What is sign slip completing the square for a circle?

Remember to subtract the squared term, so x^2 - 6x = (x - 3)^2 - 9.

EnglandMathsSyllabus dot point

How do you describe lines, circles and curves with equations and use them?

Coordinate geometry in the x and y plane including straight lines, the equation of a circle, tangents, chords and parametric equations of curves.

A focused answer to the Edexcel A-Level Mathematics coordinate geometry content, covering the straight line, gradient and midpoint, parallel and perpendicular lines, the equation of a circle, tangents and chords, and parametric equations of curves.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

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

Edexcel wants you to work with straight lines (gradient, midpoint, parallel and perpendicular conditions), the equation of a circle and its geometric properties (tangents, chords, the angle in a semicircle), and parametric equations of curves, including converting between parametric and Cartesian form. These tools appear across mechanics, calculus and vectors, so fluency pays off widely.

The answer

The straight line

A line is fixed by a gradient and a point, or by two points.

The perpendicular gradient is the negative reciprocal: if a line has gradient $m = \dfrac{3}{4}$ then any perpendicular line has gradient $-\dfrac{4}{3}$ . To find where two lines meet, solve their equations simultaneously.

The circle

The equation of a circle comes directly from the distance formula.

Find the centre and radius of

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

Step 1: Complete the square in $x$

Group the $x$ terms and complete the square. For $x^2 - 6x$ , half of $-6$ is $-3$ , so:

x^2 - 6x = (x - 3)^2 - 9.

Step 2: Complete the square in $y$

Similarly, for $y^2 + 4y$ , half of $4$ is $2$ , so:

y^2 + 4y = (y + 2)^2 - 4.

Step 3: Rewrite in standard form and read off the centre and radius

Substitute both results back and collect the constants to the right-hand side:

(x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \implies (x - 3)^2 + (y + 2)^2 = 25.

Comparing with $(x - a)^2 + (y - b)^2 = r^2$ , the centre is $(3, -2)$ and the radius is $r = \sqrt{25} = 5$ .

Final answer: Centre $(3, -2)$ , radius $5$ .

Parametric equations

A parametric curve gives $x$ and $y$ as functions of a parameter $t$ . You convert to Cartesian form by eliminating $t$ .

Examples in context

Try this

Q1. Find the equation of the line through $(1, 2)$ perpendicular to $y = 2x + 5$ . [3 marks]

Cue. Perpendicular gradient is $-\frac{1}{2}$ , so $y - 2 = -\frac{1}{2}(x - 1)$ .

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

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

Q3. Convert the parametric equations $x = 3\cos t$ , $y = 3\sin t$ to Cartesian form. [2 marks]

Cue. $x^2 + y^2 = 9\cos^2 t + 9\sin^2 t = 9$ , a circle of radius $3$ .

Exam-style practice questions

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

Edexcel 20186 marksThe circle

C

has equation

x^2 + y^2 - 10x + 4y + 13 = 0

. Find the centre and radius of

C

, and show that the point

P(2, 1)

lies outside

C

Show worked answer →

Complete the square in $x$ and $y$ (M1): $(x - 5)^2 - 25 + (y + 2)^2 - 4 + 13 = 0$ .

This gives $(x - 5)^2 + (y + 2)^2 = 16$ (A1), so the centre is $(5, -2)$ and the radius is $r = 4$ (A1).

Find the distance from the centre to $P(2, 1)$ (M1): $d = \sqrt{(5 - 2)^2 + (-2 - 1)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24$ (A1).

Since $d = \sqrt{18} > 4 = r$ , the point $P$ lies outside the circle (A1).

Markers reward correct completing the square, the centre and radius, the distance calculation, and the comparison $d > r$ .

Edexcel 20225 marksA curve is given parametrically by

x = 2t + 1

and

y = t^2 - 3

. Find a Cartesian equation of the curve, and hence determine the coordinates of the point where the curve crosses the

y

-axis.

Show worked answer →

Make $t$ the subject of the first equation: $t = \dfrac{x - 1}{2}$ (M1).

Substitute into $y = t^2 - 3$ (M1): $y = \left(\dfrac{x - 1}{2}\right)^2 - 3 = \dfrac{(x - 1)^2}{4} - 3$ (A1).

The curve crosses the $y$ -axis where $x = 0$ (M1): $y = \dfrac{(0 - 1)^2}{4} - 3 = \dfrac{1}{4} - 3 = -\dfrac{11}{4}$ , so the point is $\left(0, -\dfrac{11}{4}\right)$ (A1).

Markers reward eliminating $t$ , the correct Cartesian form, setting $x = 0$ , and the final coordinates.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)