Circles

Intuition

Every wheel keeps the same promise: spin it through exactly one full turn, and the distance it rolls compares to its own width in exactly the same way every time, for a bicycle tire or a truck tire, a small wheel or a huge one. That link between how far a circle reaches around and how far it reaches across is what makes circles different from the straight-sided shapes in the last lesson, where perimeter was just a sum of side lengths. A circle has no straight sides to add up, so measuring the distance around it, and the space inside it, takes a new idea: a single number, called pi, that connects every circle’s size to its boundary. Once you have it, both the distance around a circle and the area inside it follow from formulas you can derive yourself. A circle itself is simply the set of all points that sit the same distance from one fixed point, the center, and holding that one distance constant is exactly what makes a wheel roll smoothly and a clock face round.

Learning goals

Name the radius, diameter and center, and use $d = 2r$
Define $\pi$ as circumference over diameter, the same for every circle
Find a circumference with $C = \pi d$ or $C = 2\pi r$
Derive $A = \pi r^2$ by rearranging thin sectors into a parallelogram
Separate a semicircle's area from its perimeter, which adds the diameter
Split a composite figure into circles and polygons, then add

The parts of a circle

Every measurement of a circle is built from a few named parts, so it is worth learning the vocabulary first. The picture below shows every part named in the list underneath it.

A circle with center O. The radius r runs from the center to the edge; the diameter d crosses through the center; the dashed segment is a chord joining two edge points without passing through the center.

The center is the fixed point that every point on the circle is the same distance from. A circle is often named by its center, like circle $O$ .
A radius is a straight segment from the center to any point on the circle. Because every point on the circle is the same distance from the center, every radius of a circle has the same length. We call that length $r$ .
A diameter is a straight segment that passes through the center and joins two points on the circle. It is a radius extended straight across to the far side, so a diameter is made of two radii laid end to end. Its length is written $d$ .
A chord is any straight segment joining two points on the circle. A diameter is the special, longest chord, the one that happens to pass through the center; a chord in general does not.
An arc is a piece of the circle itself, a curved section of the boundary between two points.
The circumference is the distance all the way around the circle. It is the circle’s version of a perimeter: the total length of its curved boundary.

The most useful relationship here is the simplest one. Since a diameter is exactly two radii in a straight line, the diameter is always twice the radius:

d = 2r \qquad \text{and equivalently} \qquad r = \frac{d}{2}.

Keep this conversion ready, because some facts about circles are stated with the radius and others with the diameter. Slipping between the two is the source of most circle mistakes.

Check your understanding

A circle is drawn with a point $O$ marked exactly in the middle, the same distance from every point on the circle. Segment $PQ$ passes through $O$ and joins two points on the circle. Segment $OR$ joins $O$ to one point on the circle. Segment $ST$ joins two other points on the circle without passing through $O$ . What is point $O$ , and what are $PQ$ , $OR$ , and $ST$ ?

O

is the center;

PQ

is a diameter;

OR

is a radius;

ST

is a chord that is not a diameter. B

O

is the center;

PQ

is a radius;

OR

is a diameter;

ST

is a chord. C

O

is just a point on the circle;

PQ

OR

, and

ST

are all chords. D

O

is the center;

PQ

OR

, and

ST

are all radii.

What pi really is

Here is the question that unlocks everything. Take any circle and walk all the way around it: that distance is the circumference $C$ . Now measure straight across it through the center: that is the diameter $d$ . How do those two lengths compare?

For every circle, no matter how big or small, the circumference is the same number of times the diameter. Roll a coin and a dinner plate each through one full turn. In both cases the distance covered is just over three times the width you rolled across. That fixed ratio of circumference to diameter is the number called pi, written with the Greek letter $\pi$ :

\pi = \frac{C}{d}.

Both $C$ and $d$ are lengths measured in the same unit, so dividing one by the other cancels that unit away entirely: $\pi$ is a bare number, not a distance.

The value of $\pi$ is about $3.14159$ , a little more than $3$ . It is not a fraction of two whole numbers, and its decimals never end and never repeat. So in practice we use a rounded value: $\pi \approx 3.14$ , or sometimes the fraction $\tfrac{22}{7}$ , which is close. The fraction is especially handy whenever the radius or diameter is a multiple of $7$ , since the $7$ then cancels cleanly and leaves a whole number: for $r = 14$ cm,

C \approx 2 \times \frac{22}{7} \times 14 = 2 \times 22 \times 2 = 88 \text{ cm},

with no decimal at all. Because we round $\pi$ , any number we compute from it is approximate, even on the step where the arithmetic itself is exact once $\pi$ has already been replaced by its rounded value. That is why a circle answer is reported as “about” a certain amount rather than as an exact one.

Unrolling a circle's boundary into a straight bar shows the diameter laid off along it. The diameter fits three whole times plus a short leftover, about 0.14 of a diameter, so the circumference is about 3.14 diameters. That ratio is pi.

Why the ratio is the same for every circle

It would be a strange coincidence if a coin and a planet happened to share the same ratio by luck. They do not: the ratio is forced to be constant, because all circles have the same shape. A small circle is just a scaled-down copy of a large one, and scaling is the key.

Why $C \div d$ is the same constant for every circle#

Take any circle and imagine enlarging it by some factor $k$ , the way a photocopier blows up an image. Every length in the figure is multiplied by the same $k$ . The radius becomes $k$ times as long, and the diameter becomes $k$ times as long. The whole boundary is scaled uniformly as well, so the circumference also becomes $k$ times as long. Scaling stretches every single part of the picture by the same amount. So scaling cannot stretch the boundary by a different factor than it stretches the width across.

Now look at the ratio of circumference to diameter for the enlarged circle. The circumference is $k \times C$ and the diameter is $k \times d$ , so their ratio is

\frac{k \times C}{k \times d} = \frac{C}{d}.

The factor $k$ cancels top and bottom. The enlarged circle has exactly the same circumference-to-diameter ratio as the one you started with. Since every circle is a scaled copy of every other circle, they must all share that one ratio. That single shared number is what we name $\pi$ .

So $\pi$ is not a property of one particular circle; it is a property of circles as a shape. That is exactly why a formula written with $\pi$ works for every circle at once.

Check your understanding

A small circle has circumference $6.28$ cm and diameter $2$ cm. A much larger circle has circumference $31.4$ m and diameter $10$ m. What does comparing $C \div d$ for each circle show?

A Both ratios come out to about

3.14

, because

\pi

is the same fixed ratio for every circle, whatever its size. B The larger circle has a bigger ratio, because it is bigger. C The ratios cannot be compared, because the units are different (cm and m). D The smaller circle has a bigger ratio, because

\pi

shrinks for smaller circles.

Circumference: the distance around

The definition of $\pi$ does the work for us. Starting from $\pi = \dfrac{C}{d}$ and multiplying both sides by $d$ gives the circumference directly:

C = \pi \, d.

The distance around a circle is $\pi$ times its diameter. And since the diameter is twice the radius ( $d = 2r$ ), we can write the same formula with the radius instead:

C = \pi \, d = \pi (2r) = 2 \pi r.

Both forms say the same thing; use whichever matches what you are given. If you know the diameter, use $C = \pi d$ . If you know the radius, use $C = 2\pi r$ .

Worked example 1 Circumference from a radius

A circular pond has a radius of $7$ m. How far is it around the pond? Use $\pi \approx 3.14$ .

The radius is given, so use $C = 2\pi r$ :

C \approx 2 \times 3.14 \times 7.

Multiply left to right. First $2 \times 3.14 = 6.28$ , then

C \approx 6.28 \times 7 = 43.96 \text{ m}.

So the distance around the pond is about $44$ m. The answer is approximate because we rounded $\pi$ to $3.14$ ; the true value is a little different.

Check your understanding

A circle has a diameter of $10$ cm. What is its circumference? Use $\pi \approx 3.14$ .

31.4 \text{ cm}

15.7 \text{ cm}

62.8 \text{ cm}

78.5 \text{ cm}

Area: the space inside

Circumference measured the boundary. Area measures the flat space the disk covers, counted in square units just as it was for polygons. The formula is

A = \pi r^2,

read as “pi r squared.” Notice it uses the radius, and the radius is squared, so this is genuinely a measure of square units, not a length. The surprise is that the same constant $\pi$ that governs the boundary also governs the area. We can see exactly why by cutting the disk, the flat region a circle encloses, into pieces and rebuilding it into a shape we already know how to measure.

A disk cut into 12 equal pie-slice sectors. Each sector has two straight edges of length r (the radius) and one curved edge along the circumference.

Slice the disk into many thin equal sectors, like the slices of a pie. Each slice is almost a thin triangle: two straight edges of length $r$ and a slightly curved base along the circle. Now pull the slices apart and lay them in a row, pointing up and down in turn, so they interlock into a long strip.

The same twelve sectors, rearranged to point up and down in turn, interlock into a strip. Its height is close to the radius r, and its base is close to half the circumference, pi times r; using more, thinner sectors would bring both closer still.

That strip is built entirely from pieces of the disk, and rearranging pieces never changes how much area they cover. So the strip has exactly the area of the disk, no matter how many sectors you use. The diagram simplifies each sector to straight edges throughout, to keep the picture easy to read; a real sector’s base is a gentle curve rather than a straight line, and using more, thinner sectors brings that curve closer and closer to straight, so the strip closes in on a true parallelogram. The proof below reasons about that limit directly, measuring the ever-more-parallelogram-like strip one dimension at a time, which pins down the disk’s area too.

Why the area of a circle is $\pi r^2$ #

So all that is left is to measure that ever-more-parallelogram-like strip, one dimension at a time.

The height of the strip is the straight edge of a sector. As the sectors get thinner, those straight edges stand up straighter, so the perpendicular height of the strip gets closer and closer to the radius $r$ .

Its base is built from the curved edges of the sectors. Half of the sectors point upward and lay their curved edges along the top. The other half point downward and lay their curved edges along the bottom instead. The full circumference is $2\pi r$ , and exactly half of it forms the bottom edge. As the sectors get thinner, each curved edge flattens toward a straight segment, so the bumpy base gets closer and closer to the straight length

\text{base} = \frac{1}{2}(2\pi r) = \pi r.

The more sectors you use, the closer the strip comes to an actual parallelogram of base $\pi r$ and perpendicular height $r$ . A parallelogram’s area is base times height, so the area of the disk is

A = \text{base} \times \text{height} = (\pi r)(r) = \pi r^2.

The base contributes one factor of $r$ through $\pi r$ , and the height contributes another factor of $r$ , which is why the radius ends up squared.

Check your understanding

In the sector-to-parallelogram argument, the strip's base gets closer to $\pi r$ and its height gets closer to $r$ as the sectors get thinner. Why does the radius end up squared in $A = \pi r^2$ ?

A The base contributes one factor of

r

(inside

\pi r

) and the height contributes a second, independent factor of

r

, so multiplying base times height multiplies those two factors together. B

\pi

itself must be squared to get an accurate area. C It is the diameter, not the radius, that gets squared. D The squaring only happens because the sectors are curved rather than straight.

So for a circle,

A = \pi r^2 \qquad (r \text{ is the radius}).

If a problem hands you the diameter instead, do not put it where the radius belongs. Halve it first to get the radius, then square. This is the single most common circle error, and the next section comes back to it.

Worked example 2 Area from a radius

A circular tabletop has a radius of $5$ cm. Find its area. Use $\pi \approx 3.14$ .

Use $A = \pi r^2$ . Square the radius first, then multiply by $\pi$ :

r^2 = 5^2 = 25.

Now multiply by $\pi$ :

A \approx 3.14 \times 25 = 78.5 \text{ cm}^2.

The tabletop covers about $78.5$ square centimeters. The unit is square centimeters because area counts unit squares, and the radius was squared in the formula.

Worked example 3 Area from a diameter (halve first)

A circular plate has a diameter of $12$ cm. Find its area. Use $\pi \approx 3.14$ .

The formula needs the radius, but we were given the diameter, so halve it first:

r = \frac{d}{2} = \frac{12}{2} = 6 \text{ cm}.

Now square the radius and multiply by $\pi$ :

A = \pi r^2 \approx 3.14 \times 6^2 = 3.14 \times 36 = 113.04 \text{ cm}^2.

So the plate covers about $113 \text{ cm}^2$ . A frequent mistake is to square the diameter by accident, $3.14 \times 12^2 = 452.16$ . That answer is four times too large: the diameter is twice the radius, so squaring it squares that factor of $2$ as well, turning it into a factor of $2^2 = 4$ .

Check your understanding

A circle has a radius of $10$ m. What is its area? Use $\pi \approx 3.14$ .

314 \text{ m}^2

62.8 \text{ m}^2

31.4 \text{ m}^2

100 \text{ m}^2

Semicircles

A semicircle is half a circle, cut along a diameter. Its area is simply half the area of the full circle:

A_{\text{semicircle}} = \frac{1}{2}\pi r^2.

Its boundary needs more care. The curved part is half the circumference, $\tfrac{1}{2}(2\pi r) = \pi r$ , but the boundary of a semicircle is not only the curve. To close the shape you also travel back along the straight diameter, which has length $d = 2r$ . So the full distance around a semicircle is the curved half plus the straight diameter:

P_{\text{semicircle}} = \pi r + 2r.

Forgetting the straight edge is a classic slip: a semicircle’s perimeter is not just half the circle’s circumference. That is because cutting a circle in half adds a new straight side that was not there before.

A semicircle's full boundary is the curved half of the circle, pi r, plus the straight diameter left behind by the cut, 2r. Tracing only the curve leaves the shape open.

Worked example 4 Perimeter of a semicircle

A semicircle has a radius of $4$ cm. Find the distance around it. Use $\pi \approx 3.14$ .

The boundary is the curved half plus the straight diameter. The curved half is $\pi r$ :

\text{curve} \approx 3.14 \times 4 = 12.56 \text{ cm}.

The straight edge is the diameter $d = 2r = 8$ cm. Add the two parts:

P \approx 12.56 + 8 = 20.56 \text{ cm}.

So the distance around the semicircle is about $20.56$ cm. Leaving out the straight diameter would give only $12.56$ cm, which traces just the curved part and never closes the shape.

Check your understanding

A semicircle has a diameter of $12$ cm. What is the distance around it? Use $\pi \approx 3.14$ .

30.84 \text{ cm}

18.84 \text{ cm}

37.68 \text{ cm}

24.84 \text{ cm}

Composite figures

Real shapes often combine a circle or semicircle with the polygons from the previous lesson. The method is the same one you used for composite polygons. Break the figure into parts you know, find each part’s area, and add them (or subtract a part that has been cut away). Area is additive, so the whole is the sum of its non-overlapping pieces.

For example, take a running track shaped like a rectangle with a semicircle capping each end. That track’s area is equal to the rectangle plus two semicircular ends. Two semicircles of the same radius together make one full circle, so the total area is the rectangle’s area plus a single circle’s area. Keep the two kinds of measurement straight throughout. Lengths and circumferences are in linear units, while areas are in square units, exactly as in the perimeter and area lesson.

The window splits into a 4 m by 5 m rectangle and a semicircle whose diameter is the shared 4 m edge, so its radius is half of that, 2 m.

Worked example 5 A rectangle with a semicircle on top

A window is a $4$ m wide rectangle standing $5$ m tall, topped by a semicircle whose diameter is the $4$ m width. Find the total area of the window. Use $\pi \approx 3.14$ .

Split the window into the rectangle and the semicircle, find each area, then add.

The rectangle is $4$ m wide and $5$ m tall:

A_{\text{rect}} = 4 \times 5 = 20 \text{ m}^2.

The semicircle sits on the $4$ m width, so that width is its diameter, and the radius is half of it:

r = \frac{4}{2} = 2 \text{ m}.

A semicircle is half a circle, so its area is

A_{\text{semi}} = \frac{1}{2}\pi r^2 \approx \frac{1}{2} \times 3.14 \times 2^2 = \frac{1}{2} \times 3.14 \times 4 = 6.28 \text{ m}^2.

Add the two non-overlapping pieces:

A \approx 20 + 6.28 = 26.28 \text{ m}^2.

So the window covers about $26.28$ square meters. The radius for the semicircle was the half-width $2$ m, not the full $4$ m width, which is the same halve-the-diameter step from before.

Check your understanding

A running track is a rectangle $10$ m by $6$ m, capped by a semicircle on one end whose diameter is the $6$ m width. What is the total area? Use $\pi \approx 3.14$ .

74.13 \text{ m}^2

60 \text{ m}^2

88.26 \text{ m}^2

116.52 \text{ m}^2

Common mistakes

Confusing radius and diameter. The diameter is twice the radius ( $d = 2r$ ). Both area and the $2\pi r$ form of circumference use the radius, so when a problem gives the diameter, halve it before substituting.
Squaring the diameter in the area formula. $A = \pi r^2$ squares the radius. Using the diameter by mistake makes the answer four times too big. That is because squaring turns the factor of $2$ between $d$ and $r$ into a factor of $4$ .
Mixing up circumference and area. Circumference $C = 2\pi r$ is a length, in linear units like cm. Area $A = \pi r^2$ is a count of unit squares, in square units like $\text{cm}^2$ . One has $r$ to the first power, the other has $r$ squared.
Treating a rounded $\pi$ as exact. Since $\pi \approx 3.14$ is rounded, any answer built from it is approximate. Write it with " $\approx$ ", and do not round $\pi$ partway and then claim an exact result.
Forgetting the straight edge of a semicircle. A semicircle’s perimeter is the curved half plus the diameter ( $\pi r + 2r$ ), because slicing a circle in half exposes a new straight side.

Key takeaways

A circle is all points a fixed distance from a center. A radius $r$ reaches from the center to the edge; a diameter $d$ crosses through the center, and $d = 2r$ .
Pi is the ratio of any circle’s circumference to its diameter, $\pi = \dfrac{C}{d} \approx 3.14$ . It is the same constant for every circle because all circles are scaled copies of one another.
Circumference: $C = \pi d = 2\pi r$ , straight from the definition of $\pi$ . It is a length.
Area: $A = \pi r^2$ , derived by slicing the disk into thin sectors that rearrange into a parallelogram with base $\pi r$ and height $r$ . It is a count of square units.
A semicircle has area $\tfrac{1}{2}\pi r^2$ and perimeter $\pi r + 2r$ (the curve plus the diameter).
For a composite figure, split it into circles, semicircles, and polygons, find each area, and add (or subtract a removed piece).

Circles

Learning goals

The parts of a circle

What pi really is

Why the ratio is the same for every circle

Circumference: the distance around

Area: the space inside

Semicircles

Composite figures

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice