*Lance Cain*

# How to Find the Equation of a Circle

## Circles

If you swing a ball attached to the end of a rope over your head like a cowboy's lasso, the ball travels in a circle with your hand at the center and the length of the rope as the radius, which you can see in the figure below.

We can use the location of your hand (the center) and the length of the rope (radius) to write an equation for the circular path of the ball. Here's how.

When we trace the circle on a grid like in the figure that we just saw, with the origin (0,0) at the center, we can see that any point (x,y) on the circle can be defined by drawing a right triangle and using the Pythagorean Theorem, which, remember, is (a^2 + b^2 = c^2).

The side 'a' is the horizontal distance (x); side 'b' is the vertical distance (y); and the hypotenuse of the triangle is the radius of the circle (r). Substituting these distances from the circle into the Pythagorean Theorem, we obtain the **standard form** of the equation for a circle:

x^2 + y^2 = r^2

However, what if the circle is not centered on the origin? What if the center of the circle is at point (9,5) instead of (0,0)? No sweat, a small adjustment to our equation will allow for the center to be anywhere. Simply subtract the x-value and y-value of the center point from the x and y variables in our equation. For example, to find points on a circle centered at (9,5), we would re-write the equation as:

(x-9)^2 + (y-5)^2 = r^2

## Standard Form vs. General Form

If we assume a generic center point like (s,t), we write our equation in what is known as the standard form equation for a circle, which, as you can see, is (x - s)^2 + (y - t)^2 = r^2. With the use of a little algebra to expand and combine like terms, we can convert the standard form into the following, known as the general form of the equation for a circle, which, as you can see, is:

In the general form of the equation, the coefficients of the x and y terms - A and B, respectively - can be any real numbers. Both, the standard and general forms of the equation produce a circle when graphed; the information you have been given will determine which equation is best to use.

## Some Examples

Here is Example 1:

Write the standard form of the equation for a circle centered at point (3 , 7) and a radius of 10.

*Solution:*

You need to insert x and y values and set equal to the radius squared:

(x - 3)^2 + (y - 7)^2 = 10^2 ===> (x - 3)^2 + (y - 7)^2 = 100

And convert the standard form equation to general form as follows.

First we expand:

(x - 3)^2 + (y - 7)^2 = 100 ===> x^2 - 6x + 9 + y^2 -14y + 49 = 100

Then we rearrange terms:

x^2 + y^2 - 6x -14y + 9 +49 = 100 ===> x^2 + y^2 - 6x -14y + 58 = 100

And then we set equal to zero:

x^2 + y^2 - 6x -14y + 58 - 100 = 100 - 100 ===> *x^2 + y^2 - 6x -14y - 42 = 0*

Here is Example 2:

What is the radius and center point of the circle in the figure below? We first write the equation for the circle in general form.

*Solution:*

radius = 5

center point = (3 , -4)

To write the equation for this circle in general form, we first write down the standard form of the equation, which is:

(x - 3)^2 + (y + 4)^2 = 25

Expand the equation and combine like terms.

Expand:

(x - 3)^2 *and* (y + 4)^2

x^2 - 6x +9 +y^2 + 8y + 16= 25

Now you arrange terms:

x^2 + y^2 - 6x + 8y + 25 = 25

And set equal to zero:

x^2 + y^2 - 6x + 8y + 25 - 25 = 25 - 25 ===> *x^2 + y^2 - 6x + 8y = 0*

Now here's Example 3:

What is the radius and center point of the circle in the figure on your screen below? Write the equation for this circle in general form.

*Solution:*

radius = 3

center point = (-4 , 2)

To write the equation for this circle in general form, we first write down the standard form of the equation, which is: (x + 4)^2 + (y - 2)^2 = 9

Expand the equation and combine like terms.

Expand:

(x + 4)^2 and (y - 2)^2

x^2 + 8x +16 + y^2 - 4y + 4 = 9

Now you arrange terms:

x^2 + y^2 + 8x - 4y + 16 + 4 = 9 ===> x^2 + y^2 + 8x - 4y + 20 = 9

And set equal to zero:

x^2 + y^2 + 8x - 4y + 20 - 9 = 9 - 9 ===> *x^2 + y^2 + 8x - 4y + 11 = 0*

Finally, let's look at Example 4:

Convert the following equation for a circle from general form to standard form:

x^2 + y^2 - 6x - 12y + 29 = 0

*Solution:*

Group x and y terms together. Move the constant to the other side of the equation.

x^2 - 6x + y^2 - 12y + 29 - 29 = 0 - 29 ===> x^2 - 6x + y^2 - 12y = -29

Complete the squares, adding values to both sides of the equation.

(x^2 - 6x + 9) + (y^2 - 12y + 36) = -29 + 9 + 36

(x^2 - 6x + 9) + (y^2 - 12y + 36) = 16

Now, factor:

(x - 3)^2 + (y - 6)^2 = 16

And, using the standard form definition of a circle, (x - s)^2 + (y - t)^2 = r^2, we know:

Center point = (3 , 6)

Radius = sqrt(16) = 4

## Lesson Summary

Let's take a few moments to review what we've learned about finding the equation of a circle. When many points are arranged so that they are all the same distance (radius) from a single point (center) in every direction, those points form a circle. There are two common equations to describe a circle:

- The
**standard form**, easy to formulate if you already know the center point and radius. - The
**general form**, set equal to zero. The general form is, if you'll recall, x^2 + y^2 + Ax + By + C = 0.

Using algebraic techniques of factoring and completing the squares, you can convert between the two forms.

Here is an image of a very large circle operated by the scientific organization CERN. The Large Hadron Collider (LHC) highlighted in yellow in this picture on your screen was constructed hundreds of feet below the border of France and Switzerland near Geneva. This particle collider is made up of more than 1,600 superconductive magnets all located 2.67 miles from its center, forming a circle 5.34 miles across (diameter). If we imagine the center of LHC at the origin (0,0) the equation for this circle in standard form is:

x^2 + y^2 = (2.67)^2

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