AS Level · Pure Mathematics 2 WMA12

Circles

Revise Circles for Pure Mathematics 2 WMA12 (AS Level) — revision notes and instant AI marking.

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Edexcel IAL Pure 2

Circles

Master the equation of a circle, find its centre and radius, and understand the geometric properties that make circles special — all from one key relationship.

What You'll Learn

Circles are defined by a centre and a radius. This single relationship unlocks everything: the equation, how to find the centre from an equation, the special properties of chords and tangents, and why angles in a semicircle are always right angles.

The Big Idea: Once you know the centre (a, b) and radius r, the circle's equation is simply (x − a)² + (y − b)² = r². Everything else flows from this.

Topics Covered

  • Equation of a Circle — standard form and general form
  • Finding the Centre & Radius — completing the square
  • Bisection of Chords — perpendicular bisector theorem
  • Angle in a Semicircle — circumcircles and right angles
  • Radius & Tangent — perpendicularity and tangent equations

Equation of a Circle

The Standard Form

A circle with centre (a, b) and radius r has the equation:

Standard Form
(x − a)² + (y − b)² = r²

This comes straight from the definition of a circle: the set of all points at a fixed distance (the radius) from a centre. If a point (x, y) is on the circle, its distance from (a, b) is r, so:

distance = √[(x − a)² + (y − b)²] = r Square both sides: (x − a)² + (y − b)² = r²
Crucial tip: The numbers in the brackets are the opposite signs of the centre coordinates. If the centre is (−3, 2), the equation is (x + 3)² + (y − 2)² = r², not (x − 3)² + (y + 2)².

Example: Finding the Equation from Centre & Radius

Worked Example

Find the equation of a circle with centre (−5, −7) that passes through (1, 1).

Step 1: Find the Radius
The radius is the distance from the centre to any point on the circle. Use the distance formula:
r = √[(1 − (−5))² + (1 − (−7))²]
r = √[(6)² + (8)²]
r = √[36 + 64] = √100 = 10
Step 2: Write the Equation
Centre is (−5, −7), radius is 10. Substitute into (x − a)² + (y − b)² = r²:
(x − (−5))² + (y − (−7))² = 10²
(x + 5)² + (y + 7)² = 100

The General Form

When you expand the standard form, you get the general form:

General Form
x² + y² + 2fx + 2gy + c = 0

For example, expanding (x + 3)² + (y − 4)² = 25:

(x + 3)² + (y − 4)² = 25 x² + 6x + 9 + y² − 8y + 16 = 25 x² + y² + 6x − 8y + 25 = 0 So: 2f = 6 → f = 3, 2g = −8 → g = −4, c = 0
Why two forms? The standard form immediately tells you the centre and radius. The general form is what you get when you expand and simplify — you'll need to convert back to standard form (by completing the square) to find the centre and radius.
Try It
A circle has centre (2, −3) and radius 5. Write its equation in standard form.

Finding the Centre & Radius

From the Standard Form

If you're given an equation like (x − 3)² + (y + 2)² = 16, reading the centre and radius is direct:

(x − 3)² + (y + 2)² = 16 Centre: (3, −2) [opposite signs of the numbers in brackets] Radius: r = √16 = 4
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Also in the full note
  • What to Memorise
  • Exam Tips & Common Mistakes
  • In Summary
  • From the General Form: Completing the Square
  • The Key Property
  • Finding a Circle's Equation from Three Points
  • The Theorem
  • Why Does This Work?
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