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How do equations describe straight lines, circles and the shapes of common graphs?

The equation of a straight line and conditions for parallel and perpendicular lines, the equation of a circle and its key properties, and sketching and transforming standard curves.

A CCEA A-Level Mathematics answer on the equation of a straight line, gradients of parallel and perpendicular lines, the equation of a circle and its tangent and chord properties, and sketching and transforming standard curves.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to work fluently with straight lines and circles in the coordinate plane: find gradients, midpoints and lengths, use the conditions for parallel and perpendicular lines, recognise and manipulate the equation of a circle, apply the circle's geometric properties (the tangent is perpendicular to the radius, the perpendicular from the centre bisects a chord), and sketch and transform standard curves. These ideas recur in calculus and vectors.

The answer

The straight line

Two lines are parallel when their gradients are equal, $m_1 = m_2$ . They are perpendicular when the product of their gradients is $-1$ , that is $m_1 m_2 = -1$ , so each gradient is the negative reciprocal of the other.

The circle

Three geometric properties are tested directly. The tangent to a circle is perpendicular to the radius at the point of contact. The perpendicular from the centre to a chord bisects that chord. And the angle in a semicircle is a right angle, so if a triangle is inscribed with a diameter as one side, the opposite angle is $90^\circ$ .

Sketching and transforming curves

You should recognise the shapes of $y = x^2$ , $y = x^3$ , $y = \frac{1}{x}$ , $y = \sqrt{x}$ and similar. The transformations of $y = f(x)$ are:

$y = f(x) + a$ : translation up by $a$ .
$y = f(x + a)$ : translation left by $a$ .
$y = af(x)$ : vertical stretch, scale factor $a$ .
$y = f(ax)$ : horizontal stretch, scale factor $\frac{1}{a}$ .
$y = -f(x)$ : reflection in the $x$ -axis; $y = f(-x)$ : reflection in the $y$ -axis.

Worked example: a tangent meeting a circle

Find where a line is tangent to a circle

Show that the line $y = x + 1$ is a tangent to the circle $x^2 + y^2 = 1$ , and find the point of contact.

step Substitute the line into the circle

x^2 + (x + 1)^2 = 1.

step Form a quadratic in x

x^2 + x^2 + 2x + 1 = 1 \Rightarrow 2x^2 + 2x = 0 \Rightarrow 2x(x + 1) = 0.

step Interpret the discriminant

This factorises but gives $x = 0$ and $x = -1$ , two points, so this particular line is a secant. Re-checking with the genuine tangent condition: for tangency the quadratic must have a repeated root, that is discriminant zero. For $y = x + c$ , substituting gives $2x^2 + 2cx + (c^2 - 1) = 0$ with discriminant $4c^2 - 8(c^2 - 1) = 0$ , so $c^2 = 2$ and $c = \pm\sqrt{2}$ .

step State the tangent

The tangents parallel to $y = x$ are $y = x \pm \sqrt{2}$ , touching at one point each. The discriminant being zero is the algebraic signature of a tangent.

Examples in context

Example 1. Finding a tangent line. To find the tangent to a circle at a given point on it, compute the gradient of the radius from the centre to that point, then take the negative reciprocal as the tangent's gradient and pass it through the point. The perpendicularity property turns a geometric problem into one line equation.

Example 2. A shifted parabola. The graph $y = (x - 2)^2 + 3$ is $y = x^2$ translated $2$ right and $3$ up, so its minimum is at $(2, 3)$ . Recognising transformations lets you sketch without plotting points, and links directly to completing the square.

Try this

Q1. Find the gradient of the line perpendicular to $3x + y = 7$ . [2 marks]

Cue. The line has gradient $-3$ , so the perpendicular gradient is $\tfrac{1}{3}$ .

Q2. Write down the centre and radius of $(x + 1)^2 + (y - 4)^2 = 9$ . [2 marks]

Cue. Centre $(-1, 4)$ , radius $3$ .

Q3. Describe the transformation taking $y = x^2$ to $y = x^2 - 5$ . [1 mark]

Cue. A translation $5$ units down.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20206 marksA circle has equation

x^2 + y^2 - 6x + 4y - 12 = 0

. Find the coordinates of its centre and its radius.

Show worked answer →

Complete the square in $x$ and in $y$ :

$x^2 - 6x = (x - 3)^2 - 9$ and $y^2 + 4y = (y + 2)^2 - 4$ .

Substituting,

$(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0,$

so $(x - 3)^2 + (y + 2)^2 = 25$ .

The centre is $(3, -2)$ and the radius is $\sqrt{25} = 5$ .

Markers reward completing the square in both variables, collecting the constants correctly, and reading the centre and radius from the standard form.

CCEA 20185 marksThe points

A(1, 2)

and

B(5, 10)

lie on a line. Find the equation of the perpendicular bisector of

AB

Show worked answer →

The midpoint of $AB$ is $\left(\frac{1 + 5}{2}, \frac{2 + 10}{2}\right) = (3, 6)$ .

The gradient of $AB$ is $\frac{10 - 2}{5 - 1} = \frac{8}{4} = 2$ .

The perpendicular gradient is the negative reciprocal, $-\frac{1}{2}$ .

Through $(3, 6)$ : $y - 6 = -\frac{1}{2}(x - 3)$ , so $y = -\frac{1}{2}x + \frac{15}{2}$ , or $x + 2y - 15 = 0$ .

Markers reward the midpoint, the gradient of $AB$ , the negative-reciprocal perpendicular gradient, and the line through the midpoint.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)