Circles and other simple relations | Year 11

01 / LEARN

The idea, step by step

A circle is a fixed-distance relation

A circle is the set of points at a fixed positive distance r from its centre (h,k). Apply the distance formula and square both sides to obtain its standard equation. The constants inside brackets have the opposite sign to the centre coordinates: x-2 means the centre has x-coordinate 2.

A circle usually has two y-values for one interior x-value, so its full graph is a relation rather than a function of x. Its leftmost and rightmost x-values are h-r and h+r; its lowest and highest y-values are k-r and k+r. Use these bounds to draw an appropriately scaled sketch.

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

Select a branch and check other relations

Solving for y gives an upper and a lower branch. Choosing one sign makes a semicircle function with the same restricted x-domain. A horizontal line, vertical line and parabola provide useful comparisons: do they give one output for every permitted x, and what are their intercepts or boundaries?

For expanded circle equations, completing the square recovers the standard form. That technique is taught in the following quadratics chapter, so revisit expanded forms afterwards rather than guessing the radius from a constant term. Never infer a circle from an x² term alone: the coefficients and signs of the squared coordinates matter.

A common trap

Reading the radius as r² or reading the centre with the signs unchanged.

Instead: Match the brackets to x-h and y-k, then take the positive square root of the right-hand side.

02 / WORKED EXAMPLES

See the thinking

Example 1

Read a circle

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

Compare with standard form: h=-2, k=1 and r²=16.
Take the positive radius r=4, rather than calling 16 the radius.

Centre (-2,1), radius 4.

Example 2

Write a restricted branch

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

The circle is (x-1)²+(y+2)²=9.
The upper branch chooses the positive square root.
The square root is real when |x-1|≤3, giving the stated domain.

y = - 2 + \sqrt{9 - {(x - 1)}^{2}}

- 2 \leq x \leq 4

Check your understanding

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

Show answer

Centre (0,-3), radius 5.

The x-bracket is x-0; the y-bracket is y-(-3); r²=25.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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 r²=25.
${(x + 2)}^{2} + {(y - 3)}^{2} = 25$ .
Does (3,4) lie on $x^{2} + y^{2} = 25$ ? Explain.
2 marks
3²+4²=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=±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 5²+3²=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 ↗

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