equation has only one solution, so there is no other collision.
2	Enters at $t=2$ hours and leaves at $t=6$ hours $(5,5)$ and $(13,13)$ $4$ hours	5
(5 marks)		5
Notes
Inside the corridor, $t^2-6t+13\leq2t+1$ , so $t^2-8t+12\leq0$ . Since $(t-2)(t-6)\leq0$ , this holds for $2\leq t\leq6$ , within the modelled interval. The boundary positions are $(5,5)$ at $t=2$ and $(13,13)$ at $t=6$ . The time inside is $6-2=4$ hours.
3	$s=3$ , $u=4$ Intersection $(7,7)$ $Q$ must start $1$ minute before $P$	5
(5 marks)		5
Notes
At the route intersection, $2s+1=11-u$ and $s+4=2u-1$ . These equations are $2s+u=10$ and $s-2u=-5$ , giving $u=4$ and $s=3$ . Substitution gives the common position $(7,7)$ . Robot $Q$ needs $4$ minutes to reach it while $P$ needs $3$ minutes, so $Q$ must start $1$ minute before $P$ for them to arrive together.
4	$\dfrac{\pi}{6}\leq t\leq\dfrac{\pi}{2}$ Entry $\left(4\sqrt3,3\right)$ ; exit $(0,6)$ Fraction $=\dfrac16$	6
(6 marks)		6
Notes
The condition $y\geq3$ gives $6\sin t\geq3$ , so $\sin t\geq1/2$ and therefore $\pi/6\leq t\leq5\pi/6$ on the stated interval. The condition $x\geq0$ gives $\cos t\geq0$ , so $0\leq t\leq\pi/2$ or $3\pi/2\leq t\leq2\pi$ . Intersecting these sets gives $\pi/6\leq t\leq\pi/2$ . Substitution gives entry position $(4\sqrt3,3)$ and exit position $(0,6)$ . The duration is $\pi/3$ hours out of $2\pi$ hours, so the fraction is $1/6$ .
5	Communication starts at $t=\dfrac{36-\sqrt{321}}{13}$ and ends at $t=\dfrac{36+\sqrt{321}}{13}$ Total time $=\dfrac{2\sqrt{321}}{13}$ hours Route: $3x-2y+2=0$ , where $-6\leq x\leq6$	7
(7 marks)		7
Notes
The squared distance from the relay is $D^2=(-6+2t)^2+(-8+3t)^2=13t^2-72t+100$ . Communication is possible when $D^2\leq25$ , so $13t^2-72t+75\leq0$ . The boundary roots are $t=(36\pm\sqrt{321})/13$ , both in the modelled interval, and the inequality holds between them. Their difference is $2\sqrt{321}/13$ hours. From $x=-6+2t$ , $t=(x+6)/2$ . Substitution into $y=-8+3t$ gives $y=3x/2+1$ , or $3x-2y+2=0$ . As $0\leq t\leq6$ , the corresponding restriction is $-6\leq x\leq6$ .