A circle is the set of points a fixed distance from the centre. Write that distance with Pythagoras and the equation of a circle falls out in one line.
Today's hook, The equation of a circle is not a new fact to memorise. It is Pythagoras applied to every point on the curve at once, and the syllabus asks you to derive it that way.
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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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A point $(x, y)$ sits 5 units from the origin. Draw the right-angled triangle formed by $x$, $y$ and that distance. What does Pythagoras give you?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Pythagoras, applied to every point at once
Work through the core explanation before applying it.
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Pythagoras, applied to every point at once
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Take any point $(x, y)$ on a circle of radius $r$ centred at the origin. Dropping a perpendicular to the $x$-axis makes a right-angled triangle with legs $x$ and $y$ and hypotenuse $r$. Pythagoras gives $x^2 + y^2 = r^2$, and that holds for every point on the circle.
$x^2 + y^2 = r^2$ centre $(0,0)$, radius $r$ domain $[-r, r]$, range $[-r, r]$
The right side is $r^2$
$x^2 + y^2 = 25$ has radius 5, not 25. Take the square root to get the radius.
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 run from $-r$ to $r$.
Four easy points
The circle passes through $(r,0)$, $(-r,0)$, $(0,r)$ and $(0,-r)$. Plot those and the sketch is nearly done.
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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 of radius $r$ centred at the origin has equation $x^2 + y^2 = r^2$.
The equation comes from Pythagoras applied to the triangle with legs $x$ and $y$.
The radius is the square root of the right-hand side.
Domain and range are both $[-r, r]$, and the circle is a relation, not a function.
Understand
Concepts
Why the equation is a statement about distance rather than an arbitrary formula.
Why the same relationship holds for every point on the circle simultaneously.
Why a circle cannot be a function.
Can do
Skills
Derive $x^2 + y^2 = r^2$ from Pythagoras.
Graph a circle from its equation, showing the radius and the four axis points.
State the domain and range of a circle centred at the origin.
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Key terms
CircleThe set of all points a fixed distance from a centre. Like this: every point on $x^2 + y^2 = 25$ is exactly 5 units from the origin.
RadiusThe fixed distance from the centre to the curve, the square root of the right-hand side. Like this: $x^2 + y^2 = 49$ has radius 7.
RelationA rule linking two variables where one input may give more than one output. Like this: on $x^2+y^2=25$, $x = 0$ gives both $y = 5$ and $y = -5$.
Vertical line testThe check for whether a graph is a function; a circle fails it. Like this: a vertical line through the middle of a circle cuts it twice.
Axis pointsThe four easy points where a circle meets the axes. Like this: $x^2+y^2=9$ passes through $(3,0)$, $(-3,0)$, $(0,3)$ and $(0,-3)$.
Domain of a circleThe $x$-values the circle occupies, from $-r$ to $r$ inclusive. Like this: $x^2 + y^2 = 16$ has domain $[-4, 4]$.
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Deriving the equation from Pythagoras
Work through the core explanation before applying it.
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Deriving the equation from Pythagoras
core concept
Let $(x, y)$ be any point on a circle of radius $r$ centred at the origin. Drop a perpendicular from that point to the $x$-axis.
That makes a right-angled triangle whose horizontal leg has length $x$, whose vertical leg has length $y$, and whose hypotenuse is the radius $r$, since the radius joins the origin to the point.
Pythagoras gives $x^2 + y^2 = r^2$. Because the point was arbitrary, the relationship holds for every point on the circle, so it is the equation of the circle.
The syllabus asks you to derive it. The dot-point says "derive the equation of a circle of radius $r$ with centre at the origin by considering Pythagoras theorem", so the three-line argument is examinable, not just the formula.
Quick check: what is the radius of the circle $x^2 + y^2 = 36$?
Any point $(x,y)$ on the circle forms a right-angled triangle with legs $x$ and $y$ and hypotenuse $r$, so Pythagoras gives $x^2 + y^2 = r^2$. The point was arbitrary, so this holds for the whole circle. The radius is the square root of the right-hand side.
Pause, copy the derivation in three lines with the triangle drawn, and the reminder that the right side is $r^2$, into your book.
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Graphing a circle from its equation
core concept
We just saw where the equation comes from. That raises a question: given an equation, how do you draw the circle quickly and accurately? This card answers it → take the square root for the radius, then plot the four axis points.
From $x^2 + y^2 = 25$, the radius is $\sqrt{25} = 5$. The circle is centred at the origin with radius 5.
Plot the four points where it meets the axes: $(5,0)$, $(-5,0)$, $(0,5)$ and $(0,-5)$. Those are found by setting $y = 0$ then $x = 0$.
Draw a smooth circle through the four points. Marking the radius on the sketch is usually worth a mark.
Not every such equation is a circle. $x^2 + y^2 = 0$ is the single point at the origin, and $x^2 + y^2 = -4$ has no points at all, since a sum of squares cannot be negative.
Fill the blank: the circle $x^2 + y^2 = 81$ meets the positive $x$-axis at $x = $ .
Take the square root of the right-hand side for the radius, then plot $(r,0)$, $(-r,0)$, $(0,r)$ and $(0,-r)$ and draw through them. A zero right-hand side gives a single point and a negative one gives no graph at all.
Pause, copy the graphing method with the worked $x^2+y^2=25$, the four axis points, and the two degenerate cases, into your book.
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Domain, range, and why a circle is not a function
core concept
We just saw how to draw the circle. That raises a question: how do you describe it in the language of functions and relations? This card answers it → it is a relation, with domain and range both running from $-r$ to $r$.
The circle reaches as far left as $x = -r$ and as far right as $x = r$, so its domain is $[-r, r]$. By the same symmetry its range is $[-r, r]$.
Both are closed intervals, using square brackets, because the endpoints are genuinely on the circle: $(r, 0)$ and $(-r, 0)$ are points of the curve.
A circle fails the vertical line test, since a vertical line through its interior cuts it twice. So it is a relation but not a function. That is why the syllabus extends the definitions of domain and range to relations.
One input, two outputs. On $x^2 + y^2 = 25$, putting $x = 3$ gives $y^2 = 16$, so $y = 4$ or $y = -4$. Two outputs from one input is exactly what disqualifies it as a function.
Which statement about $x^2 + y^2 = 16$ is FALSE?
Domain and range are both $[-r, r]$, closed because the endpoints lie on the curve. A circle fails the vertical line test, since one $x$-value inside the domain gives two $y$-values, so it is a relation and not a function.
Pause, copy the domain and range in interval notation, and the worked $x = 3$ giving $y = \pm 4$ as the reason a circle is not a function, 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 · DERIVING THE EQUATION
Derive the equation of a circle of radius $r$ centred at the origin.
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Let $(x, y)$ be any point on the circle, and drop a perpendicular to the $x$-axis
This creates a right-angled triangle.
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The legs have lengths $x$ and $y$, and the hypotenuse is the radius $r$
The radius joins the origin to the point.
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By Pythagoras, $x^2 + y^2 = r^2$, and since $(x,y)$ was arbitrary this holds for every point
That is the equation of the circle.
PROBLEM 2 · GRAPHING
Graph $x^2 + y^2 = 49$, stating the radius, domain and range.
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Radius $= \sqrt{49} = 7$
The right-hand side is $r^2$.
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Axis points $(7,0)$, $(-7,0)$, $(0,7)$, $(0,-7)$
Set $y = 0$, then $x = 0$.
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Domain $[-7, 7]$, range $[-7, 7]$
One radius either side of the centre in each direction.
PROBLEM 3 · TESTING A POINT
Does the point $(3, -4)$ lie on the circle $x^2 + y^2 = 25$?
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Substitute: $3^2 + (-4)^2$
Test whether the coordinates satisfy the equation.
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$= 9 + 16 = 25$
Squaring removes the negative sign.
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It equals 25, so yes, $(3,-4)$ lies on the circle
A point is on the curve exactly when it satisfies the equation.
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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 radius of $x^2 + y^2 = 100$.
Where does $x^2 + y^2 = 36$ meet the negative $y$-axis?
Does $(5, 12)$ lie on $x^2 + y^2 = 169$?
State the domain of $x^2 + y^2 = 9$.
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Revisit your triangle
Run the quick drill and copy the summary into your book.
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Revisit your triangle
At the start you drew the triangle for a point 5 units from the origin and applied Pythagoras. Write the equation you obtained, and explain in one sentence why the same equation describes every point on the circle rather than only that one.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
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Short answer
UnderstandBand 43 marks
Q1. Derive the equation of a circle of radius $r$ centred at the origin, using Pythagoras theorem. Include a description of the triangle you use. (3 marks)
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ApplyBand 43 marks
Q2. For the circle $x^2 + y^2 = 64$, state the radius, the domain and the range, and give the coordinates of the four points where it meets the axes. (3 marks)
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UnderstandBand 32 marks
Q3. Explain why a circle centred at the origin is a relation but not a function, using a specific value of $x$ in your answer. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: 10. Practice 2: $(0, -6)$. Practice 3: $25 + 144 = 169$, so yes. Practice 4: $[-3, 3]$.
Q1 (3 marks): Let $(x,y)$ be any point on the circle, and drop a perpendicular from it to the $x$-axis [1]. This forms a right-angled triangle with horizontal leg $x$, vertical leg $y$ and hypotenuse equal to the radius $r$ [1]. By Pythagoras $x^2 + y^2 = r^2$, and since the point was arbitrary this holds for every point on the circle [1].
Q2 (3 marks): Radius $= \sqrt{64} = 8$ [1]. Domain $[-8, 8]$ and range $[-8, 8]$ [1]. Axis points $(8,0)$, $(-8,0)$, $(0,8)$ and $(0,-8)$ [1].
Q3 (2 marks): A relation is a function only if each input gives exactly one output [1]. On $x^2 + y^2 = 25$, substituting $x = 3$ gives $y^2 = 16$, so $y = 4$ or $y = -4$: one input produces two outputs, the graph fails the vertical line test, and the circle is therefore a relation but not a function [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Circle Check
earn bronze · silver · gold
Read radii, plot axis points and test whether points lie on a circle. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.