# Circles

### Topics

## Introduction to :

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 are 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 length equal to the circle's radius. In this case, it's 5 units.

Pythagoras tells us that *x*^{2} + *y*^{2} = 5^{2}.

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 *x*^{2} + *y*^{2} = *r*^{2}.

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.

### Sample Problem

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 *x*^{2} + *y*^{2} = *r*^{2}. 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* = 2π*r*, we can plug in 25.1 inches for *C* and solve for *r* = 4 inches. Our final equation, then, is *x*^{2} + *y*^{2} = 4^{2} or *x*^{2} + *y*^{2} = 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} = 5^{2}.

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} = *r*^{2}

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.

### Sample Problem

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} = *r*^{2}. 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} = 18^{2}, or (*x* – 6)^{2} + (*y* + 2)^{2} = 324 when simplified.