Circumferences are defined as a set of points that are at a constant distance from a fixed point. The fixed point is called the center, while the constant distance is called the radius. We can define a circumference using three equations depending on whether its vertex is located at the origin or outside the origin. Also, we can rewrite these equations in their general form.

Here, we will look at the different ways to write equations of circumferences along with some examples.

## Circumference with center at the origin

We can use the Pythagorean theorem to derive an equation for a circumference centered at the origin. We draw a circle in the Cartesian plane along with a right triangle:

The center of this circle is (0, 0). The coordinates (*x, y*) represent a general point that is part of the circumference. The *x*-coordinate of the point forms the base of the right triangle and the *y*-coordinate forms the height of the triangle.

Also, the hypotenuse of the triangle is formed by the radius of the circumference. Using the Pythagorean theorem, we can write the following equation:

$latex {{x}^2}+{{y}^2}={{r}^2}$ |

This equation represents a circle that is centered at the origin, where *r* is the radius and (*x, y*) are the coordinates of a point that is located on the circle.

### EXAMPLE 1

What is the equation of a circle that is centered at the origin and has a radius of 8?

**Solution:** We use the equation obtained with the radius $latex r = 8$. Therefore, we have:

$latex {{x}^2}+{{y}^2}={{r}^2}$

$latex {{x}^2}+{{y}^2}={{8}^2}$

$latex {{x}^2}+{{y}^2}=64$

### EXAMPLE 2

What is the equation of a circle that is centered at the origin and passes through the point (3, 5)?

**Solution:** Since the point (3, 5) is part of the circumference, we have the coordinates $latex x=3$ and $latex y=5$. Therefore, we plug these values into the equation for the circumference and find the radius:

$latex {{x}^2}+{{y}^2}={{r}^2}$

$latex {{3}^2}+{{5}^2}={{r}^2}$

$latex 9+25={{r}^2}$

$latex {{r}^2}=36$

Now, we use this value in the general equation to obtain the equation for this circumference:

$latex {{x}^2}+{{y}^2}={{r}^2}$

$latex {{x}^2}+{{y}^2}=36$

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## Circumference with center outside the origin

We find the equation of a circle that is not centered at the origin by using the equation of a circle centered at the origin and applying vertical and horizontal translations. If we rewrite the equation obtained above using the center, we would have $latex {{(x-0)}^2}+{{(y-0)}^2}={{r}^2}$.

Now, let’s look at the following circumference:

This circle has its center at (*h, k*). Then, applying the translation of *h* units on the *x*-axis and *k* units on the *y*-axis, we have:

$latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$ |

where, *r* is the radius, (*x, y*) are the coordinates of the points on the circle, and (*h, k*) represents the center of the circle.

### EXAMPLE 1

What is the radius and center of the circle $latex {{(x-4)}^2}+{{(x-5)}^2}=16$?

**Solution:** We know that the equation of a circle is $latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$. In this equation, the center is (*h, k*) and the radius is *r*. Comparing this equation with the given equation, we have:

$latex {{r}^2}=16$

$latex r=4$

$latex h=4$

$latex k=5$

Therefore, the center of the circle is (4, 5) and the radius is 4.

### EXAMPLE 2

Determine the equation of the circumference that has a radius of 8 units and the center at (-2, 5).

**Solution: **We have the values $latex h= -2, k = 5$ and $latex r = 8$. We substitute these values in the general equation $latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$. Therefore, we have:

$latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$

$latex {{(x-(-2))}^2}+{{(y-5)}^2}={{8}^2}$

$latex {{(x+2)}^2}+{{(y-5)}^2}=64$

## Equation of a circumference in general form

We can expand the equation $latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$ to get:

$${{x}^2}+{{y}^2}-2hx-2ky+{{h}^2}+{{k}^2}-{{r}^2}=0$$

Now, we can make the substitutions $latex A=-2h$, $latex B=-2k$, $latex C={{h}^2}+{{k}^2}-{{r}^2}$. Therefore, we have:

$latex {{x}^2}+{{y}^2}+Ax+Bx+C=0$ |

This is the equation of the circumference written in general form. Using the following substitutions, we can determine the radius and the coordinates of the center:

$latex A=-2h$ $latex B=-2k$ $latex C={{h}^2}+{{k}^2}-{{r}^2}$

We solve these expressions for *h, k* and *r* and we have:

$latex h=-\frac{A}{2}$

$latex k=-\frac{B}{2}$

$latex {{r}^2}={{h}^2}+{{k}^2}-C$

$latex {{r}^2}={{(-\frac{A}{2})}^2}+{{(-\frac{B}{2})}^2}-C$

$latex {{r}^2}=\frac{{{A}^2}+{{B}^2}-4C}{4}$

Since in the original equation the center of the circle is $latex (h, k)$, we know that the center of the circle in its general form is $latex (-\frac{A}{2}, -\frac{B}{2})$. Furthermore, the radius is given by $latex r=\sqrt{\frac{{{A}^2}+{{B}^2}-4C}{4}}$.

### EXAMPLE 1

What is the center and radius of the circle $latex {{x}^2} + {{y}^2} -2x + 4y-4 = 0$?

**Solution:** The center of a circle in its general form is given by:

$$(-\frac{A}{2}, -\frac{B}{2})=(-\frac{-2}{2}, -\frac{4}{2})=(1, -2)$$

The radius is given by:

$latex r=\sqrt{\frac{{{A}^2}+{{B}^2}-4C}{4}}$

$latex =\sqrt{\frac{{{(-2)}^2}+{{4}^2}-4(-4)}{4}}$

$latex =\sqrt{\frac{4+16+16}{4}}$

$latex =\sqrt{\frac{36}{4}}$

$latex =\frac{6}{2}=3$

The center of the circle is (1, -2) and the radius is 3.

### EXAMPLE 2

Determine the equation of the circle that has a center at (3, -2) and a radius of 4.

**Solution:** We use the expressions given above to find the constants A, B and C:

$latex A=-2h=-2(3)=-6$

$latex B=-2k=-2(-2)=4$

$latex C={{h}^2}+{{k}^2}-{{r}^2}$

$latex C={{3}^2}+{{(-2)}^2}-{{4}^2}$

$latex C=9+4-16=-3$

We substitute these values in the general equation:

$latex {{x}^2}+{{y}^2}+Ax+Bx+C=0$

$latex {{x}^2}+{{y}^2}-6x+4x-3=0$

## See also

Interested in learning more about equations of a circumference? Take a look at these pages:

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