A circle is the one graph in this course that is not a function, but every transformation rule you have learned still applies to it. Translating the unit circle is all you need.
Today's hook, You already know that replacing $x$ by $x - h$ slides a graph right by $h$. Apply that to $x^2 + y^2 = r^2$ and you can place a circle anywhere on the plane without learning a single new idea.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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The circle $x^2 + y^2 = 25$ is centred at the origin with radius 5. Where do you think $(x-3)^2 + (y+2)^2 = 25$ is centred, and why? Watch the sign carefully.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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A circle is the unit circle, translated and scaled
Work through the core explanation before applying it.
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A circle is the unit circle, translated and scaled
+5 XP to read
In $(x-h)^2 + (y-k)^2 = r^2$ the centre is $(h, k)$ and the radius is $r$. The signs in the equation are opposite to the signs in the centre, so $(x+2)^2$ means the centre has $x$-coordinate $-2$.
$(x-h)^2 + (y-k)^2 = r^2$ centre $(h, k)$, radius $r$ domain $[h-r,\ h+r]$ range $[k-r,\ k+r]$
The sign flips
$(x - 3)^2$ gives centre $x = 3$; $(x + 3)^2$ gives centre $x = -3$. Read the sign inside the bracket and reverse it.
It is $r^2$, not $r$
The right-hand side is the radius squared. $(x-1)^2 + y^2 = 9$ has radius 3, not 9.
A circle is not a function
It fails the vertical line test, so it is a relation. Domain and range are still defined, and both are closed intervals.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
A circle with centre $(h, k)$ and radius $r$ has equation $(x-h)^2 + (y-k)^2 = r^2$.
The signs of $h$ and $k$ in the equation are the reverse of the coordinates of the centre.
The domain is $[h-r,\ h+r]$ and the range is $[k-r,\ k+r]$.
The general form $x^2 + y^2 + ax + by + c = 0$ becomes centre-radius form by completing the square twice.
Understand
Concepts
Why the translation rules for functions apply unchanged to a circle.
Why the sign in the bracket is opposite to the coordinate of the centre.
Why a circle has a closed interval for both domain and range.
Can do
Skills
State the centre, radius, domain and range from an equation in centre-radius form.
Convert general form to centre-radius form by completing the square twice.
Graph a circle from its equation, and write the equation from a graph.
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Key terms
Centre-radius formThe equation of a circle written so the centre and radius can be read straight off. Like this: $(x-2)^2 + (y+1)^2 = 16$ has centre $(2, -1)$ and radius 4.
General formThe expanded equation of a circle, where the centre is hidden until you complete the square. Like this: $x^2 + y^2 - 4x + 2y - 11 = 0$ is the same circle as above.
RadiusThe distance from the centre to any point on the circle, which is the square root of the right-hand side. Like this: if the equation ends in $= 25$, the radius is 5.
RelationA rule linking two variables where an input may give more than one output, so it need not be a function. Like this: a circle is a relation because $x = 0$ on $x^2+y^2=25$ gives both $y = 5$ and $y = -5$.
Domain of a circleThe set of $x$-values the circle occupies, running one radius either side of the centre. Like this: centre $(2,-1)$ with radius 4 gives domain $[-2, 6]$.
Range of a circleThe set of $y$-values the circle occupies, running one radius above and below the centre. Like this: centre $(2,-1)$ with radius 4 gives range $[-5, 3]$.
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Reading the centre and radius from the equation
Work through the core explanation before applying it.
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Reading the centre and radius from the equation
core concept
Start from $x^2 + y^2 = r^2$, a circle centred at the origin. Replacing $x$ by $x - h$ slides it right by $h$, and replacing $y$ by $y - k$ slides it up by $k$. That is the same rule you used for parabolas.
So $(x-h)^2 + (y-k)^2 = r^2$ has centre $(h, k)$ and radius $r$. Read the signs carefully: $(x+2)^2$ is $(x - (-2))^2$, so the centre has $x$-coordinate $-2$.
The right-hand side is the radius **squared**. If the equation reads $= 9$, the radius is 3.
Domain and range come free. The circle reaches exactly one radius either side of the centre, so the domain is $[h-r,\ h+r]$ and the range is $[k-r,\ k+r]$. Both are closed intervals, because the endpoints are on the circle.
Quick check: what is the centre of $(x+3)^2 + (y-1)^2 = 16$?
$(x-h)^2 + (y-k)^2 = r^2$ has centre $(h,k)$ and radius $r$. The sign inside each bracket is the reverse of the coordinate. The right-hand side is $r^2$, not $r$. Domain is $[h-r, h+r]$ and range is $[k-r, k+r]$.
Pause, copy the centre-radius form, the sign-reversal warning with $(x+2)^2$ giving centre $-2$, and the domain and range formulas, into your book.
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Completing the square twice
core concept
We just saw how to read a circle straight off its centre-radius form. That raises a question: what if the equation arrives expanded, as $x^2 + y^2 - 4x + 2y - 11 = 0$, with no brackets to read? This card answers it → complete the square on the $x$ terms and again on the $y$ terms, which rebuilds the brackets.
Group the $x$ terms and the $y$ terms, and move the constant to the right: $x^2 - 4x + y^2 + 2y = 11$.
Complete the square on each group separately, using the same halve-and-square method as for quadratics. For $x^2 - 4x$, half of $-4$ is $-2$, so add $4$. For $y^2 + 2y$, half of 2 is 1, so add 1.
Whatever you add on the left must be added on the right: $(x-2)^2 + (y+1)^2 = 11 + 4 + 1 = 16$. So the centre is $(2, -1)$ and the radius is 4.
Add to both sides. The most common error is completing the square on the left and forgetting to add the same amounts on the right, which changes the radius.
Fill the blank: to complete the square on $x^2 - 6x$ you add .
From general form, group the $x$ terms and the $y$ terms and move the constant right. Complete the square on each group by halving and squaring the coefficient, adding the same amount to both sides. The result is centre-radius form.
Pause, copy the two-group method and the fully worked $x^2 + y^2 - 4x + 2y - 11 = 0$ becoming $(x-2)^2 + (y+1)^2 = 16$, into your book.
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Graphing a circle, and reading its equation from a graph
core concept
We just saw how to recover the centre and radius from any form of the equation. That raises a question: how do you turn that into an accurate sketch, and how do you go the other way from a picture? This card answers it → plot the centre, step out one radius in four directions, and reverse the process to read an equation off a graph.
To graph: plot the centre, then mark the four points one radius directly left, right, above and below it. Draw the circle through those four points.
To go the other way, read the centre and radius off the diagram and substitute into $(x-h)^2 + (y-k)^2 = r^2$. If the centre is $(-1, 2)$ and the radius is 3, the equation is $(x+1)^2 + (y-2)^2 = 9$.
If the radius is not marked, find it from the distance between the centre and any point the circle passes through.
Check with the intercepts. A circle centred at $(2,-1)$ with radius 4 must reach $x = -2$ and $x = 6$. If your sketch does not, the radius or the centre is wrong.
Which is NOT true of the circle $(x-1)^2 + (y+2)^2 = 25$?
Graph by plotting the centre and stepping one radius left, right, up and down. To read an equation from a graph, take the centre and radius off the diagram and substitute into the centre-radius form. Check the sketch against the domain and range.
Pause, copy the four-point graphing method, one worked equation-from-graph such as centre $(-1,2)$ radius 3 giving $(x+1)^2+(y-2)^2=9$, and the intercept check, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · READING CENTRE, RADIUS, DOMAIN AND RANGE
For $(x-2)^2 + (y+1)^2 = 16$, state the centre, radius, domain and range.
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Centre is $(2, -1)$
Reverse the sign inside each bracket: $x-2$ gives 2, and $y+1$ gives $-1$.
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Radius is $\sqrt{16} = 4$
The right-hand side is $r^2$, so take the square root.
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Domain $[-2, 6]$, range $[-5, 3]$
One radius either side of each coordinate of the centre: $2 \pm 4$ and $-1 \pm 4$.
PROBLEM 2 · COMPLETING THE SQUARE TWICE
Find the centre and radius of $x^2 + y^2 - 4x + 2y - 11 = 0$.
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$x^2 - 4x + y^2 + 2y = 11$
Group the $x$ terms and the $y$ terms; move the constant right.
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$(x-2)^2 - 4 + (y+1)^2 - 1 = 11$
Half of $-4$ is $-2$; half of 2 is 1. Subtract what you added.
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$(x-2)^2 + (y+1)^2 = 16$, so centre $(2,-1)$ and radius 4
Collect the constants on the right.
PROBLEM 3 · EQUATION FROM A GRAPH
A circle has centre $(-1, 2)$ and passes through the point $(2, 2)$. Find its equation.
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Radius is the distance from $(-1,2)$ to $(2,2)$
The two points share a $y$-coordinate, so the distance is the difference in $x$.
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$r = 2 - (-1) = 3$
Radius 3, so $r^2 = 9$.
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$(x+1)^2 + (y-2)^2 = 9$
Substitute $h=-1$, $k=2$ into the centre-radius form, reversing the signs.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
State the centre and radius of $(x-5)^2 + (y-3)^2 = 49$.
State the centre and radius of $(x+2)^2 + y^2 = 9$.
Find the centre and radius of $x^2 + y^2 - 6x = 0$.
Write the equation of the circle with centre $(0, -4)$ and radius 6.
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Revisit your prediction
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Revisit your prediction
At the start you predicted the centre of $(x-3)^2 + (y+2)^2 = 25$. Confirm it now, and explain in one sentence why the sign in the bracket is the reverse of the coordinate.
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Multiple choice
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Multiple choice
+5 XP per correct · +25 XP all-correct
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. For the circle $(x+4)^2 + (y-1)^2 = 36$, state the centre, the radius, the domain and the range. (3 marks)
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ApplyBand 54 marks
Q2. Find the centre and radius of $x^2 + y^2 + 6x - 8y + 9 = 0$ by completing the square. (4 marks)
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UnderstandBand 32 marks
Q3. Explain why a circle is a relation but not a function, and state what its domain and range are in terms of the centre and radius. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: centre $(5,3)$, radius 7. Practice 2: centre $(-2,0)$, radius 3. Practice 3: $(x-3)^2 + y^2 = 9$, centre $(3,0)$, radius 3. Practice 4: $x^2 + (y+4)^2 = 36$.
Q1 (3 marks): Centre $(-4, 1)$ [1]. Radius $\sqrt{36} = 6$ [1]. Domain $[-10, 2]$ and range $[-5, 7]$ [1].
Q3 (2 marks): A vertical line through the interior of a circle cuts it twice, so one $x$-value gives two $y$-values and it fails the vertical line test [1]. Domain is $[h-r,\ h+r]$ and range is $[k-r,\ k+r]$, where $(h,k)$ is the centre and $r$ the radius [1].
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Review and finish
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Boss battle · Circle Sweep
earn bronze · silver · gold
Read centres, radii and equations of circles at speed, including from general form. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.