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Let's take a look at (surprise, surprise) a circle.
We don't know anything about this circle. Sure, we've spent almost a whole chapter learning about properties of circles in general, but we don't know anything about this circle in particular. What's its favorite color? How does it spend its leisure time? Does it like artificial banana flavor or not?
For example, we learned at the beginning of the chapter that two things define a circle:
1. Where it is (center)
2. How big it is (radius)
But without any context, we don't know where this circle is or how big it is. We know that it has some center and some radius, and for many mathematicians that's enough. But if we're practical people who want to put this circle to work, we need to get some specifics (and probably a résumé of some kind).
We need to know whether the circle is in Chile or China. We need to know if it's the right size to be a wheel on a shopping cart or a fence around the Smiths' disk-shaped ranch. Maybe it's not even on Earth. Maybe it's as big as Earth. After all, scale is pretty important.
Fortunately, in this case, since we simply made up the circle, we can make up its characteristics. Let's say its radius is 5 km and it's located at a point 4 km east and 3 km north of the center of Sydney, Australia.
Now if we want to give that information to somebody else, we can tell them using those cumbersome English words, or we can take a leaf from Descartes's sketch pad and use a coordinate plane with appropriate labels.
Isn't that nice? And we didn't even have to pay for airfare.
We can describe a circle in the coordinate plane with an equation. But before we go there, we'll make things a little easier. Let's start by considering a circle with its center at the origin and radius 5 units. Here it is, on the coordinate plane.
Notice that the circle goes through the points (5, 0), (0, 5), (-5, 0), and (0, -5).
We're trying to find an equation relating the coordinates of a generic point on the circle (x, y). Notice that by nature of the Cartesian system, we can easily draw a right triangle based on any point (x, y).
One leg is horizontal and has length x. The other leg is vertical and has length y. The hypotenuse connects (x, y) and the origin. If (x, y) is on the circle and the circle's center is at the origin, then the hypotenuse has a length equal to the circle's radius. In this case, it's 5 units.
Pythagoras tells us that x2 + y2 = 52.
But the radius of a circle centered at the origin doesn't have to be 5 units. It could be 6, or 7, or 50, or even a million units. Let's cover all our bases and call it r units. So we have the equation x2 + y2 = r2.
There you have it. We now have an equation that relates the x- and y-coordinates of any point on a circle with radius r centered at the origin. Wasn't that easy?
You may notice that we left the x and y terms on the same side of the equation, rather than solving for y as we like to do for linear equations. (Remember those from algebra?) A formula with all the variables on one side is called implicit, while a formula that has been solved for one variable in terms of the other (such as y = mx + b) is called explicit.
Go ahead and try solving our implicit equation for y. You'll end up with a messy plus-or-minus square root. We'll stick with the nice, clean, simple equation we've got here. Besides, this is a family website. We don't want any explicit content here.
What is the formula for a circle centered at the origin with a circumference of 25.1 inches?
We know the formula for a circle is x2 + y2 = r2. All we need to complete it is the length of the radius, which we can find using the circumference. If we just recall the formula C = 2πr, we can plug in 25.1 inches for C and solve for r ≈ 4 inches. Our final equation, then, is x2 + y2 = 42 or x2 + y2 = 16.
Our formula is certainly nice, clean, and simple, but it's only useful if our circle is at the origin. How do we describe a circle like that one outside of Sydney?
We can still draw a right triangle, so Pythagoras can still help us out. The hypotenuse still has length equal to the radius of the circle, since its endpoints are the center of the circle and a point on the circle. We know the Sydney Circle's radius is 5 km.
But the lengths of the legs aren't just plain old x and y anymore. We have to adjust for the "shifting" of the center of the circle from the origin. So the lengths of the horizontal and vertical legs of our right triangle are x – 4 and y – 3, respectively. Now we can plug these lengths into the Pythagorean Theorem to get (x – 4)2 + (y – 3)2 = 52.
And there's our formula for the Sydney Circle, with x and y in kilometers (because Aussies, like the rest of the world, use the metric system).
In general, the implicit formula for a circle with center (h, k) and radius r is:
(x – h)2 + (y – k)2 = r2
Now we have all the information we need to define any circle in a nice, compact formula. You can use it to describe anything from the orbit of a spacecraft to a cookie on a cookie sheet. That's right: the mathematical technology necessary to put a satellite into orbit is no more complex than that necessary to make snickerdoodles at home. More or less.
What is the equation for a circle with center (6, -2) and radius 18 units?
The implicit formula for a circle with center (h, k) and radius r is (x – h)2 + (y – k)2 = r2. We're given the center and radius, so all we need to do is plug the information into the right places. We should end up with (x – 6)2 + (y – (-2))2 = 182, or (x – 6)2 + (y + 2)2 = 324 when simplified.
We just learned how to represent a circle with an equation. We've known for a long time how to represent a line with an equation. So let's use equations to examine what happens when circles and lines interact.
We represent a generic circle with (x – h)2 + (y – k)2 = r2and a generic line with y = mx + b. Let's put a line and a circle together in the same system of equations.
(x – h)2 + (y – k)2 = r2
y = mx + b
Don't get too worried (or excited). We won't make you solve systems like this, even though you could do it. The algebra is a little tough and we're trying to focus on the geometry here. Besides, focusing on the geometry might actually help us understand the algebra.
Remember that the solution of a system of two equations like this one is the set of all points (x, y) that make both equations true at the same time. In other words, the solution of that system is the set of all points that are both on the circle and on the line. In other other words, the solution of that system is the set of points where the circle and the line intersect.
We already know that a line can intersect a circle at two, one, or zero points (it can be a secant of the circle, a tangent to the circle, or neither). Therefore, this system of equations must always have two, one, or zero solutions, depending on whether the line is a secant, a tangent, or neither.
As we said before, we won't make you solve a system like this. But we will make you do some algebra, because it's fun. And maybe we're a little sadistic.
We've talked before about cutting up a pizza, but never in Australia. Each cut made by a pizza knife stretches from one side of the pizza to the other, intersecting the crust of the pizza at two points. So each cut is a secant. Hopefully, each cut is also a diameter, so that the pizza ends up looking like this:
and not like this:
Suppose we put a coordinate grid over that pizza so you can see the coordinates of the points where the sloppy pizza man's cut intersects the edge of the pizza as well as find the equation of the circle.
Could you find the equation of the line made by the cut? Of course you could. You actually don't need to consider the circle at all: it's just finding the equation of a line given two points on the line, which is algebra-level stuff.
But let's say the pizza man is really sloppy and makes a cut that's not a secant of the pizza, but rather a tangent to the pizza. He's basically cutting the box, only touching the tiniest nub of pizza. Could you find the equation of the line then?
Let's draw a diagram of the situation. The pizza has a radius of 25 cm. Let's consider the center of the pizza to be at the origin. This sloppy pizza man, who is seriously liable to damage himself, his co-workers, or his place of employment, makes a cut tangent to the pizza at the point (7, 24).
We know one point on the line already, so let's see if we can find the slope of the line. Then we'll have enough information to write the equation of the line.
We know that the tangent line is perpendicular to the radius of the circle at the point of tangency. So if we find the slope of the radius, we can take the negative reciprocal and we have the slope of the tangent line. Ingenious!
What is the equation of the tangent line that intersects a circle centered at the origin with radius 25 at the point (7, 24)?
In this case, the radius is the line segment with endpoints (0, 0) and (7, 24). The slope of the radius is
Taking the negative reciprocal, we get a slope of for the tangent line. We now have enough to write the equation of the tangent line in point-slope form:
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