What do we mean by *the equation of a circle*? A circle is a nice, simple, round geometric object; while equations are — well — *equations*! In this post, you’ll learn about some of the connections that exist between the geometric world and the realm of algebra.

You’ll see how an algebraic equation can describe any circle. And then we’ll see what algebra can tell us about the area of a circle formula.

_{Image by kalhh}

## What Is the Equation of a Circle?

People often get confused when talking about “the equation of a circle.” Some may think that we’re talking about area or circumference, but that’s not it. There are formulas that compute area and other quantities, but *formulas* are not quite the same as *equations*.

In fact the **equation of a circle** is not for finding area, but instead provides an algebraic way to *describe* a circle. The equation tells you where the center of the circle is in the *xy*-plane, and what its radius is.

Think of this equation as the *master template* for every possible equation of a circle.

In this equation, the letters *h*, *k*, and *r* are **constants** — in other words, their values are fixed. On the other hand, the letters *x* and *y* are **variables** — they do not have a particular value, and usually just stick around as “*x*” and “*y*” in the equation.

### Example — Graphing a Circle from its Equation

To illustrate how the equation of a circle works, let’s graph the circle whose equation is:

(*x* — 1)^{2} + (*y* + 2)^{2} = 25

First, compare the given equation with the *master template* above. It looks almost the same, except that there is a plus (+) in the second group, instead of a minus (–).

These small details are very important!

Your job is to make the given equation look *exactly* like the template. The trick here is to change the *plus* into the equivalent *minus negative*.

(*x* — 1)^{2} + (*y* — –2)^{2} = 25

Also, you have to make the constant 25 match *r*^{2} from the template. In this case, that’s pretty easy, because we all know that 25 = 5^{2}.

(*x* — 1)^{2} + (*y* — –2)^{2} = 5^{2}

Now we can read off the values of *h*, *k*, and *r* from the equation.

*h*= 1*k*= –2*r*= 5

Therefore, the circle represented by this equation has center at (1, –2) and radius 5. Now we can graph it!

If you have graph paper and a compass handy, start by drawing and labeling the coordinate axes. Place one point of the compass on the center (1, –2), and the drawing point of the compass on a point that is 5 units away, such as (1 + 5, –2) = (6, –2). Then sweep out a circle!

### Example — Determining Whether Points Are on a Circle

A point (*x*, *y*) in the plane is on a circle if and only if that point satisfies the equation of the circle. Let’s see how this works by example.

Consider the circle whose equation is (*x* — 3)^{2} + *y*^{2} = 169. Determine whether each point is on the circle.

- (3, 0)
- (–2, 12)
- (–6, –10)

Ok, so what do we do?

We have the equation of the circle, and we need to test each point in that equation by plugging in for *x* and *y*. If the (*x*, *y*) pair satisfies the equation, then the point is on the circle.

#### Solutions

- (3, 0): Plug in
*x*= 3 and*y*= 0 into the left side of the equation and simplify.(3 — 3)

^{2}+ 0^{2}= 0^{2}+ 0^{2}= 0The result of 0 clearly doesn’t match the right side of the equation (169). Thus, (3, 0) is

*not*on the circle.(By the way, did you recognize that (3, 0) is the center of the circle? Points interior to a circle are not considered to be

*on*the circle.) - (–2, 12): The method is the same. Just plug into the left side of the equation and check the value against the constant on the right side.

Be careful, though: squaring a negative results in a positive.(–2 — 3)

^{2}+ 12^{2}= 25 + 144 = 169Thus (–2, 12) is definitely on the circle.

- (–6, –10):

Again, simply plug in and verify the value.

(–6 — 3)^{2} + (–10)^{2} = 81 + 100 = 181

Because 181 does not equal 169, you know that (–6, –10) is not on the circle.

In fact, you can say that (–6, –10) is *outside* of the circle, because 181 > 169.

This has to do with the fact that the equation of a circle tells us something about distance from the center. More on that below!

## Completing the Square

Sometimes you may see an equation that doesn’t immediately look like the equation of a circle.

There may be “extra” terms that don’t belong. But if the equation has both *x*^{2} and *y*^{2} showing up on the same side of the equation, then chances are that you can use algebra to put the equation into the standard form of a circle equation.

The method to do this is called **completing the square**. You may have seen this important technique when studying quadratic equations.

Let’s just see how it works in the case of a circle equation.

### Example — Completing the Square to Find Center and Radius

Find the center and radius of the circle whose equation is:

*x*^{2} + *y*^{2} + 6*x* — 10*y* — 2 = 0

#### Solution

First, rewrite the equation so that the two *x* terms are next to one another, the two *y* terms are next to each other, and the constant is on the other side. In this case, we have to add 2 to both sides in order to get the constant on the correct side.

*x*^{2} + 6*x* + *y*^{2} — 10*y* = 2

Next, take a look at the coefficient of *x*. Here, it’s *b* = 6. To **complete the square**, you have to add (*b*/2)^{2} to both sides of the equation. So in this case, we have to add (6/2)^{2} = 3^{2} = 9.

*x*^{2} + 6*x* **+ 9** + *y*^{2} — 10*y* = 2 **+ 9** = 11

Then do the same thing with the *y* coefficient, *b* = –10. For this example, you’d have to add (–10/2)^{2} = 25 to both sides.

*x*^{2} + 6*x* + 9 + *y*^{2} — 10*y* **+ 25** = 11 + **25** = 36

Finally, *factor* each trinomial (i.e., three terms in a group). If you’ve done the previous steps correctly, then each group will factor as a *perfect square*.

(*x* + 3)^{2} + (*y* — 5)^{2} = 36

Now we can answer the question! The center is at (–3, 5), and the radius is 6.

## The Circle Equation and Distance

How do you define a circle? The mathematical definition for a circle is simply *the set of points lying at a constant distance (the radius) from a fixed point (the center)*.

In fact anything spinning around a point at a specific distance from that point will look circular. It’s just how geometry works!

_{Image by theartofsounds2001}

So it may come as no surprise that the notion of “fixed distance from a point” is already baked into the equation of a circle.

First, let’s review distance.

The distance between two points (*a*, *b*) and (*x*, *y*) in the plane is given by the formula,

Already, I bet you can spot a few similarities between this and the equation of a circle. For starters, they both have terms like (*something* — *something*)^{2}.

### Coming Around Full Circle

Now let’s dig a little deeper!

Just as above, let’s label the center of a circle by (*h*, *k*), and its radius by *r*. Now any point (*x*, *y*) on a circle must be exactly *r* units away from the center, right?

So set *a* = *h* and *b* = *k* in the formula, and because we know that the distance must be *r*, we can also replace *d* by *r*.

Finally, if you square both sides of this equation, you end up with exactly the equation of a circle! Yes, the *r*^{2} is on the “wrong” side, but just swap the left and right sides of the equation and there you have it!

## The Area of a Circle Formula

Finally, this article would not be complete if I didn’t spend some time talking about the **Area of a Circle Formula**.

Many times students confuse this formula with the equation of a circle. However, the area formula does one thing and one thing only: it computes the **area** within a circle.

Area is like the amount of paint it would take to cover a region.

_{Image by boladenavidad}

So, for example, a circle of radius 7 would have an area of *A* = π(7)^{2} = 49π That works out to roughly 153.94 square units of area.

### Other Formulas Involving Circles

The **circumference** of a circle is like the length of fence that it would take to go all the way around a circular garden. You can find circumference *C* if you know the radius of the circle.

For example, what is the circumference of the circle of radius 7? Just plug *r* = 7 into the formula!

*C* = 2π(7) = 14π. The circumference is roughly 43.98 units.

There are other formulas that compute lengths of arcs, areas of sectors, and segments of the circle, etc. But this article is not the place to delve into those.

## Summary

So remember, the **equation of a circle** of radius *r* centered at (*h*, *k*) is:

To find **area**, use the area of a circle formula: .

**Circumference** has its own formula:

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