Circles — Interactive Learning Module

Singapore Additional Mathematics Syllabus (O-Level)

Welcome to the Circles module. In this interactive lesson, you will learn the mathematical foundations of circles on the Cartesian plane. Choose a section below to begin.

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1. Learn

Understand standard and general forms, and completing the square.

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2. Practice

Test your understanding with an interactive quiz.

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3. Explore

Interact with the live graphing visualiser.

Understanding the Equation of a Circle

Stage 1: Deriving the Standard Form

A circle is defined as the locus of all points $P(x, y)$ that are at a constant distance, the radius $r$, from a fixed centre point $C(a, b)$.

C(a,b) P(x,y) r x - a y - b

If we drop perpendiculars to form a right-angled triangle, the horizontal length is $(x - a)$ and the vertical length is $(y - b)$.

Applying Pythagoras' Theorem gives us the fundamental equation:

Standard Form:
$$(x - a)^2 + (y - b)^2 = r^2$$
  • Centre = $(a, b)$
  • Radius = $r$

Stage 2: Expanding to General Form

Let's take the standard form and expand the brackets algebraically.

Start with: $$(x - a)^2 + (y - b)^2 = r^2$$

Expand the squares:

$$x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2$$

Rearrange to group the $x^2$ and $y^2$ terms, the $x$ and $y$ terms, and the constants:

$$x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - r^2) = 0$$

To make this cleaner, mathematicians substitute new letters: Let $2g = -2a$, $2f = -2b$, and $c = a^2 + b^2 - r^2$.

General Form:
$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Notice that the coefficients of $x^2$ and $y^2$ are identical (usually 1), and there is no $xy$ term!

Stage 3: Converting Back via Completing the Square

In exams, you are often given the general form and must find the centre and radius. You do this by completing the square for $x$ and $y$ separately.

Worked Example: Find the centre and radius of $x^2 + y^2 - 6x + 4y - 3 = 0$.

Step 1: Group the $x$ and $y$ terms, move constant to the right.
$$(x^2 - 6x) + (y^2 + 4y) = 3$$

Step 2: Complete the square. Add $(\frac{b}{2})^2$ to both sides.
$$(x - 3)^2 - (-3)^2 + (y + 2)^2 - (2)^2 = 3$$
$$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 3$$

Step 3: Simplify into standard form.
$$(x - 3)^2 + (y + 2)^2 = 3 + 9 + 4$$
$$(x - 3)^2 + (y + 2)^2 = 16$$

From this, we extract: Centre = $(3, -2)$ and Radius = $\sqrt{16} = 4$.

Summary: Comparing Both Forms

Feature Standard Form General Form
Equation $(x - a)^2 + (y - b)^2 = r^2$ $x^2 + y^2 + 2gx + 2fy + c = 0$
Centre $(a, b)$ $(-g, -f)$
Radius $r$ $\sqrt{g^2 + f^2 - c}$
Best For... Quickly graphing or identifying geometry Solving intersections with straight lines

You've completed the lesson! Head to the Quiz to test your understanding.

Quiz: Test Your Understanding

Question 1 of 8

Question 1

Identify the form of the following equation:
$(x - 4)^2 + (y + 1)^2 = 25$

Question 2

Identify the form of the following equation:
$x^2 + y^2 + 6x - 2y - 15 = 0$

Question 3

Identify the form of the following equation:
$2x^2 + 3y^2 - 4x + 6y = 12$

Question 4

Extract the centre and radius from the equation:
$(x - 5)^2 + (y + 2)^2 = 49$

Centre X: Centre Y: Radius:

(Type integers only)

Question 5

Extract the centre and radius from the equation:
$(x + 3)^2 + y^2 = 16$

Centre X: Centre Y: Radius:

(Careful with the signs!)

Question 6

Extract the centre and radius from the equation:
$x^2 + y^2 = 9$

Centre X: Centre Y: Radius:

Question 7

Write the Standard Form equation for a circle with Centre $(2, -4)$ and Radius $6$.

Question 8

Convert the standard equation $(x - 1)^2 + (y - 2)^2 = 9$ into General Form.

Quiz Complete!

Interactive Graph Explorer

Adjust Parameters

Standard Form:
x² + y² = 9

General Form:
x² + y² - 9 = 0

Guiding Questions:
  • What happens to the circle when you increase $r$?
  • How does changing $a$ affect the position?
  • Can you set the circle so it passes through $(0,0)$?