Circles and other simple relations | Year 11

01 / LEARN

Understand the idea.

A circle is a fixed-distance relation

A circle consists of points a fixed positive distance $r$ from a centre $(h, k)$ . Applying the distance formula gives ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ . Read each bracket as a coordinate difference, rather than copying its written sign.

The extreme coordinates are $h - r$ and $h + r$ horizontally, and $k - r$ and $k + r$ vertically. These bounds guide a sketch. Interior inputs generally have two outputs, so the full circle is a relation rather than a function of x.

{(x - h)}^{2} + {(y - k)}^{2} = r^{2}

A true circle centred at (1,-1) with radius 3 on an equal-scale grid from -5 to 5 on both axes. The gold centre and its supplied coordinates are labelled.

Select a branch and check other relations

Isolating the squared vertical difference gives $y = k \pm \sqrt{r^{2} - {(x - h)}^{2}}$ . The plus sign selects the upper semicircle and the minus sign the lower. Each branch is a function on $h - r \leq x \leq h + r$ .

An expanded equation may need completing the square before its centre and radius become visible. The next chapter develops that technique. Squared terms alone do not identify a circle: their equal coefficients, signs and resulting radius must also be checked.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Read a circle

State the centre and radius of ${(x + 2)}^{2} + {(y - 1)}^{2} = 16$ .

Worked solution

1
Match the equation to circle form.
$\begin{aligned} {(x - (- 2))}^{2} + {(y - 1)}^{2} & = 4^{2} \end{aligned}$
Standard form uses differences from the centre and the square of the radius.
2
Read the centre coordinates.
$(h, k) = (- 2, 1)$
The brackets vanish at the centre’s coordinates; the plus two corresponds to horizontal coordinate negative two.
3
Take the positive radius.
$r = \sqrt{16} = 4$
A radius is a positive length, so sixteen is its square rather than its value.

Answer

Centre (-2,1), radius 4.

EXAMPLE 02

Write a restricted branch

Write the upper semicircle of a circle centred at (1,-2) with radius 3, including its domain.

Worked solution

1
Write the full circle from its centre and radius.
${(x - 1)}^{2} + {(y + 2)}^{2} = 9$
The centre is one horizontally and negative two vertically, with radius three.
2
Isolate the upper branch.
$\begin{aligned} {(y + 2)}^{2} & = 9 - {(x - 1)}^{2} \\ y & = - 2 + \sqrt{9 - {(x - 1)}^{2}} \end{aligned}$
The upper semicircle has y at least negative two, so the positive square-root branch is required.
3
Require a real square root.
$\begin{aligned} {(x - 1)}^{2} & \leq 9 \\ - 3 & \leq x - 1 \leq 3 \\ - 2 & \leq x \leq 4 \end{aligned}$
The radicand is non-negative precisely between the circle’s leftmost and rightmost points, including both endpoints.

Answer

$y = - 2 + \sqrt{9 - {(x - 1)}^{2}}$ , $- 2 \leq x \leq 4$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What are the centre and radius of $x^{2} + {(y + 3)}^{2} = 25$ ?

Show answer & reasoning

Centre (0,-3), radius 5.

Write the equation as ${(x - 0)}^{2} + {(y - (- 3))}^{2} = 5^{2}$ . Its centre is $(0, - 3)$ and its radius is $5$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

State the centre and radius of ${(x - 4)}^{2} + {(y + 1)}^{2} = 9$ .
2 marks
Write the equation of a circle centred at $(- 2, 3)$ with radius 5.
2 marks
Does $(3, 4)$ lie on $x^{2} + y^{2} = 25$ ? Explain.
2 marks
Find the x-intercepts of ${(x - 1)}^{2} + y^{2} = 9$ .
2 marks
Write the lower semicircle of $x^{2} + y^{2} = 16$ , including domain and range.
2 marks
A circular sprinkler reaches 6 m from a nozzle at $(2, 1)$ on a metre grid. Give its boundary equation and decide whether $(7, 4)$ lies inside.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

State the centre and radius of ${(x - 4)}^{2} + {(y + 1)}^{2} = 9$ .
2 marks
Read the bracket signs and take the positive square root of 9.
Centre $(4, - 1)$ , radius 3.
Write the equation of a circle centred at $(- 2, 3)$ with radius 5.
2 marks
Insert $h = - 2$ , $k = 3$ and $r^{2} = 25$ .
${(x + 2)}^{2} + {(y - 3)}^{2} = 25$ .
Does $(3, 4)$ lie on $x^{2} + y^{2} = 25$ ? Explain.
2 marks
$3^{2} + 4^{2} = 9 + 16 = 25$ .
Yes.
Find the x-intercepts of ${(x - 1)}^{2} + y^{2} = 9$ .
2 marks
Set $y = 0$ : ${(x - 1)}^{2} = 9$ , so $x - 1 = \pm 3$ .
$(- 2, 0)$ and $(4, 0)$ .
Write the lower semicircle of $x^{2} + y^{2} = 16$ , including domain and range.
2 marks
Select the negative square-root branch.
Its x-values run across the diameter, while its outputs run from -4 to 0.
$y = - \sqrt{16 - x^{2}}$ ; domain $[- 4, 4]$ , range $[- 4, 0]$ .
A circular sprinkler reaches 6 m from a nozzle at $(2, 1)$ on a metre grid. Give its boundary equation and decide whether $(7, 4)$ lies inside.
2 marks
Its squared distance from the nozzle is $5^{2} + 3^{2} = 34 < 36$ .
Thus its distance is less than 6 m.
Boundary ${(x - 2)}^{2} + {(y - 1)}^{2} = 36$ ; $(7, 4)$ is inside.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Topic 1: Functions and graphs. Prerequisite algebra and coordinate geometry are embedded as supporting lessons.

Official curriculum source ↗

In this lesson

Understand the idea.

A circle is a fixed-distance relation

Select a branch and check other relations

See the method in action.

Read a circle

Match the equation to circle form.

Read the centre coordinates.

Take the positive radius.

Write a restricted branch

Write the full circle from its centre and radius.

Isolate the upper branch.

Require a real square root.

Try it yourself.

Check your understanding

Keep practising