Circles and Translations Revision Notes for HSC SSCE Mathematics Advanced

Circles and Translations

Introduction

This topic explores how circles can be represented algebraically using equations, and how translations affect these equations. Understanding circle equations allows us to identify key features such as the centre, radius, domain, and range, and to sketch accurate graphs of circles on the coordinate plane.

After studying this content, you will be able to:

Identify the centre and radius of a circle from its equation in standard form
Determine the domain and range of a circle from its centre and radius
Convert the equation of a circle from general form to standard form by completing the square
Find the equation of a circle given its centre and radius
Sketch the graph of a circle from its equation

Circle equations

Standard form at the origin

When a circle is centred at the origin, its equation in standard form is:

$x^2 + y^2 = r^2$

where $r$ is the radius of the circle, and $r > 0$ .

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This equation represents all points that are exactly $r$ units away from the origin $(0, 0)$ . The radius measures the distance from the centre to any point on the circle, which is why it must always be positive.

Translated circles in standard form

When a circle is translated (moved) from the origin, the equation changes to account for the new position. The equation of a circle translated horizontally by $a$ units and vertically by $b$ units is given in standard form as:

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

Understanding the parameters:

$a$ $a$ represents the horizontal translation
- If $a > 0$ , the circle moves right
- If $a < 0$ , the circle moves left
$b$ $b$ represents the vertical translation
- If $b > 0$ , the circle moves up
- If $b < 0$ , the circle moves down
$r$ is the radius of the circle, where $r > 0$

Identifying the centre and radius:

From the standard form equation $(x - a)^2 + (y - b)^2 = r^2$ :

The centre is located at $(a, b)$
The radius is $r$

chatImportant

When evaluating $r$ from $r^2$ , only the positive root is taken, since the radius represents a distance, and distance is always defined as a positive scalar quantity.

Domain and range of circles

Once you know the centre $(a, b)$ and radius $r$ of a circle, you can determine its domain and range:

Domain: $[a - r, a + r]$

The domain represents the horizontal extent of the circle. It extends $r$ units left and right from the centre's x-coordinate.

Range: $[b - r, b + r]$

The range represents the vertical extent of the circle. It extends $r$ units down and up from the centre's y-coordinate.

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Understanding Domain and Range:

Think of domain and range as the "shadow" the circle would cast on the x-axis and y-axis respectively. The domain tells you how far left and right the circle extends, while the range tells you how far up and down it extends.

General form

The general form of a circle's equation is:

$x^2 + y^2 + ax + by + c = 0$

where $a$ , $b$ , and $c$ are constants that define the circle's position and size.

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Critical Distinction:

The variables $a$ and $b$ in the general form equation $x^2 + y^2 + ax + by + c = 0$ are different constants from the $a$ and $b$ used to denote the centre $(a, b)$ in the standard form. The context will always make clear which definition is being used.

Converting from general form to standard form

To convert a circle equation from general form to standard form, you need to complete the square for both the $x$ and $y$ terms. Completing the square allows you to rewrite the equation so that the centre and radius become easily identifiable.

The process:

Rearrange the equation so that $x$ terms and $y$ terms are grouped together
Complete the square for the $x$ terms
Complete the square for the $y$ terms
Simplify to identify the standard form $(x - a)^2 + (y - b)^2 = r^2$

Once in standard form, the centre is easily identified, and the radius can be calculated directly.

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Completing the Square Reminder:

To complete the square for a term like $x^2 + px$ :

Take half of the coefficient of $x$ (which is $p$ ): $\frac{p}{2}$
Square this value: $\left(\frac{p}{2}\right)^2$
Add this value inside the bracket and to the other side of the equation to maintain equality

Worked examples

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Worked Example 1: Analysing a circle from its standard form equation

Given: A circle has equation $(x - 2)^2 + (y + 1)^2 = 9$ .

Part a: Determine the centre

Strategy:

Compare the given equation to the standard form $(x - a)^2 + (y - b)^2 = r^2$ to identify the values of $a$ and $b$ .

Solution:

$(x - 2)^2 + (y + 1)^2 = 9$

This can be rewritten as:

$(x - 2)^2 + (y - (-1))^2 = 9$

Therefore, the centre is at (2, -1).

Part b: Determine the radius

Strategy:

Extract the value of $r$ from $r^2 = 9$ .

Solution:

$r^2 = 9$ , so $r = 3$ .

The radius is 3 units.

Part c: Determine the domain and range

Strategy:

Use the formulas:

Domain: $[a - r, a + r]$
Range: $[b - r, b + r]$

Solution:

The domain is $[2 - 3, 2 + 3] = [-1, 5]$ .

The range is $[-1 - 3, -1 + 3] = [-4, 2]$ .

Part d: Sketch the circle, labelling the centre and key points

Strategy:

Plot the centre, then plot points at the domain and range boundaries. These points lie at the ends of the horizontal and vertical diameters of the circle. Then draw the circle passing through these points.

Solution:

The sketch shows a circle with:

Centre at $(2, -1)$
Radius 3
Key points at $(2, 2)$ (top), $(5, -1)$ (right), $(2, -4)$ (bottom), and $(-1, -1)$ (left)

Reflection:

The sketch confirms the centre at $(2, -1)$ and radius 3, with translations right by 2 units and down by 1 unit from the origin.

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Worked Example 2: Converting from general form to standard form

Given: Convert $x^2 + y^2 - 4x + 6y - 3 = 0$ to standard form.

Part a: Determine the standard form of the equation

Strategy:

Complete the square for both $x$ and $y$ terms to rewrite the equation in standard form.

Solution:

x^2 + y^2 - 4x + 6y - 3 = 0 x^2 - 4x + y^2 + 6y = 3 (x^2 - 4x + 4) + (y^2 + 6y + 9) = 3 + 4 + 9 (x - 2)^2 + (y + 3)^2 = 16

The standard form is $(x - 2)^2 + (y + 3)^2 = 16$ .

Working explanation:

First, we rearranged the equation to group $x$ and $y$ terms
To complete the square for x: take half of the coefficient of $x$ (which is $-4$ ), giving $-2$ , then square it to get $4$ . Add this inside the bracket
To complete the square for y: take half of the coefficient of $y$ (which is $6$ ), giving $3$ , then square it to get $9$ . Add this inside the bracket
Add the same values ( $4 + 9 = 13$ ) to the right side to maintain equality
Factorise to get the standard form

Part b: Determine the centre

Strategy:

Identify $a$ and $b$ from the standard form $(x - a)^2 + (y - b)^2 = r^2$ .

Solution:

The centre is at (2, -3).

Part c: Determine the radius

Strategy:

Extract $r$ from $r^2 = 16$ .

Solution:

$r^2 = 16$ , so $r = 4$ .

The radius is 4 units.

Part d: Sketch the circle, labelling the centre and key points

Strategy:

Plot the centre and points at the domain and range boundaries, then draw the circle.

Solution:

The sketch shows a circle with:

Centre at $(2, -3)$
Radius 4
Key points at $(2, 1)$ (top), $(6, -3)$ (right), $(2, -7)$ (bottom), and $(-2, -3)$ (left)

Reflection:

The sketch verifies the centre and radius, aligning with the standard form obtained by completing the square.

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Worked Example 3: Finding the equation from centre and radius

Given: A circle has centre $(1, -2)$ and radius 5.

Part a: Determine the equation of the circle

Strategy:

Use the standard form $(x - a)^2 + (y - b)^2 = r^2$ with $a = 1$ , $b = -2$ , and $r = 5$ .

Solution:

(x - a)^2 + (y - b)^2 = r^2 (x - 1)^2 + (y - (-2))^2 = 5^2 (x - 1)^2 + (y + 2)^2 = 25

The equation is $(x - 1)^2 + (y + 2)^2 = 25$ .

Working explanation:

We wrote the general formula for a circle in standard form
We substituted the known values: $a = 1$ , $b = -2$ , and $r = 5$
We evaluated $r^2 = 25$
We simplified the expression for $y - b$ to get $y + 2$

Part b: Sketch the circle, labelling the centre and key points

Strategy:

Plot the centre and the points at the boundaries of the circle, then draw the circle passing through these points.

Solution:

The sketch shows a circle with:

Centre at $(1, -2)$
Radius 5
Key points at $(1, 3)$ (top), $(6, -2)$ (right), $(1, -7)$ (bottom), and $(-4, -2)$ (left)

Reflection:

The sketch confirms the equation and helps us visualise the circle's position and size.

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Key Concepts Summary

A circle's equation in the form $(x - a)^2 + (y - b)^2 = r^2$ has:

Centre at (a, b)
Radius r (where $r > 0$ )

To convert from general form $x^2 + y^2 + ax + by + c = 0$ to standard form:

Complete the square for both $x$ and $y$ terms
Rearrange to identify the centre and radius

The domain and range of a circle can be determined from its centre and radius:

Domain: $[a - r, a + r]$
Range: $[b - r, b + r]$

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Remember!

The standard form $(x - a)^2 + (y - b)^2 = r^2$ immediately reveals the centre $(a, b)$ and radius $r$ of a circle
Pay careful attention to signs: $(x - 2)$ means the centre has x-coordinate $+2$ , while $(y + 3)$ means the centre has y-coordinate $-3$
The radius is always positive; when finding $r$ from $r^2$ , take only the positive square root
To convert from general form to standard form, complete the square for both variables
Domain and range can be found by adding and subtracting the radius from the centre coordinates: domain uses the x-coordinate, range uses the y-coordinate

Circles and Translations (HSC SSCE Mathematics Advanced): Revision Notes