Circles are one of nature's and geometry's most perfect creations. A circle, in plane or **Euclidean** geometry, is completely described by giving the (x, y) location of its **center** and giving the length of its **radius**.

The formal definition of a circle is "*The locus of all points equidistant from a fixed point called the center.*" To translate, that means all points (an infinite number) that are the same distance from the center. Think of it this way: Put a pin in the table and loop a loop of string around it and the tip of a pencil. Stretch the string tight and draw until you come back around to where you began. That's a circle.

A **circle** is a collection of all points located a distance, **r** (called the **radius**) from a central point called the **center**.

We define the **radius** of a circle to be the distance from the center to *any* point on the circle. We further define the **diameter** to be twice the radius, or the distance across the circle through its center. A diameter **bisects** a circle into two **semicircles**. An infinite number of **radii** (plural of radius) and diameters can be drawn.

The **circumference** of a circle is really just the length of its **perimeter**. It's the distance around the circle.

Early geometers noticed an interesting thing about diameter and perimeter of circles: No matter what the size of a circle, when we divide the circumference by the diameter, the result is always the same: 3.141592653589 ... or **π**. It's the very definition of **π**, a number that appears in a great many applications of mathematics to the description of nature. **π** is often called a transcendental number.

The formula for **circumference (c)** is really the definition of **π ** rearranged to solve for **c**:

**Notice**: *It's difficult to measure the circumference of a circle – you need a flexible measuring device to wrap around it. But we don't need to measure circumference if we know the radius or diameter, which are much easier to measure.*

Circles are closed figures, meaning that they have an inside and an outside that are separate. Figures in plane geometry are either open or closed. Other examples of closed figures are squares and triangles. Unlike those, however, circles have no angles, nor do they have straight sides.

In other sections we'll discuss **ellipses**. Ellipses can be thought of as circles with two "centers," which elongates them along a particular diameter. A circle is just an ellipse in which these to centers, or **foci** (plural of **focus**) are on top of one another.

Often we need to work with only a part of a circle, a part intercepted by two **radii** (plural of radius). When two radii intercept, or mark out a part of the area of a circle, the resulting wedge or slice is called a **sector**.

When two radii intersect the circle, the intersections mark the endpoints of an **arc**. Here is a picture.

A **central angle** is one like the blue sector in the figure above. The vertex of the angle is the center of the circle.

We use two kinds of units to measure the angles of a circle, degrees and radians. There are 360 degrees (360˚) around a circle. Here they are counted out in quarter-circles. It's customary to start counting the angle on the right side of a circle, moving counter-clockwise.

Radians are a measure of angle based on **π**. There are 2**π** radians around a circle, so halfway around is **π **(180˚), and a quarter of the way around is **π/2** (90˚). We generally write **π rad** or **π/2 rad**.

Radians are the unit of angle measure most often used in science and mathematics because, as we shall see below, calculations can be greatly simplified using them.

Take a deeper dive into angular measurement units here.

The **area** of a circle is a little more complicated to calculate than the area of, say, a quadrilateral. There are many methods, and each requires some imagination – the kind we will use in calculus. One such method is shown below.

The circle on the left is divided into 16 equal sectors. Each has a "height" of **r** if we stack them up on the right, as shown. Each also has an arc end of measure 1/16 of the circumference, or **πr / 8**.

By stacking those sectors as shown, we approximate a rectangle of dimensions length = **πr** and height **r**. Multiplying length by width to get area, we find: **A = πr ^{2}**.

Notice that this is an approximation of the area of a circle, because the arc of each sector is less and less noticeable all the time. In the limit at which we use an infinite number of sectors, we find the exact area: **A = πr ^{2}**.

The area of a circle is the square of the radius multiplied by π.

Now let's consider how we can use the angle of a sector to calculate both the area of that sector and the length of the arc of the circle it intercepts.

It's actually pretty easy to calculate the area of a sector of a circle. All we need to know is the radius and the central angle, and we do it with **proportions**. Take the example below, a **23˚** sector of a circle of radius **r = 2 cm**. Now we just need to think about the proportions and write them down: The __area__ of the sector is to the area of the whole circle (**πr ^{2} = 4π**) as the

Then solving for the area of the sector, **a**, we get:

$$a = \frac{4(23)\pi}{360} = 0.8 \; cm^2$$

We use the same strategy to calculate the length of the arc intercepted by that 23˚ angle. The ratio of the angle to the total number of degrees in a circle (23˚/360˚), is equal to the length of the arc, **s**, divided by the circumference of the circle, **c = 2πr**. It looks like this:

$$\frac{23˚}{360˚} = \frac{s}{2 \pi(2)}$$

We rearrange to find the arc length:

$$s = \frac{4 \pi (23)}{360}= 0.8 \; cm$$

Well, this is embarrassing ... it's an accident that the arc length and area of this sector are numerically the same (though the units are different). They usually aren't, and the calculation is just fine.

Now lets do some similar area and arc length calculations, but this time we'll use radians as our measure of angle. You'll see that the calculations are simplified.

Here's a similar circle and sector. This time the angle is in radians, but we'll just call it **x** instead of specifying it.

The **π**'s cancel as shown, then we can rearrange to solve for the area: **a = πr ^{2}/2**

$$a = \frac{x r^2}{2}$$

Now we'll do the same calculation for the length of the arc intercepted by the sector.

$$\frac{x}{2 \pi} = \frac{s}{2 \pi r}$$

This time the **2π**'s cancel on both sides, and the arc length **s = rx**. That means that when we're using radians, the arc length is just the angle of the sector in radians times the length of the radius.

$$s = rx$$

These calculations are a bit easier when we measure angles in radians.

Symmetry is an important consideration in all areas of mathematics and the sciences, and circles are highly symmetric figures.

Take a look at a regular hexagon and compare it to a circle of the same size.

The hexagon has quite a few symmetry elements. The two sets of dashed lines, black and red, are **two-fold axes of symmetry**. The hexagon is symmetric by 180˚ rotation about any of these six axes, and each also denotes a line of **mirror-image symmetry**. Additionally, the hexagon has a **six-fold axis of rotational symmetry**, poking out of your screen. Rotation of the figure in the direction of the blue arrow by 60˚ gets us right back to the same hexagon. And the hexagon has a **center of inversion symmetry**. If we drag the whole figure through the center point, moving each point an equal distance to the opposite side of the center, we just get the same hexagon back. That's a lot of symmetry.

Now consider the circle. It has an **infinite** number of rotation symmetry axes and lines of reflection symmetry. Rotation around the axis perpendicular to your screen by any amount gives you a circle indistinguishable from the original one, and the circle also has inversion symmetry.

Because this is one of the most difficult of the circle proofs (most are pretty easy), let's tackle it first. The result will also be helpful later in other circle proofs.

We begin by drawing the figure, circle **X** with tangent segment $\overline{AC}$ and point of tangency **A**. Remember that we can always postulate that a tangent to a curve – a point where the line just touches the curve – exists.

We'll do this proof **contradiction**** not** perpendicular to $\overline{AC}.$ Now let's draw segment $\overline{XC},$ which intersects the circle at point

Now if $\angle XCA$ is a right angle, then $\angle AXB$ and $\angle XAC$ must be acute because the non-right angles of a right triangle must be acute. That is, 180˚ - 90˚ = 90˚, so each remaining angle must have a measure less than 90˚. Now remember that the side opposite the largest angle of a triangle is the longest side, which means that $\overline{XA} \gt \overline{XC}.$

But, that can't be true: $\overline{XA}$ is congruent to $\overline{XB}$ because both are radii of circle **X** and $\overline{XC}$ is the radius plus a little extra piece, $\overline{BC}.$ So our original assumption, that $\overline{XA}$ is not perpendicular to $\overline{AC}$ is contradicted, and $\overline{AX} \perp \overline{AC}.$

My students call this the *ice cream cone theorem*. Here's why:

$\overline{CA}$ and $\overline{CB}$ are tangent to circle **X**. Now we can draw radius $\overline{XB}$ and radius $\overline{XA},$ and segment $\overline{XC}.$

Now drawing on the theorem we just proved above, $\angle A$ and $\angle B$ are right angles. We also know that $\overline{XA} \cong \overline{XB}$ because radii of a circle are congruent (that just comes from the definition of a circle), and that $\overline{XC}$ is congruent to itself (reflextive property). That makes $\Delta AXC$ congruent to $\Delta BXC.$ Therefore $\overline{AC} \cong \overline{BC}$ because corresponding parts of congruent triangles are congruent (CPCTC).

We'll actually prove to theora in one here, first that congruent chords of the same circle intercept congruent arcs, and second, that they intercept congruent central angles. We'll start with the basic figure, a circle with two congruent chords, $\overline{AB}$ and $\overline{CD}.$

Now we'll draw radii to **A**, **B**, **C** and **D**. Note that all radii are congruent, so we're really forming two congruent triangles, $\angle ABX$ and $\angle CDX,$ by SSS.

Now $\angle AXB \cong \angle CXD$ because CPCTC, so the central angles are congruent. And because we define the arc by the central angle, arcs **AB** and **CD** are also congruent.

These ideas will work in reverse, too: congruent central angles will intercept congruent arcs and chords. We'll prove those another way below.

This theorem is the converse of the one above. We begin by drawing circle x with two congruent central angles, $\Delta AXB$ and $\Delta CXD.$

Now we'll draw chords $\overline{AB}$ and $\overline{CD},$ and note that the two triangles we've formed are congruent by SAS.

The two chords, $\overline{AB}$ and $\overline{CD}$ are congruent because CPCTC, so we've proved part of our theorem. The arcs are congruent by the definition of the arc: The angle measure of an arc is the measure of the central angle that intercepts it.

Remember that **equidistant** means "same distance," and that the **distance** from a line to a point is the length of the line segment that extends from the point to the segment and is perpendicular to the segment. First we'll set up the problem:

Chords $\overline{AB}$ and $\overline{CD}$ are congruent. Next (below) we'll draw radii $\overline{AX}, \; \overline{BX}, \; \overline{CX},$ and $\overline{DX},$ and the altitudes shown in each of the triangles, $\Delta ABX$ and $\Delta CDX.$ Notice that these triangles are congruent by SSS.

Now if we notice that $\Delta ABX$ and $\Delta CDX$ are isoscelese triangles, and we recall that the altitude of an isosceles triangle bisects the base, we end up with four congruent triangles (by SSS), one of which is shown in green below.

Now it's easy to see that these altitudes are congruent, which proves our assertion.

Notice that we might have just noted that altitudes of congruent triangles are congruent. I just thought this proof might be a little more thorough.

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