Which point on the major arc between X and Y results in largest angle? Is it angle
∠XAY,∠XBY,∠XCY, or are they all the same?
To answer these questions, we need to improve our understanding of circles. (Well, I suppose you don’t need circles to read the answers.
The answers are B and D, respectively. But in order to appreciate why these answers are true, we need to understand circles.) A circle can be described by a point O and a positive number r so that every point on the circle has a distance r away from O, as shown on the opposite page.
The point O is called the center of the circle. The distance r is called the radius of the circle. As a mathematical convenience, a line segment OP from O to a point P on the circle is also called a radius.
Circumference and Area
For any circle, its diameter D is defined to be twice its radius, and is the distance across the circle. That is,
P
O r
Q
r
A circle with centerO, radiusr, and diameterD=2r
D=2r
The perimeter of a circle (its distance around) is called the circle’s circumference, denoted by C. From the picture, it is clear that C is bigger than 2D, since the distance along the circle from P to Q is bigger than D and the distance from Q back to P is also bigger than D. Consequently, C> 2D. If you eyeball it, you might even convince yourself that C is a little bigger than 3D. (But to see it clearly, you might need to wear 3-D glasses. Sorry.)
Now, if you wanted to go about comparing a circular object’s cir-cumference to its diameter, you might wrap a string around the circum-ference. Then divide the length you measured by the diameter. You’ll find, regardless of whether you are measuring a coin, the base of a glass, a dinner plate, or a giant hula hoop, that
C/D≈3.14
We define the numberπ (pi, pronounced “pie”; the Greek letter for the “p” sound) to be the exact constant that represents the ratio of a circle’s circumference to its diameter. That is,
π =C/D
andπ is the same for every circle! Or if you prefer, you can write this as a formula for the circumference of any circle. Given the diameter D or radius r of any circle, we have
C=πD or
C =2πr The digits ofπ begin as follows:
π =3.14159. . .
We will provide more digits of π and discuss some of its numerical properties later in this chapter.
Aside
Interestingly, the human eye is not so good at estimating circumferences.
For example, take any large drinking glass. What do you think is bigger, its height or its circumference? Most people think the height is bigger, but it’s usually the circumference. To convince yourself, put your thumb and middle finger on opposite sides of a glass to determine its diameter.
You will likely see that your glass is less than 3 diameters tall.
We can now answer question 1 from the beginning of the chapter. If we think of the equator of the Earth as a perfect circle with circumfer-ence C=25,000 miles, then its radius must be
r= 2Cπ = 25,0006.28 ≈4000 miles
But we don’t actually need to know the value of the radius to answer this question. All we really need to know is how much the radius will change if we increase the circumference by 10 feet. Adding 10 feet to the circumference will create a slightly larger circle with a radius that is larger by exactly 10/2π = 1.59 feet. Hence there would be enough space beneath the rope that you could crawl under it (but not walk under it, unless you were quite a limbo dancer!). What is especially surprising about this problem is that the answer of 1.59 feet does not depend at all on the Earth’s actual circumference. You would get the same answer with any planet or with a ball of any size! For example, if we have a circle with circumference C = 50 feet, then its radius is 50/(2π) ≈ 7.96. If we increase the circumference by 10 feet, then the new radius is 60/(2π) ≈9.55, which is bigger by about 1.59 feet.
Aside
Here’s another important fact about circles.
Theorem: Let X and Y be opposite points on a circle. Then for any other point P on the circle,∠XPY=90◦.
For example, in the figure below, angles∠XAY,∠XBY, and∠XCY are all right angles.
A B
C
Y
X r O r
Proof: Draw the radius from O to P, and suppose that∠XPO = x and∠YPO=y. Our goal is to show that x+y=90◦.
y x
Y X
P
O
x y
r
r r
Since OX and OP are radii of the circle, they both have length r, and therefore triangle XPO is isosceles. By the isosceles triangle theorem,
∠OXP= ∠XPO=x. Similarly, OY is a radius and∠OYP= ∠YPO=y.
Since the angles of triangle XYP must sum to 180◦, we have 2x+2y= 180◦, and therefore x+y=90◦, as desired. The theorem above is a special case of one of my favorite theorems from geometry — the central angle theorem, described in the next aside.
Aside
The answer to question 2 from the beginning of the chapter is revealed by the central angle theorem. Let X and Y be any two points on the circle.
The major arc is the longer of the two arcs connecting X and Y. The shorter arc is called the minor arc. The central angle theorem says that the angle∠XPY will be the same for every point P on the major arc between X and Y. Specifically, angle∠XPY will be half of the central angle∠XOY.
If Q is on the minor arc from X to Y, then∠XQY=180◦− ∠XPY.
O X
Y
P 100°
50°
Q 130°
For example, if∠XOY=100◦, then every point P on the major arc from X to Y has∠XPY =50◦, and every point Q on the minor arc from X to Y has∠XQY=130◦.
Once we know the circumference of a circle, we can derive the im-portant formula for the area of a circle.
Theorem: The area of a circle with radius r isπr2.
This is a formula that you probably had to memorize in school, but it is even more satisfying to understand why it is true. A perfectly rig-orous proof requires calculus, but we can give a pretty convincing ar-gument without it.
Proof 1:Think of a circle as consisting of a bunch of concentric rings, as pictured on the following page. Now cut the circle from the top to its center, as shown, then straighten out the rings to form an object that looks like a triangle. What is the area of this triangular object?
2¼r
r r
r
The area of a circle with radiusrisπr2
The area of a triangle with base b and height h is 12bh. For the triangle-like structure the base is 2πr (the circumference of the circle) and the height is r (the distance from the center of the circle to the bot-tom). Since the peeled circle becomes more and more triangular as we use more and more rings, then the circle has area
1
2bh = 12(2πr)(r) =πr2
as desired.
For a theorem so nice, let’s prove it twice! The last proof treated the circle like an onion. This time we treat the circle like a pizza.
Proof 2: Slice the circle into a large number of equally sized slices, then separate the top half from the bottom half and interweave the slices. We illustrate with 8 slices, then with 16 slices, on the next page.
As the number of slices increases, the slices become more and more like triangles with height r. Interlacing the bottom row of triangles (think stalagmites) with the top row of triangles (stalactites) gives us an object that is very nearly a rectangle with height r and base equal to half the circumference, namely πr. (To make it look even more like a rectangle, instead of a parallelogram, we can chop the leftmost stalac-tite in half and move half of it to the far right.) Since the sliced circle
r
r
¼r
¼r
Another proof (by pizza pi?) that the area isπr2
becomes more and more rectangular as we use more and more slices, the circle has area
bh= (πr)(r) =πr2
as predicted.
We often want to describe the graph of a circle on the plane. The equation to do so for a circle of radius r centered at(0, 0)is
x2+y2=r2
as seen in the graph on the next page. To see why this is true, let(x, y) be any point on the circle, and draw a right triangle with legs of length x and y and hypotenuse r. Then the Pythagorean theorem immediately tells us that x2+y2 =r2.
(x,y)
y x r
A circle of radiusrcentered at(0, 0)has formulax2+y2=r2and areaπr2 When r = 1, the above circle is called the unit circle. If we “stretch”
the unit circle by a factor of a in the horizontal direction and by a factor of b in the vertical direction, then we get an ellipse, like the one below.
a b
{ a
{ b
The area of an ellipse isπab
Such an ellipse has the formula x2
a2 +yb22 =1
and has area πab, which makes sense, because the unit circle has area π and the area has been stretched by ab. Notice that when a = b = r,