Math 6

What a circle is

A triangle has three sides and a square has four corners, but a circle has no sides and no corners at all. So what exactly makes a shape a circle?

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Where you are 15 in Two-dimensional shapes

Geometry11

Two-dimensional shapes15

▴ 1 above Polygons3 Position words: above, below, beside, in front of, behind

What a circle is

Composing and decomposing shapes Quadrilaterals7 ▾ Show all 15 (9 more)

Why this exists

Shapes like triangles and squares are named by counting: how many straight sides, how many corners. A circle gives you nothing to count. Its edge is one smooth curve that never turns a corner, so the usual way of sorting shapes has nothing to hold on to.

Yet everyone knows a circle when they see one, and a wheel shows why the idea matters. A wheel rolls smoothly because the axle is the same distance from every part of the rim. Make the rim even a little lopsided and the cart bumps up and down as it rolls.

That one fact is the whole idea. Tie a string to a peg, pull it tight, and walk around the peg: the path you trace keeps exactly the length of the string between you and the peg. A compass draws a circle the same way, with a fixed gap between its point and its pencil. So the honest description of a circle is not about sides at all. It is about distance from one point.

Definition

A circle is the set of all points in a flat plane that are the same distance from one fixed point. The fixed point is the center, and the fixed distance is the radius.

A circle is named by its center. The circle with center O is written \odot O and read “circle O”.

Three kinds of segment come with every circle (Figure 1).

A radius is a segment from the center to a point on the circle, such as \overline{OA}. The word also means its length, written r.
A chord is a segment whose two endpoints are both on the circle, such as \overline{DE}.
A diameter is a chord that passes through the center, such as \overline{BC}. The word also means its length, written d.

Figure 1A radius, a diameter and a chord. Every point named on the curve is the same distance from O.

The circle is only the curve. The center is not on the circle, and neither is any point between the center and the curve. Those points are inside the circle, and the points beyond the curve are outside it. When a problem means the curve together with everything inside it, the region is called a disc. A pizza is a disc; its crust is the circle.

Circles that share a center but have different radii are called concentric, like the rings on a target.

Intuition

The definition turns every question about a circle into a question about one distance: how far is the point from the center? Write OP for the distance from the center O to a point P. There are only three possibilities, and each one is a place.

Figure 2Compare each distance with the radius 5. Less is inside, equal is on, more is outside.

In Figure 2 the radius is 5 cm. The point A is 3 cm from O, so it is closer than the curve: inside. B is exactly 5 cm away, so it is on the circle. C is 7.5 cm away, past the curve: outside.

The same picture explains the other two segments. A diameter is just two radii laid end to end in a straight line through the center, so it is twice as long as a radius. A chord that misses the center cuts across the circle without going through the middle, so it is shorter than the diameter: the diameter is the longest straight walk from one point of the circle to another.

Results

Result 1 (all radii are equal). Every radius of a circle has the same length, r.

Result 2 (diameter and radius). Every diameter is twice the radius:

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

Result 3 (the center halves every diameter). The center of a circle is the midpoint of every diameter.

Result 4 (the longest chord). Every chord of a circle is at most 2r long. A chord is exactly 2r long only when it is a diameter.

Result 5 (inside, on or outside). For a circle with center O and radius r, and any point P in its plane:

Distance from the center	Where P is
OP < r	inside the circle
OP = r	on the circle
OP > r	outside the circle

Result 6 (same radius, same circle). Two circles with the same radius are the same size and shape: slide one so that its center lands on the other’s, and the two curves match exactly.

Why it is true

Results 1 and 5 are the definition itself, read carefully. Every point on the circle is at distance r from the center, so every radius has length r; and a point is on the circle exactly when its distance is r, so any other distance puts it inside or outside.

Results 2 and 3. Let \overline{BC} be a diameter of \odot O. It passes through O, so O lies on the segment between B and C, and the whole is the sum of its parts:

BC = BO + OC

Both \overline{BO} and \overline{OC} join the center to a point on the circle, so both are radii and both have length r. So BC = r + r = 2r. And since BO = OC, the point O splits \overline{BC} into two equal halves: it is the midpoint.

Result 4. Take a chord \overline{DE} that does not pass through O, and draw the radii \overline{OD} and \overline{OE} (Figure 3).

Figure 3The chord is one side of a triangle whose other two sides are radii, so it is shorter than r + r.

The points O, D and E do not lie on one line. (A straight line through the center meets the circle at the two ends of a diameter, so if O were on line DE the chord would be a diameter.) So they are the corners of a triangle, and in any triangle one side is shorter than the other two put together. That is the triangle inequality, and it gives

DE < OD + OE = r + r = 2r

So every chord that misses the center is shorter than 2r, and by Result 2 every chord through the center is exactly 2r.

Result 6. Sliding a shape does not stretch it, so every distance stays the same. After the slide both circles have the same center, and each is the set of points at distance r from that center. They are the same set of points, so they are the same circle.

These are among the oldest facts in mathematics. Euclid opens the Elements by defining the circle in the same spirit, by equal distance from one point inside (his circle is the region the curve encloses, where ours is only the curve), and his very first construction (Example 7 below) uses nothing else.

Worked examples

Example 1

Naming the parts

Figure 4 shows \odot K with points P, Q, R, S and T on it and a point M inside it. Name every radius, every chord and every diameter that is drawn, and say whether \overline{KM} is a radius.

Figure 4Name each segment by what its two ends are.

Solution

Name each segment by where its two ends are.

\overline{KP} runs from the center to a point on the circle, so it is a radius.

\overline{QS} has both ends on the circle and passes through K, so it is a diameter. A diameter is a chord as well, so it counts in both lists. Its two halves, \overline{KQ} and \overline{KS}, each run from the center to the circle, so they are radii too.

\overline{RT} has both ends on the circle but does not pass through K, so it is a chord and not a diameter.

\overline{KM} starts at the center, but M is inside the circle, not on it. So \overline{KM} is not a radius, and it is shorter than one.

The radii are \overline{KP}, \overline{KQ} and \overline{KS}; the chords are \overline{QS} and \overline{RT}; the only diameter is \overline{QS}.

Example 2

From diameter to radius and back

(a) A round rug measures 3.6 m straight across through its center. Find its radius. (b) A circle has radius 2\tfrac{3}{4} inches. Find its diameter.

Solution

(a) A measurement straight across through the center is a diameter, so d = 3.6 m. By Result 2 the radius is half of it:

r = \frac{d}{2} = \frac{3.6}{2} = 1.8 \text{ m}

(b) The diameter is two radii end to end, so double it. It is easiest to double the whole number and the fraction separately:

\begin{aligned} d = 2r &= 2 \times 2\tfrac{3}{4} \\ &= 4 + \tfrac{6}{4} \\ &= 4 + 1\tfrac{1}{2} \\ &= 5\tfrac{1}{2} \text{ inches} \end{aligned}

Check. Half of 5\tfrac{1}{2} is 2\tfrac{3}{4}, the radius we started with.

Example 3

Inside, on or outside

\odot O has radius 5 cm. Points A, B, C and D have OA = 3 cm, OB = 5 cm, OC = 7.5 cm and OD = 10 cm. Say where each point is, and where the center O itself is.

Solution

Compare each distance with the radius, as Result 5 says.

OA = 3 < 5, so A is inside.
OB = 5, exactly the radius, so B is on the circle.
OC = 7.5 > 5, so C is outside.
OD = 10 > 5, so D is outside. It is tempting to say “on”, because 10 cm is the diameter, but the diameter is a distance across the circle. The distance that decides is always the one from the center, and D is twice as far from O as the curve is.
The distance from O to itself is 0, and 0 < 5, so the center is inside the circle, not on it.

Example 4

A circle on a grid

On a coordinate grid, a circle has center C(2, 1) and passes through A(6, 1). (a) Find its radius. (b) Find three more points on the circle whose coordinates are both integers. (c) Find the other end of the diameter that starts at A. (d) Is D(5, 1) inside, on or outside the circle?

Solution

(a) A and C are on the same horizontal line, y = 1, so the distance between them is the difference of the x-coordinates: 6 - 2 = 4. A is on the circle, so the radius is 4.

(b) Every point 4 units from C is on the circle. Moving 4 along a grid line from C gives three more: 4 left to (-2, 1), 4 up to (2, 5) and 4 down to (2, -3).

(c) By Result 3 the center is the midpoint of the diameter, so the far end is as far past C as A is before it. From A to C is 4 to the left; another 4 to the left reaches (-2, 1). That is the point already found in (b).

(d) D is also on the line y = 1, and CD = 5 - 2 = 3. Since 3 < 4, D is inside.

The circle passes through many more points than these four, but the others are not on grid lines through C, and measuring a slanted distance needs the distance formula.

Example 5

What length can a chord be?

(a) A circle has radius 6 cm. Could it have a chord 13 cm long? 12 cm long? 8 cm long? (b) A chord of some circle is 10 cm long. What can you say about the radius?

Solution

(a) By Result 4 no chord is longer than the diameter, d = 2 \times 6 = 12 cm.

13 cm is longer than 12 cm, so no chord of this circle can be 13 cm.
12 cm is exactly the diameter, so a chord of that length must pass through the center: it is a diameter.
8 cm is less than 12 cm, so it is possible. Such chords miss the center.

(b) Read Result 4 backwards. The chord can be no longer than the diameter, so 10 \le 2r, which gives r \ge 5. The radius is at least 5 cm, and it is exactly 5 cm only if the chord is a diameter.

Example 6

Two circles with one center

Two concentric circles have center O and radii 3 cm and 7 cm. (a) A point P has OP = 5 cm. Where is it? (b) How wide is the ring between the circles? (c) A straight line through O crosses the large circle at A and B and the small circle at C and D, in the order A, C, D, B. Find AB, CD and AC.

Solution

(a) 5 > 3, so P is outside the small circle; 5 < 7, so it is inside the large one. It lies in the ring between them.

(b) Along any line from the center, the small circle is 3 cm out and the large one 7 cm out, so the ring is 7 - 3 = 4 cm wide.

(c) The line passes through the center, so \overline{AB} is a diameter of the large circle and \overline{CD} a diameter of the small one:

AB = 2 \times 7 = 14 \text{ cm}

CD = 2 \times 3 = 6 \text{ cm}

\overline{AC} runs from the large circle in to the small one along a line through the center, so AC = OA - OC = 7 - 3 = 4 cm.

Check. The two end pieces \overline{AC} and \overline{DB} are equal, and together with \overline{CD} they make up \overline{AB}: \displaystyle \frac{14 - 6}{2} = 4.

Example 7

A triangle from two circles

Draw a circle with center A and radius 5 cm. Mark a point B on it, and draw a second circle with center B and the same radius. The circles cross at P (Figure 5). Show that triangle ABP has three equal sides, and find its perimeter.

Figure 5Each side of the shaded triangle is a radius of one circle or the other.

Solution

Each side of the triangle is a radius of one of the circles, so Result 1 does all the work.

B is on the circle with center A, so AB is a radius of it: AB = 5 cm.
P is on the circle with center A, so AP = 5 cm.
P is also on the circle with center B, so BP = 5 cm.

All three sides are 5 cm, so the triangle is equilateral, and its perimeter is 3 \times 5 = 15 cm.

Nothing was measured. The two circles force the three lengths to match, which is why this is how an equilateral triangle is constructed with a compass and straightedge. It is the first proposition in Euclid’s Elements.

Boundaries

A circle lives in a flat plane. In space, the points at one fixed distance from a center do not form a curve but a whole hollow ball, a sphere. The rule is the same; the setting changes the shape.

The radius must be more than zero. The only point at distance 0 from O is O itself, so “radius 0” gives a single point, not a circle.

The circle is the curve, not the region. The center and every other inside point are not on the circle. A question about the region means the disc, and it should say so.

A segment’s ends decide its name. A segment from the center to a point that is not on the circle is not a radius. A segment with only one end on the circle is not a chord. A chord is a diameter only if it really passes through the center, which a drawing alone cannot promise: it has to be stated or marked.

Round is not enough. An oval looks round, but its middle is closer to some parts of its edge than to others, so it is not a circle. There are even curved shapes, such as one made from three arcs around a triangle, that are the same width in every direction and still are not circles. What defines a circle is equal distance from one center, not equal width across.

Distance means straight-line distance. If distance were measured only along the lines of a grid, like a taxi driving on city blocks, the points 4 blocks from a corner would form a tilted square, not a circle. Every result here assumes the ordinary straight-line distance.

Where this goes next

Everything about circles grows from this one definition. The link between the two lengths you met here, and how to use them, is radius and diameter. Measuring around the curve leads to the number \pi and the circumference, and measuring the disc inside leads to the area of a circle.

Chords open up a world of their own: chords, secants and tangents, the arcs they cut off in arcs and central angles, and theorems such as a radius at right angles to a chord cuts it in half and a tangent meets the radius at a right angle. All of them are gathered in Circles.

The definition is also the first example of a locus, a shape described by a rule its points obey, and the compass that draws it is the main tool of compass and straightedge constructions. Written with coordinates, the same rule becomes the equation of a circle. And the circles that form the two ends of a can are where Cylinders and cones begin.

Exercises

Naming the parts

Use Figure 6 for Exercises 1 to 5.

Figure 6For Exercises 1 to 5.

Name the center and the diameter that is drawn.
Name every radius that is drawn.
Name every chord that is drawn.
Which of the labeled points lie on the circle, which inside it and which outside it?
Is \overline{OF} a radius? Explain.

Radius and diameter

A circle has radius 9 cm. Find its diameter.
A circle has diameter 15 m. Find its radius.
A circle has radius 4.25 inches. Find its diameter.
A circle has diameter 3\tfrac{1}{3} feet. Find its radius.
A round table top is 1.2 m across through its center. How far is it from the center of the table to its edge?

Inside, on or outside

For Exercises 11 to 13, \odot O has radius 6 cm. Say whether each point is inside, on or outside the circle.

OP = 4 cm
OQ = 6 cm
OR = 6.5 cm
A circle with center O has diameter 10 cm, and OS = 5 cm. Is S inside, on or outside the circle?
On a grid, a circle has center (3, 2) and passes through (3, 7). Find its radius, then say whether each point is inside, on or outside: (a) (-2, 2), (b) (3, -1), (c) (9, 2).

Chords and diameters

A circle has radius 8 cm. Which of these could be the length of one of its chords: 3 cm, 16 cm, 17 cm? Explain each.
A chord of a circle is 14 cm long. What is the smallest the radius could be?
On a grid, a circle has center C(1, 2), and one end of a diameter is A(4, 6). Find the other end.
Two concentric circles have radii 4 cm and 9 cm. A point P is 7 cm from their center. Where is P, and how wide is the ring between the circles?

Using the definition

Two circles each have radius 3 cm. Their centers A and B are 4 cm apart, and the circles cross at P and Q. Find the perimeter of the four-sided shape APBQ.
A goat is tied to a post by a rope 5 m long. Describe the edge of the patch of grass the goat can reach, and find the greatest distance between two points it can reach.

Reasoning

Explain why every diameter of a circle has the same length.
Explain why the center of a circle is never on the circle.
Explain why a chord that does not pass through the center is shorter than a diameter.
Points A and B are on \odot O, and the chord \overline{AB} is exactly as long as the radius. Show that triangle OAB has three equal sides.

Answers

The center is O, and the diameter is \overline{AD}: it is the only drawn segment with both ends on the circle that passes through O.
\overline{OA}, \overline{OD} and \overline{OB}. Each joins the center to a point on the circle; the first two are the halves of the diameter \overline{AD}.
\overline{AD} and \overline{CE}. Both have both ends on the circle, and a diameter is a chord too.
On the circle: A, B, C, D and E. Inside: O and F. Outside: G.
No. A radius must end on the circle, and F is inside it, so OF is less than the radius.
18 cm. The diameter is two radii: d = 2 \times 9 = 18.
7.5 m. The radius is half the diameter: r = 15 \div 2 = 7.5.
8.5 inches. d = 2 \times 4.25 = 8.5.
1\tfrac{2}{3} feet. 3\tfrac{1}{3} = \tfrac{10}{3}, and \tfrac{10}{3} \div 2 = \tfrac{5}{3}.
0.6 m. The measurement through the center is the diameter, and the center-to-edge distance is the radius: 1.2 \div 2 = 0.6.
Inside, because 4 < 6.
On the circle, because 6 is exactly the radius.
Outside, because 6.5 > 6.
On the circle. The radius is 10 \div 2 = 5 cm, and OS equals it.
The radius is 7 - 2 = 5. (a) On: it is 3 - (-2) = 5 to the left of the center. (b) Inside: it is 2 - (-1) = 3 below the center, and 3 < 5. (c) Outside: it is 9 - 3 = 6 to the right, and 6 > 5.
The diameter is 2 \times 8 = 16 cm, the longest any chord can be. 3 cm is possible, as a chord that misses the center. 16 cm is possible, but only as a diameter. 17 cm is impossible, since it is longer than the diameter.
7 cm. The chord is at most the diameter, so 14 \le 2r and r \ge 7, with r = 7 exactly when the chord is a diameter.
(-2, -2). The center is the midpoint, so step from A to C (3 left and 4 down) and take the same step again: (1 - 3, 2 - 4).
P is in the ring: 7 > 4, so it is outside the small circle, and 7 < 9, so it is inside the large one. The ring is 9 - 4 = 5 cm wide.
12 cm. P and Q are on both circles, so AP, AQ, BP and BQ are all radii of length 3 cm, and 4 \times 3 = 12. The distance AB is not a side of APBQ and is not needed.
The edge is a circle with the post as center and radius 5 m; the goat can reach the whole disc inside it. Any two reachable points are each at most 5 m from the post, so they are at most 5 + 5 m apart, and the two ends of a diameter are exactly that far apart: 2 \times 5 = 10 m.
A diameter passes through the center, so it is made of two radii end to end. Every radius has length r, so every diameter has length r + r = 2r.
Every point on the circle is at distance r from the center, and r is more than 0. The center is at distance 0 from itself, so it is not one of those points.
With the radii to the chord’s two ends, the chord makes a triangle with them, because the center is not on the chord’s line. One side of a triangle is shorter than the other two together, so the chord is shorter than r + r = 2r, the diameter.
\overline{OA} and \overline{OB} are radii, so OA = OB = r, and AB = r is given. All three sides equal r, so the triangle is equilateral.