Circles are formed by a set of points that are located at the same distance from a fixed point. The fixed point is called the center of the circumference and the distance between the points is called the radius. The equation of a circumference with center outside the origin is found using the equation of a circumference that has a center at the origin and applying vertical and horizontal translations.

Here, we will learn to find equations for these types of circumferences. Then, we will look at some practice problems.

##### PRECALCULUS

**Relevant for**…

Finding the equation of the circumference centered outside the origin.

##### PRECALCULUS

**Relevant for**…

Finding the equation of the circumference centered outside the origin.

## Determining the equation of a circumference with center outside the origin

To find the equation of a circumference centered outside the origin, we use the equation of a circumference that has a center at the origin, and then we apply vertical and horizontal translations. Recall that the equation of a circumference with the center at the origin is $latex {{x}^2}+{{y}^2}={{r}^2}$.

This equation was derived using the Pythagorean theorem. If we rewrite this equation using the center, we would have $latex {{(x-0)}^2} + {{(y-0)}^ 2} = {{r}^2}$.

Now, let’s consider the following circumference:

We can see that this circle has its center located at the point (*h, k*). Therefore, if we use the equation of the circumference with this center, we have:

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

This is the equation of the circle centered outside the origin, where *r* is the radius, (*x, y*) is any point that is located on the circle, and (*h, k*) are the coordinates of the center of the circle.

## Circumference center outside the origin – Examples with answers

The following exercises facilitate the understanding of the application of the equation of the circumference with the center outside the origin. Try to solve the exercises yourself before looking at the answer.

**EXAMPLE 1**

Find the radius and center of the circle $latex {{(x-2)}^2}+{{(x-3)}^2} = 9$.

##### Solution

The general equation for the circumference is $latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$, where (*h, k*) is the center and *r* is the radius. Comparing this equation with the given equation, we have:

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

$latex r=3$

$latex h=2$

$latex k=3$

Therefore, the radius of the circumference is 3 and the center is (2, 3).

**EXAMPLE 2**

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

##### Solution

We recall that the general equation of the circumference is $latex {{(x-h)}^2}+{{(y-k)}^2}={{r}^2}$. This equation tells us that (*h, k*) is the center and *r* is the radius of the circumference. Therefore, we can determine the following:

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

$latex r=4$

$latex h=-4$

$latex k=5$

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

**EXAMPLE 3**

Find the equation of the circumference that has the center at the point (2, -3) and has a radius of 4.

##### Solution

We plug the values $latex h = 2$, $latex k = -3$ and $latex r=4$ into the general equation $latex {{(x-h)}^2} + {{(y-k)}^2}={{r}^2}$. Therefore, we have:

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

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

$latex {{(x-2)}^2}+{{(y+3)}^2}=16$

**EXAMPLE 4**

Find the equation of the circumference that has the center at the point (-1, 2) and in which the point (2, 6) is part of the circumference.

##### Solution

In this case, we know the values $latex h=-1, k = 2$. However, we do not know the radius of the circumference. For this we can use the formula for the distance between two points since this distance represents the radius. Therefore, we have:

$latex d=\sqrt{{{(x_{2}-x_{1})}^2}+{{(y_{2}-y_{1})}^2}}$

$latex d=\sqrt{{{(2-(-1))}^2}+{{(6-2)}^2}}$

$latex d=\sqrt{{{3}^2}+{{4}^2}}$

$latex d=\sqrt{9+16}$

$latex d=\sqrt{25}$

$latex d=5$

The radius of the circumference is $latex r=5$. We use the general equation of the circumference with these values:

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

$latex {{(x-(-1))}^2}+{{(y-2)}^2}={{5}^2}$

$latex {{(x+1)}^2}+{{(y-2)}^2}=25$

## Circumference with center outside the origin – Practice problems

Solve the following problems using what you have learned about the equation of the circumference with the center outside the origin. You can look at the solved examples above in case you need help.

## See also

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

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