A-Level Mathematics – Chapter 2: Coordinate Geometry
Coordinate geometry brings together algebra and geometry by representing shapes, lines, and curves using equations in the $xy$-plane. In this chapter you will extend your GCSE knowledge of straight lines to include circles and, in Year 2, parametric equations. These ideas underpin calculus, mechanics, and statistics, so a confident understanding here pays dividends across the whole course.
Specification Note
Content labelled Year 2 is A2-level only and is not required for AS Mathematics. All other content is required for both AS and the full A-Level.
2.1 Straight Lines
Gradient and the Equation of a Line
Definition — Gradient
The gradient of a straight line passing through two distinct points $(x_1,\, y_1)$ and $(x_2,\, y_2)$ is
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
A positive gradient means the line rises from left to right; a negative gradient means it falls. A horizontal line has gradient $0$; a vertical line has undefined gradient.
Once you know the gradient $m$ and any single point $(x_1,\, y_1)$ on the line, the equation is given by the point–slope form:
$$y - y_1 = m(x - x_1)$$
Rearranging gives the slope–intercept form $y = mx + c$, where $c$ is the $y$-intercept, or the general form $ax + by + c = 0$.
Figure 2.1 — A straight line $y = mx + c$ with gradient $m$ and $y$-intercept $c$. Note how the line pivots about the $y$-axis as $m$ changes.
Parallel and Perpendicular Lines
Parallel and Perpendicular Gradients
Parallel lines have equal gradients: $m_1 = m_2$.
Perpendicular lines have gradients whose product is $-1$:
$$m_1 \times m_2 = -1 \quad \Longleftrightarrow \quad m_2 = -\frac{1}{m_1}$$
A line with gradient $\frac{p}{q}$ is perpendicular to any line with gradient $-\frac{q}{p}$ (the negative reciprocal).
Example 2.1 — Gradient between two points
Find the gradient of the line passing through $A(1,\, 3)$ and $B(4,\, 9)$.
Step 1 Apply the gradient formula:
$$m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$$
The gradient is $\mathbf{2}$.
Example 2.2 — Equation of a line through two points
Find the equation of the line through $P(2,\,-1)$ and $Q(5,\, 5)$, giving your answer in the form $ax + by + c = 0$ where $a$, $b$, $c$ are integers.
Step 1 Gradient: $m = \dfrac{5 - (-1)}{5 - 2} = \dfrac{6}{3} = 2$.
Step 2 Point–slope form using $P(2,\,-1)$:
$$y - (-1) = 2(x - 2) \implies y + 1 = 2x - 4 \implies \boxed{2x - y - 5 = 0}$$
Example 2.3 — Perpendicular lines
Line $\ell$ has equation $y = 3x + 1$. Find the equation of the line perpendicular to $\ell$ that passes through $(6,\, 2)$.
Step 1 Gradient of $\ell$ is $3$. Perpendicular gradient: $-\dfrac{1}{3}$.
Step 2 Equation through $(6,\, 2)$:
$$y - 2 = -\frac{1}{3}(x - 6) \implies y = -\frac{1}{3}x + 2 + 2 \implies \boxed{y = -\frac{1}{3}x + 4}$$
Example 2.4 — Perpendicular bisector
Points $A(-2,\, 1)$ and $B(4,\, 7)$ are the endpoints of a line segment. Find the equation of the perpendicular bisector of $AB$.
Step 1 Midpoint: $M = \left(\dfrac{-2+4}{2},\; \dfrac{1+7}{2}\right) = (1,\, 4)$.
Step 2 Gradient of $AB$: $m = \dfrac{7-1}{4-(-2)} = 1$. Perpendicular gradient: $-1$.
Step 3 Equation through $M(1,\, 4)$ with gradient $-1$:
$$y - 4 = -(x - 1) \implies \boxed{y = -x + 5}$$
Distance Between Two Points
Definition — Distance Formula
The distance between $(x_1,\, y_1)$ and $(x_2,\, y_2)$ is
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
This is Pythagoras’ theorem applied to the horizontal separation $|x_2 - x_1|$ and the vertical separation $|y_2 - y_1|$.
Example 2.5 — Using the distance formula
The points $C(0,\, 2)$ and $D(k,\, 6)$ satisfy $CD = \sqrt{20}$. Find the two possible values of $k$.
Step 1 Set up the equation:
$$\sqrt{k^2 + (6-2)^2} = \sqrt{20} \implies k^2 + 16 = 20$$
Step 2 $k^2 = 4 \implies k = \pm 2$.
Exam Tip — Straight Lines
Always check which form the mark scheme expects: $y = mx + c$, $y - y_1 = m(x-x_1)$, or $ax + by + c = 0$. Converting costs only seconds, but omitting it can cost marks. Also, verify perpendicular gradients by multiplying them — they must equal $-1$ exactly.
2.2 Circles
The Equation of a Circle
Definition — Standard Form of a Circle
A circle with centre $(a,\, b)$ and radius $r > 0$ has equation
$$(x - a)^2 + (y - b)^2 = r^2$$
A circle centred at the origin simplifies to $x^2 + y^2 = r^2$. The equation encodes the fact that every point on the circle is exactly $r$ units from the centre.
Circles are also often given in general form: $x^2 + y^2 + 2fx + 2gy + c = 0$. To recover the centre and radius, complete the square in $x$ and in $y$ separately.
Figure 2.2 — A circle with centre $(4,\, 2)$, radius $3$. The dashed radius runs to the point $P$ on the circumference, and the tangent at $P$ is perpendicular to this radius.
Example 2.6 — Centre and radius by completing the square
Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 3 = 0$.
Step 1 Collect and complete the square:
$$(x^2 - 6x) + (y^2 + 4y) = 3$$
$$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 3$$
Step 2 Rearrange: $(x-3)^2 + (y+2)^2 = 16$.
Centre $\mathbf{(3,\,-2)}$, radius $\mathbf{4}$.
Tangents and Normals to a Circle
Theorem — Tangent–Radius Perpendicularity
The tangent to a circle at a point $P$ on its circumference is perpendicular to the radius $CP$ at that point. Therefore, if the gradient of $CP$ is $m$, the gradient of the tangent at $P$ is $-\dfrac{1}{m}$.
Example 2.7 — Equation of a tangent to a circle
Circle $C$: $(x-2)^2 + (y-1)^2 = 25$. Point $P(5,\, 5)$ lies on $C$. Find the equation of the tangent at $P$.
Step 1 Verify: $(5-2)^2 + (5-1)^2 = 9 + 16 = 25$ ✓
Step 2 Gradient of radius $CP$, $C = (2,\,1)$: $m_{CP} = \dfrac{5-1}{5-2} = \dfrac{4}{3}$.
Step 3 Tangent gradient: $-\dfrac{3}{4}$. Equation through $P(5,\,5)$:
$$y - 5 = -\frac{3}{4}(x - 5) \implies 4y - 20 = -3x + 15 \implies \boxed{3x + 4y = 35}$$
Example 2.8 — Line and circle intersections
Determine whether the line $y = 2x + 5$ is a tangent to, intersects, or does not meet the circle $x^2 + y^2 = 10$.
Step 1 Substitute $y = 2x + 5$:
$$x^2 + (2x+5)^2 = 10 \implies 5x^2 + 20x + 25 = 10 \implies 5x^2 + 20x + 15 = 0$$
$$x^2 + 4x + 3 = 0 \implies (x+1)(x+3) = 0$$
Step 2 Discriminant $\Delta = 16 - 12 = 4 > 0$: two distinct real roots, so the line intersects the circle at two points: $(-1,\,3)$ and $(-3,\,-1)$.
Example 2.9 — Finding the tangent condition
The line $y = mx + 3$ is a tangent to the circle $x^2 + y^2 = 5$. Find the possible values of $m$.
Step 1 Substitute: $x^2 + (mx+3)^2 = 5 \implies (1+m^2)x^2 + 6mx + 4 = 0$.
Step 2 For a tangent, $\Delta = 0$: $(6m)^2 - 4(1+m^2)(4) = 0$.
$$36m^2 - 16 - 16m^2 = 0 \implies 20m^2 = 16 \implies m^2 = \frac{4}{5} \implies m = \pm\frac{2}{\sqrt{5}}$$
Chord Properties
Perpendicular from Centre to Chord
The perpendicular from the centre of a circle to any chord bisects the chord. This is the basis for finding chord midpoints and lengths using Pythagoras’ theorem in the right triangle formed by the centre, the midpoint, and an endpoint of the chord.
Example 2.10 — Length of a chord
A circle has centre $O(0,\, 0)$ and radius $5$. A chord $AB$ has its midpoint at $M(3,\, 0)$. Find the length of chord $AB$.
Step 1 $OM = 3$ (distance from centre to midpoint).
Step 2 By Pythagoras in triangle $OMA$ (right-angled at $M$):
$$AM = \sqrt{r^2 - OM^2} = \sqrt{25 - 9} = 4$$
Step 3 $AB = 2 \times AM = \mathbf{8}$.
Exam Tip — Circles
When completing the square, remember $(x^2 + 2fx)$ becomes $(x + f)^2 - f^2$. A very common error is adding $f^2$ instead of subtracting it. After finding the centre and radius, always verify by substituting a known point on the circle back into your equation.
2.3 Parametric Equations Year 2
Introduction to Parametric Equations
A parametric equation describes a curve by expressing both $x$ and $y$ as functions of a third variable $t$, called the parameter. Rather than relating $x$ and $y$ directly with a single Cartesian equation, you write $x = f(t)$ and $y = g(t)$. As $t$ varies over a specified interval, the point $(f(t),\, g(t))$ traces the curve.
Definition — Parametric Equations
A curve is defined parametrically by $x = f(t)$ and $y = g(t)$ for $t$ in some domain. The corresponding Cartesian equation is obtained by eliminating $t$ between the two equations. The domain of $t$ may restrict the curve to a portion of the full Cartesian curve.
Converting Between Parametric and Cartesian Forms
Example 2.11 — Line: parametric to Cartesian
A line is given parametrically by $x = 1 + 2t$, $y = 3 - t$. Find the Cartesian equation.
Step 1 From $x = 1 + 2t$: $t = \dfrac{x-1}{2}$.
Step 2 Substitute into $y = 3 - t$:
$$y = 3 - \frac{x-1}{2} = \frac{7 - x}{2} \implies \boxed{x + 2y = 7}$$
Example 2.12 — Parabola: parametric to Cartesian
A curve has parametric equations $x = t^2$, $y = 2t$. Show that the Cartesian equation is $y^2 = 4x$.
Step 1 From $y = 2t$: $t = \dfrac{y}{2}$.
Step 2 Substitute: $x = \left(\dfrac{y}{2}\right)^2 = \dfrac{y^2}{4}$, so $y^2 = 4x$. This is the standard form of a parabola.
Parametric Equations of a Circle
Parametric Form of a Circle
The circle with centre $(a,\, b)$ and radius $r$ has parametric equations
$$x = a + r\cos\theta, \quad y = b + r\sin\theta, \quad 0 \le \theta < 2\pi$$
This follows from $\cos^2\theta + \sin^2\theta = 1$: substituting gives $(x-a)^2 + (y-b)^2 = r^2$ immediately.
Example 2.13 — Circle: parametric to Cartesian
A curve has parametric equations $x = 3\cos\theta - 1$, $y = 3\sin\theta + 2$. Find the Cartesian equation, and state the centre and radius.
Step 1 Rearrange: $\cos\theta = \dfrac{x+1}{3}$, $\sin\theta = \dfrac{y-2}{3}$.
Step 2 Apply $\cos^2\theta + \sin^2\theta = 1$:
$$\frac{(x+1)^2}{9} + \frac{(y-2)^2}{9} = 1 \implies (x+1)^2 + (y-2)^2 = 9$$
Centre $(-1,\, 2)$, radius $3$.
Example 2.14 — Finding coordinates and the parameter
A curve is defined by $x = t^2 + 1$, $y = t^3 - t$ for $t \in \mathbb{R}$. Find the coordinates when $t = 2$, and find the value(s) of $t$ at the point $(5,\, 6)$.
Step 1 At $t = 2$: $x = 5$, $y = 6$. The point is $(5,\, 6)$.
Step 2 At $(5,\,6)$: $x = 5 \implies t^2 = 4 \implies t = \pm 2$. Check $y$: at $t = 2$, $y = 6$ ✓; at $t = -2$, $y = -6 \ne 6$. So $t = 2$.
Example 2.15 — Trigonometric parametric equations
A curve is given by $x = \cos 2\theta$, $y = \sin\theta$ for $-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}$. Find the Cartesian equation and the range of $x$.
Step 1 Use the double-angle identity $\cos 2\theta = 1 - 2\sin^2\theta$:
$$x = 1 - 2y^2$$
Step 2 Since $\theta \in \left[-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right]$, $y = \sin\theta \in [-1,\, 1]$, so $x = 1 - 2y^2 \in [-1,\, 1]$.
Exam Tip — Parametric Equations
When eliminating $\theta$ from trigonometric parametric equations, look for a Pythagorean identity to exploit: $\cos^2\theta + \sin^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, or $\cot^2\theta + 1 = \csc^2\theta$. Always check whether the domain of $t$ restricts the Cartesian curve to a sub-region, and state this restriction in your answer.
Practice Problems
Problem 1
Find the equation of the straight line that is perpendicular to $3x - 2y + 1 = 0$ and passes through the point $(3,\, -1)$. Give your answer in the form $ax + by = c$.
Show solution
Rearrange: $y = \dfrac{3}{2}x + \dfrac{1}{2}$, gradient $= \dfrac{3}{2}$.
Perpendicular gradient: $-\dfrac{2}{3}$.
Through $(3,\,-1)$: $y + 1 = -\dfrac{2}{3}(x - 3) \implies 3y + 3 = -2x + 6 \implies \mathbf{2x + 3y = 3}$.
Problem 2
The points $A(-1,\, 4)$, $B(3,\, 6)$, and $C(5,\, 2)$ form a triangle. Show that the triangle is right-angled and identify the vertex at the right angle.
Show solution
$m_{AB} = \dfrac{6-4}{3-(-1)} = \dfrac{1}{2}$, $m_{BC} = \dfrac{2-6}{5-3} = -2$.
$m_{AB} \times m_{BC} = \dfrac{1}{2} \times (-2) = -1$, so $AB \perp BC$.
The right angle is at vertex $\mathbf{B}$.
Problem 3
Find the centre and radius of the circle $x^2 + y^2 + 8x - 2y - 8 = 0$.
Show solution
$(x^2 + 8x) + (y^2 - 2y) = 8$
$(x+4)^2 - 16 + (y-1)^2 - 1 = 8$
$(x+4)^2 + (y-1)^2 = 25$
Centre $\mathbf{(-4,\, 1)}$, radius $\mathbf{5}$.
Problem 4
The line $y = x + k$ is a tangent to the circle $x^2 + y^2 = 8$. Find the possible values of $k$.
Show solution
Substitute: $x^2 + (x+k)^2 = 8 \implies 2x^2 + 2kx + k^2 - 8 = 0$.
For a tangent, $\Delta = 0$: $(2k)^2 - 4 \cdot 2 \cdot (k^2 - 8) = 0$.
$4k^2 - 8k^2 + 64 = 0 \implies k^2 = 16 \implies \mathbf{k = \pm 4}$.
Problem 5
Point $P(7,\, 1)$ lies on the circle $(x-3)^2 + (y+2)^2 = 25$. Find the equation of the normal to the circle at $P$.
Show solution
The normal at $P$ passes through the centre $C(3,\,-2)$.
Gradient of $CP$: $m = \dfrac{1-(-2)}{7-3} = \dfrac{3}{4}$.
Through $P(7,\,1)$: $y - 1 = \dfrac{3}{4}(x-7) \implies 4y - 4 = 3x - 21 \implies \mathbf{3x - 4y - 17 = 0}$.
Problem 6
A circle passes through $A(0,\, 0)$, $B(4,\, 0)$, and $C(0,\, 6)$. Find the equation of the circle in the form $(x-a)^2 + (y-b)^2 = r^2$.
Show solution
Use $x^2 + y^2 + 2fx + 2gy + c = 0$:
$A(0,0)$: $c = 0$. $B(4,0)$: $16 + 8f = 0 \implies f = -2$. $C(0,6)$: $36 + 12g = 0 \implies g = -3$.
$(x-2)^2 + (y-3)^2 = 4 + 9 = 13$.
Centre $(2,\, 3)$, radius $\sqrt{13}$.
Problem 7 Year 2
A curve has parametric equations $x = 3t - 1$, $y = t^2 + 2$. Find the Cartesian equation of the curve.
Show solution
From $x = 3t - 1$: $t = \dfrac{x+1}{3}$.
Substitute: $y = \left(\dfrac{x+1}{3}\right)^2 + 2 = \dfrac{(x+1)^2}{9} + 2$.
$$\mathbf{(x+1)^2 = 9(y - 2)}$$
Problem 8 Year 2
A curve is defined by $x = 4\cos\theta$, $y = 3\sin\theta$, $0 \le \theta < 2\pi$. Find the Cartesian equation and name the curve.
Show solution
$\cos\theta = \dfrac{x}{4}$, $\sin\theta = \dfrac{y}{3}$.
$\cos^2\theta + \sin^2\theta = 1 \implies \dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.
This is an ellipse with semi-axes $a = 4$ (horizontal) and $b = 3$ (vertical), centred at the origin.
Problem 9
Find all points of intersection of the line $y = x - 1$ and the circle $(x-2)^2 + (y-3)^2 = 10$.
Show solution
Substitute $y = x - 1$:
$(x-2)^2 + (x-4)^2 = 10 \implies x^2 - 4x + 4 + x^2 - 8x + 16 = 10$
$2x^2 - 12x + 10 = 0 \implies x^2 - 6x + 5 = 0 \implies (x-1)(x-5) = 0$
$x = 1 \implies y = 0$; $x = 5 \implies y = 4$. Points: $\mathbf{(1,\, 0)}$ and $\mathbf{(5,\, 4)}$.
Problem 10 Year 2
A curve is defined parametrically by $x = \sec\theta$, $y = \tan\theta$ for $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. Find the Cartesian equation and state the domain of $x$.
Show solution
Use the identity $\sec^2\theta - \tan^2\theta = 1$:
$$x^2 - y^2 = 1$$
Since $\sec\theta \ge 1$ for $\theta \in \left(-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right)$, the domain is $x \ge 1$. This is one branch of a hyperbola.