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Cambridge International AS & A Level Mathematics 9709 · Pure Mathematics 1

Coordinate geometry

Gradients, distances and midpoints; the equation of a straight line in its three forms; parallel and perpendicular lines; the equation of a circle and its expanded form; tangents and other circle geometry; and intersections of lines with curves, using the discriminant.

1.3Paper 1: Pure Mathematics 1Free lesson

What the syllabus says

1.3 (Paper 1)

find the equation of a straight line given sufficient information
interpret and use any of the forms $y = m x + c$ , $y - y_{1} = m (x - x_{1})$ , $a x + b y + c = 0$ in solving problems
understand that the equation $(x - a)^{2} + (y - b)^{2} = r^{2}$ represents the circle with centre $(a, b)$ and radius $r$
use algebraic methods to solve problems involving lines and circles
understand the relationship between a graph and its associated algebraic equation, and use the relationship between points of intersection of graphs and solutions of equations.

Notes and examples

e.g. given two points, or one point and the gradient.
Including calculations of distances, gradients, midpoints, points of intersection and use of the relationship between the gradients of parallel and perpendicular lines.
Including use of the expanded form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
Including use of elementary geometrical properties of circles, e.g. tangent perpendicular to radius, angle in a semicircle, symmetry.
Implicit differentiation is not included.
e.g. to determine the set of values of $k$ for which the line $y = x + k$ intersects, touches or does not meet a quadratic curve.

Quoted from the Cambridge International AS and A Level Mathematics 9709 syllabus (2026-2027 and 2028-2030 editions, which share this content).

Gradients, distances and midpoints Paper 1

For two points $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ :

the gradient of $A B$ is $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , the change in $y$ for each unit increase in $x$ ;
the distance $A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ , which is Pythagoras' theorem on the right-angled triangle with $A B$ as its hypotenuse;
the midpoint of $A B$ is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ , the average of the coordinates.

Parallel lines have equal gradients. Perpendicular lines have gradients $m_{1}$ and $m_{2}$ with $m_{1} m_{2} = - 1$ , so the perpendicular gradient is the negative reciprocal: a line of gradient $\frac{2}{3}$ is perpendicular to one of gradient $- \frac{3}{2}$ .

Why $m_{1} m_{2} = - 1$ : rotating a line through $90^{°}$ turns a step of $(1, m)$ along it into a step of $(- m, 1)$ , whose gradient is $\frac{1}{- m} = - \frac{1}{m}$ .

Draw a quick sketch for every coordinate question. It catches sign errors and often shows which of two answers the question wants.

The equation of a straight line Paper 1

A straight line can be written in three standard forms, and you should move between them freely.

$y = m x + c$ : gradient $m$ and $y$ -intercept $(0, c)$ . Best for reading off the gradient.
$y - y_{1} = m (x - x_{1})$ : the line through $(x_{1}, y_{1})$ with gradient $m$ . Best for finding an equation, because it uses exactly the information you usually have.
$a x + b y + c = 0$ : the general form. Questions often ask for it with integer coefficients: multiply through to clear fractions. To find its gradient, rearrange to $y = - \frac{a}{b} x - \frac{c}{b}$ , so $m = - \frac{a}{b}$ .

Finding the equation of a line needs "sufficient information": either a point and the gradient, or two points (from which you work out the gradient first).

Where two lines meet, both equations hold, so solve them simultaneously. Where a line meets an axis, put $y = 0$ (for the $x$ -axis) or $x = 0$ (for the $y$ -axis).

A very common task is the perpendicular bisector of $A B$ : the line through the midpoint of $A B$ that is perpendicular to $A B$ . Every point on it is the same distance from $A$ as from $B$ , which is why it appears in circle questions: the centre of any circle through $A$ and $B$ lies on it.

The shortest distance from a point to a line is along the perpendicular. Find the perpendicular line through the point, find where it meets the original line (the foot of the perpendicular), and then find the distance between the two points.

Worked example 1 Paper 1

The points $A$ and $B$ have coordinates $(- 2, 5)$ and $(4, 1)$ . Find the equation of the perpendicular bisector of $A B$ , giving your answer in the form $a x + b y + c = 0$ , where $a$ , $b$ and $c$ are integers.

Gradient of $A B$ : $\frac{1 - 5}{4 - (- 2)} = \frac{- 4}{6} = - \frac{2}{3}$ .
The perpendicular gradient is the negative reciprocal: $\frac{3}{2}$ .
Midpoint of $A B$ : $(\frac{- 2 + 4}{2}, \frac{5 + 1}{2}) = (1, 3)$ .
Equation: $y - 3 = \frac{3}{2} (x - 1)$ .
Multiply by 2: $2 y - 6 = 3 x - 3$ , so $3 x - 2 y + 3 = 0$ .
Check: the midpoint $(1, 3)$ gives $3 - 6 + 3 = 0$ . Correct.

Answer: $3 x - 2 y + 3 = 0$

Check yourself Paper 1

Find the equation of the line through $(3, - 1)$ that is perpendicular to the line $2 x + 5 y = 7$ . Give your answer in the form $y = m x + c$ .

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The equation of a circle Paper 1

A point $(x, y)$ lies on the circle with centre $(a, b)$ and radius $r$ exactly when its distance from the centre is $r$ . By the distance formula, $\sqrt{(x - a)^{2} + (y - b)^{2}} = r$ , and squaring gives the equation of the circle:

$(x - a)^{2} + (y - b)^{2} = r^{2}$

For example, $(x - 3)^{2} + (y + 1)^{2} = 16$ is the circle with centre $(3, - 1)$ and radius 4. Note the signs: $(y + 1)$ means $b = - 1$ .

The expanded form. Multiplying out gives an equation of the form

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

To find the centre and radius from this, complete the square in $x$ and in $y$ : $(x + g)^{2} + (y + f)^{2} = g^{2} + f^{2} - c$ . So the centre is $(- g, - f)$ and the radius is $\sqrt{g^{2} + f^{2} - c}$ . Completing the square is safer than remembering the formula, and examiners accept either.

An equation is only a circle if the coefficients of $x^{2}$ and $y^{2}$ are equal (divide through if they are both 2, say), there is no $x y$ term, and $g^{2} + f^{2} - c > 0$ .

Is a point inside, on or outside? Substitute it into $(x - a)^{2} + (y - b)^{2}$ and compare the result with $r^{2}$ .

Worked example 2 Paper 1

Find the centre and radius of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .

Group the $x$ terms and the $y$ terms: $(x^{2} - 6 x) + (y^{2} + 4 y) = 12$ .
Complete each square: $(x - 3)^{2} - 9 + (y + 2)^{2} - 4 = 12$ .
So $(x - 3)^{2} + (y + 2)^{2} = 25$ .
The centre is $(3, - 2)$ and the radius is $\sqrt{25} = 5$ .
Check with the formula: $g = - 3$ , $f = 2$ , $c = - 12$ , so the centre is $(3, - 2)$ and $r = \sqrt{9 + 4 + 12} = 5$ .

Answer: centre $(3, - 2)$ , radius 5

Check yourself Paper 1

Find the centre and the radius of the circle $x^{2} + y^{2} + 8 x - 2 y + 8 = 0$ .

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Using circle geometry Paper 1

The circle theorems from IGCSE turn into algebra here, and questions expect you to use them.

A tangent is perpendicular to the radius at the point of contact. To find the tangent at $P$ , find the gradient of the radius $C P$ from the centre $C$ , take its negative reciprocal, and use $y - y_{1} = m (x - x_{1})$ at $P$ .
The angle in a semicircle is a right angle. If $A B$ is a diameter and $P$ is any other point on the circle, then angle $A P B = 90^{°}$ , so $A P$ and $B P$ are perpendicular. Conversely, if angle $A P B = 90^{°}$ , then $P$ lies on the circle with diameter $A B$ . The centre of that circle is the midpoint of $A B$ .
The perpendicular from the centre to a chord bisects the chord, so the perpendicular bisector of any chord passes through the centre. Two chords give two perpendicular bisectors, and they meet at the centre.
Symmetry. The highest and lowest points of a circle are directly above and below the centre; a point and its reflection in a line through the centre are both on the circle; the two tangents parallel to a given direction touch at the two ends of a diameter.
Tangent from a point outside. If $T$ is outside the circle and $T P$ touches the circle at $P$ , triangle $C P T$ has a right angle at $P$ , so $T P^{2} = C T^{2} - r^{2}$ .

Worked example 3 Paper 1

The circle $C$ has centre $(2, - 1)$ and passes through the point $P (5, 3)$ . Find the radius of the circle and the equation of the tangent to the circle at $P$ .

Radius: $C P = \sqrt{(5 - 2)^{2} + (3 - (- 1))^{2}} = \sqrt{9 + 16} = 5$ .
Gradient of the radius from the centre to $P$ : $\frac{3 - (- 1)}{5 - 2} = \frac{4}{3}$ .
The tangent is perpendicular to this radius, so its gradient is $- \frac{3}{4}$ .
Tangent: $y - 3 = - \frac{3}{4} (x - 5)$ . Multiply by 4: $4 y - 12 = - 3 x + 15$ .
So the tangent is $3 x + 4 y = 27$ . Check at $P$ : $15 + 12 = 27$ .

Answer: radius 5; tangent $3 x + 4 y = 27$

The radius $C P$ has gradient $\frac{4}{3}$ , so the tangent at $P$ has gradient $- \frac{3}{4}$ : the product of the two gradients is $- 1$ .

Worked example 4 Paper 1

The points $A (- 1, 4)$ and $B (5, - 2)$ are the ends of a diameter of a circle. (a) Find the equation of the circle. (b) Show that the point $P (5, 4)$ lies on the circle, and hence state the size of angle $A P B$ .

(a) The centre is the midpoint of $A B$ : $(\frac{- 1 + 5}{2}, \frac{4 + (- 2)}{2}) = (2, 1)$ .
The radius is the distance from the centre to $A$ : $\sqrt{(- 1 - 2)^{2} + (4 - 1)^{2}} = \sqrt{18}$ , so $r^{2} = 18$ .
The equation is $(x - 2)^{2} + (y - 1)^{2} = 18$ .
(b) Substitute $P (5, 4)$ : $(5 - 2)^{2} + (4 - 1)^{2} = 9 + 9 = 18$ . So $P$ is on the circle.
$A B$ is a diameter, so the angle in a semicircle gives angle $A P B = 90^{°}$ .
Check with gradients: $A P$ has gradient $\frac{4 - 4}{5 + 1} = 0$ (horizontal) and $B P$ is vertical ( $B$ and $P$ both have $x = 5$ ), so its gradient is not defined. They are perpendicular.

Answer: (a) $(x - 2)^{2} + (y - 1)^{2} = 18$ (b) angle $A P B = 90^{°}$

Intersections of lines and curves Paper 1

A point lies on a graph exactly when its coordinates satisfy the graph's equation. So the points where two graphs meet are the solutions of their two equations taken together, and the $x$ -coordinates of those points are the solutions of the single equation you get by eliminating $y$ . This works both ways:

to find where a line meets a curve, solve the equations simultaneously, usually by substituting the line into the curve;
to solve an equation such as $x^{2} - 4 x + 3 = c$ , you can think of where the curve $y = x^{2} - 4 x + 3$ meets the horizontal line $y = c$ .

How many times does a line meet a curve? After substituting, you have a quadratic $a x^{2} + b x + c = 0$ , and its discriminant decides the answer:

$b^{2} - 4 a c > 0$ : the line intersects the curve at two points;
$b^{2} - 4 a c = 0$ : the line touches the curve: it is a tangent, meeting it at one point;
$b^{2} - 4 a c < 0$ : the line does not meet the curve.

This is how to find the values of $k$ for which $y = x + k$ is a tangent to a parabola or a circle without any calculus. (Implicit differentiation is not part of this syllabus, and it is never needed for tangents to circles: use the radius instead.)

For a circle, the alternative is geometric: a line is a tangent exactly when its perpendicular distance from the centre equals the radius.

Worked example 5 (exam style) Paper 1

The line $y = x + k$ , where $k$ is a constant, and the curve $y = x^{2} - 3 x + 6$ are given.

(a) Find the value of $k$ for which the line is a tangent to the curve, and find the coordinates of the point of contact. [4]

(b) Find the set of values of $k$ for which the line meets the curve at two distinct points. [1]

(a) At an intersection, $x^{2} - 3 x + 6 = x + k$ , so $x^{2} - 4 x + (6 - k) = 0$ .
A tangent meets the curve at one point, so the discriminant is 0: $(- 4)^{2} - 4 (1) (6 - k) = 0$ .
$16 - 24 + 4 k = 0$ , so $4 k = 8$ and $k = 2$ .
With $k = 2$ : $x^{2} - 4 x + 4 = 0$ , so $(x - 2)^{2} = 0$ and $x = 2$ . Then $y = 2 + 2 = 4$ .
The point of contact is $(2, 4)$ . Check on the curve: $4 - 6 + 6 = 4$ .
(b) Two distinct points need the discriminant positive: $4 k - 8 > 0$ , so $k > 2$ .

Answer: (a) $k = 2$ , point $(2, 4)$ (b) $k > 2$

Check yourself Paper 1

Find the values of the constant $m$ for which the line $y = m x - 1$ is a tangent to the curve $y = x^{2} + 3$ .

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Check yourself Paper 1

The line $y = x + 2$ meets the circle $x^{2} + y^{2} = 10$ at two points. Find their coordinates.

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Key points Paper 1

Gradient $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , distance by Pythagoras, midpoint by averaging.
Parallel lines: equal gradients. Perpendicular lines: $m_{1} m_{2} = - 1$ .
Use $y - y_{1} = m (x - x_{1})$ to find a line; rearrange to $y = m x + c$ or $a x + b y + c = 0$ as asked.
$(x - a)^{2} + (y - b)^{2} = r^{2}$ has centre $(a, b)$ and radius $r$ . For $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , complete the square: centre $(- g, - f)$ , radius $\sqrt{g^{2} + f^{2} - c}$ .
Tangent perpendicular to radius; angle in a semicircle is $90^{°}$ ; the perpendicular bisector of a chord passes through the centre.
Intersections: substitute and solve. The discriminant of the resulting quadratic tells you whether a line cuts, touches or misses a curve.

Where marks are lost Paper 1

Using the reciprocal instead of the negative reciprocal for a perpendicular gradient.
Getting the signs of the centre wrong: $(x + 3)^{2} + (y - 2)^{2} = 9$ has centre $(- 3, 2)$ .
Giving $r^{2}$ as the radius, or forgetting to square the radius when writing the equation.
Writing the tangent with the gradient of the radius, instead of the perpendicular gradient.
Leaving fractions in when asked for $a x + b y + c = 0$ with integer coefficients.
After substituting a line into a curve, forgetting to find the second coordinate of each point, or pairing them wrongly.
Using $b^{2} - 4 a c > 0$ for a tangent, or mixing up the $a$ , $b$ , $c$ of the quadratic with the constants of the line.

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An original lesson written to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026 to 2027 and 2028 to 2030). The worked examples and questions are our own, not Cambridge questions. Back to every AS & A Level Mathematics 9709 lesson.