Lesson 12 ~25 min Unit 2 · Non-Linear +85 XP

Circles — $x^2 + y^2 = r^2$

From Pythagoras to the equation of a circle centred at the origin. Read radius off the equation, sketch the circle, and recognise circle equations on sight.

Today's hook: What's the radius of the circle $x^2 + y^2 = 49$? Why is the radius $7$ and not $49$?

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Think First

warm-up

A circle is the set of all points the SAME distance from the centre. Centred at the origin $(0, 0)$, a point $(x, y)$ on the circle of radius $r$ satisfies $x^2 + y^2 = r^2$ (Pythagoras). For $x^2 + y^2 = 49$: what is $r$? Where does the circle cross the $x$-axis and the $y$-axis?

Record your answer in your workbook.

The Big Idea

+5 XP

The equation $x^2 + y^2 = r^2$ is just Pythagoras' theorem in disguise. Every point $(x, y)$ at distance $r$ from the origin sits on the circle.

The red circle $x^2 + y^2 = 25$ has $r^2 = 25$, so $r = 5$. Centre $(0, 0)$. The circle crosses the axes at $(5, 0)$, $(-5, 0)$, $(0, 5)$, $(0, -5)$. Note: $r$ is the SQUARE ROOT of the number on the right — not the number itself.

$x^2 + y^2 = r^2$ — centre $(0, 0)$, radius $r$.

Right side is $r^2$

Take $\sqrt{}$ to get the actual radius.

Centre at origin

When no shifts appear, centre is $(0, 0)$.

Both terms squared

$x^2$ AND $y^2$, with equal coefficients of $+1$.

What You'll Master

objectives

Know

$x^2 + y^2 = r^2$ is a circle centred at $(0, 0)$ with radius $r$
The right side is $r^2$, so take the square root to get $r$
The circle crosses the axes at $(\pm r, 0)$ and $(0, \pm r)$

Understand

Why the equation comes from Pythagoras: $x^2 + y^2 = r^2$ measures squared distance from the origin
Why we need both $x^2$ and $y^2$ (one would give a parabola, not a circle)
How to recognise a circle equation versus a parabola equation

Can Do

Read the radius and centre from an equation
Sketch a circle given its equation
Test whether a given point lies on a circle

Words You Need

vocabulary

CircleThe set of all points a fixed distance from a fixed centre.

CentreThe fixed point inside — here, the origin $(0, 0)$.

RadiusThe distance from the centre to any point on the circle.

Pythagoras' theorem$a^2 + b^2 = c^2$ for a right-angled triangle. The circle equation is this with $a = x$, $b = y$, $c = r$.

Equation of a circle$x^2 + y^2 = r^2$ (centre at origin).

On / inside / outsidePoint $(x, y)$ is ON if $x^2 + y^2 = r^2$, INSIDE if $< r^2$, OUTSIDE if $> r^2$.

Spot the Trap

heads-up

Wrong: Saying $x^2 + y^2 = 16$ has radius 16.

Right: Right side is $r^2 = 16$, so $r = \sqrt{16} = 4$.

Wrong: Calling $y = x^2$ a circle.

Right: $y = x^2$ is a PARABOLA. A circle needs BOTH $x^2$ AND $y^2$, with the same coefficient.

Reading and Sketching

+5 XP

Given $x^2 + y^2 = r^2$, sketching is a 3-step ritual:

Find $r$: Take the square root of the right side. e.g. $x^2 + y^2 = 36 \Rightarrow r = 6$.
Plot the 4 axis points: $(\pm r, 0)$ and $(0, \pm r)$.
Draw a smooth circle through them, centred at the origin.

For a quick check: pick a point ON the circle, sub it in. e.g. $(3, 4)$: $3^2 + 4^2 = 9 + 16 = 25$, so this point lies on $x^2 + y^2 = 25$.

Find $r$ $\to$ plot 4 axis points $\to$ draw the circle.

Don't forget $\sqrt{}$

$r = \sqrt{r^2}$, not the right-side value itself.

Compass-smooth

Round and symmetric — no flat sides.

Test a point

$(3, 4)$ is on $r = 5$ because $9 + 16 = 25$.

Recognising Circle Equations

+5 XP

Three checks for "is this a circle centred at the origin?":

Both $x^2$ and $y^2$ present: yes / no?
Same coefficient on both (and positive): e.g. both $+1$? If $x^2 + y^2 = c$ with $c > 0$, it's a circle.
Right side is positive: $r^2 > 0$. If $r^2 = 0$, just the origin. If $r^2 < 0$, no real solutions.

$x^2 + y^2 = 25$: circle, $r = 5$. $y = x^2$: parabola. $x^2 - y^2 = 25$: not a circle (different signs — that's a hyperbola at extension level). $x^2 + y^2 = -4$: no real solutions.

Circle $\iff x^2 + y^2 = r^2$ with $r^2 > 0$.

$+$ in middle

$x^2 + y^2$, not $x^2 - y^2$ or $y^2 - x^2$.

Equal coefficients

$x^2 + 4y^2 = 16$ is an ellipse, not a circle.

$r^2$ must be positive

A negative right side has no real graph.

Watch Me Solve It · 3 examples

Watch Me Solve It · Sketch $x^2 + y^2 = 49$

+15 XP per step

PROBLEM

State the centre and radius of $x^2 + y^2 = 49$, then sketch it labelling the axis crossings.

1
Find $r$
$r^2 = 49 \Rightarrow r = \sqrt{49} = 7$.
2
State centre and axis points
Centre $(0, 0)$. Axis crossings: $(7, 0)$, $(-7, 0)$, $(0, 7)$, $(0, -7)$.
3
Sketch
Smooth circle of radius 7 around the origin, through the four labelled points.
Always take the square root — the radius is 7, not 49.

AnswerCentre $(0, 0)$, $r = 7$.

Watch Me Solve It · Equation from radius

+15 XP per step

PROBLEM

Write the equation of the circle centred at the origin with radius 8.

1
Use the form
Centre $(0, 0)$: $x^2 + y^2 = r^2$.
2
Square the radius
$r = 8 \Rightarrow r^2 = 64$.
3
Write the equation
$x^2 + y^2 = 64$.
Going IN to the equation, you square the radius; going OUT, you take the square root.

Answer$x^2 + y^2 = 64$.

Watch Me Solve It · On, inside, or outside?

+15 XP per step

PROBLEM

Does the point $(3, 4)$ lie on, inside, or outside the circle $x^2 + y^2 = 25$?

1
Compute $x^2 + y^2$
$3^2 + 4^2 = 9 + 16 = 25$.
2
Compare with $r^2$
$x^2 + y^2 = 25 = r^2$.
3
Conclude
Equality $\Rightarrow$ the point lies ON the circle.
$<$ would mean inside; $>$ would mean outside.

Answer$(3, 4)$ lies ON the circle.

Common Pitfalls

heads-up

Confusing $r$ and $r^2$

For $x^2 + y^2 = 36$ saying radius $= 36$.

Fix: the right side is $r^2$. Take the square root: $r = 6$.

Squaring instead of square-rooting

Asked for the equation of radius 5, writing $x^2 + y^2 = \sqrt{5}$.

Fix: $r = 5 \Rightarrow r^2 = 25$, so $x^2 + y^2 = 25$.

Mistaking $x^2 - y^2$ for a circle

Reading $x^2 - y^2 = 9$ as "a circle with $r^2 = 9$".

Fix: a circle MUST have $x^2 + y^2$ (same sign, same coefficient). Otherwise it's a different curve.

Copy Into Your Books

Circle equation

$x^2 + y^2 = r^2$
Centre $(0, 0)$
Radius $r$

Reading $r$

Right side is $r^2$
$r = \sqrt{r^2}$
Don't skip the square root

Sketch steps

Find $r$
Plot $(\pm r, 0)$, $(0, \pm r)$
Smooth circle through 4 points

Point test

$= r^2$: ON
$< r^2$: INSIDE
$> r^2$: OUTSIDE

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Circle Drills

4 problems

Four quick circle reads.

1 State the radius of $x^2 + y^2 = 100$.
$r^2 = 100$.$r = 10$
2 Write the equation of the circle centred at the origin with radius 3.
$r^2 = 9$.$x^2 + y^2 = 9$
3 Does $(0, 4)$ lie on $x^2 + y^2 = 16$?
$0 + 16 = 16 = r^2$.Yes (ON the circle)
4 Is $(1, 2)$ inside or outside $x^2 + y^2 = 9$?
$1 + 4 = 5 < 9$.INSIDE

Complete in your workbook.

Quick Check · 5 questions

The radius of the circle $x^2 + y^2 = 49$ is:

+10 XP

The equation of the circle centred at the origin with radius 6 is:

+10 XP

The point $(5, 12)$ lies on the circle $x^2 + y^2 = r^2$. The radius is:

+10 XP

The point $(-2, 1)$ relative to the circle $x^2 + y^2 = 9$ is:

+10 XP

Which equation represents a circle centred at the origin?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyEasy3 MARKS

Q6. For $x^2 + y^2 = 81$, state the centre, the radius, and the four axis crossings.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Write the equation of a circle centred at the origin which passes through the point $(6, 8)$.

Answer in your workbook.

ReasonHard3 MARKS

Q8. For the circle $x^2 + y^2 = 25$, decide whether each of these points lies on, inside, or outside: (a) $(3, 4)$, (b) $(2, 2)$, (c) $(-5, 1)$. Show your substitution working for each.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — $r = \sqrt{49} = 7$.

2. B — $r = 6 \Rightarrow r^2 = 36$. $x^2 + y^2 = 36$.

3. A — $5^2 + 12^2 = 169 = 13^2$, so $r = 13$.

4. D — $4 + 1 = 5 < 9$, inside.

5. B — only $x^2 + y^2 = 16$ has the right form with $r^2 > 0$.

Show Your Working Model Answers

Q6 (3 marks): $r^2 = 81 \Rightarrow r = 9$ [1]. Centre $(0, 0)$ [1]. Axis crossings $(9, 0)$, $(-9, 0)$, $(0, 9)$, $(0, -9)$ [1].

Q7 (3 marks): Since $(6, 8)$ lies on the circle, $r^2 = 6^2 + 8^2$ [1]. $= 36 + 64 = 100$ [1]. Equation: $x^2 + y^2 = 100$ (so $r = 10$) [1].

Q8 (3 marks): (a) $9 + 16 = 25 = r^2$, ON the circle [1]. (b) $4 + 4 = 8 < 25$, INSIDE [1]. (c) $25 + 1 = 26 > 25$, OUTSIDE [1].

Stretch Challenge · +25 XP, +10 coins

Half Circles — Upper and Lower

Rearrange $x^2 + y^2 = 25$ to make $y$ the subject. (a) Show $y = \pm\sqrt{25 - x^2}$. (b) Explain why $y = \sqrt{25 - x^2}$ alone is the UPPER half of the circle (a semicircle). (c) What restrictions on $x$ are needed for $y$ to be a real number? (d) Sketch $y = \sqrt{25 - x^2}$ — what shape do you get?

Reveal solution

(a) $y^2 = 25 - x^2 \Rightarrow y = \pm\sqrt{25 - x^2}$. (b) The $+$ root only gives non-negative $y$, so it's the UPPER semicircle. (c) Need $25 - x^2 \geq 0 \Rightarrow x^2 \leq 25 \Rightarrow -5 \leq x \leq 5$. (d) Sketch is the upper half of the circle of radius 5, like a dome.

Quick Review

Equation

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

Centre

$(0, 0)$

Radius

$r = \sqrt{r^2}$

Axis points

$(\pm r, 0)$, $(0, \pm r)$

Point test

$< r^2$: in. $= r^2$: on. $> r^2$: out.

Origin

From Pythagoras

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