# Equation of a Circle

A circle is a geometric shape in the plane that is defined as the set of all points that are equidistant from a central point in a plane. The length of the line connecting the center of the circle to any point on its circumference is called the radius, while a line segment that passes through the center and connects two points on the circumference is known as the diameter.

Assume that (x, y) are the coordinates of a point on the circle shown below. The center is at $\left(h,k\right)$ , and the radius is r.

The equation of the circle is given as

${(x\u2013h)}^{2}+{(y\u2013k)}^{2}={r}^{2}$

This is the standard form of the equation. Therefore, if we know the coordinates of the center of the circle as well as its radius, we can easily find the equation of the circle.

**Example 1**

Let's say point (1,2) is the center of the circle and the radius is equal to 4 cm. The equation of this circle will be:

${(x\u20131)}^{2}+{(y\u20132)}^{2}={4}^{2}$

$({x}^{2}\u20132x+1)+({y}^{2}\u20134y+4)=16$

${x}^{2}+{y}^{2}\u20132x\u20134y\u201311=0$

## Equation of a circle with the center at the origin

When a circle's center is at the plane's origin, the circle's equation can be shown by using only the x and y coordinates of points on the circle's edge. Consider any point P(x, y) on the circle's circumference. The radius of the circle, denoted as "r," is equal to the distance between the origin and the point P, which is calculated using the Pythagorean theorem as follows:

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

This equation represents the standard form of the equation of a circle with the center at the origin.

**Example 2**

We know that the distance between the point (x, y) and origin (h, k) can be found using the distance formula, which is equal to

$\sqrt{{({x}_{2}\u2013{x}_{1})}^{2}+{({y}_{2}\u2013{y}_{1})}^{2}}=d$

Substitute $\left(h,k\right)$ for $\left({x}_{1},{y}_{1}\right)$ , $\left(x,y\right)$ for $\left({x}_{2}-{y}_{2}\right)$ , and $r$ for $d$ .

Thus, the equation of a circle with the center at $\left(h,k\right)$ is

${(x\u2013h)}^{2}+{(y\u2013k)}^{2}={r}^{2}$

where $r$ is the radius of the circle.

## Circle translations in the coordinate plane

In the equations generated previously, a circle that is centered at the origin can be shifted to be centered at $\left(h,k\right)$ by modifying the x and y terms in the equation. The x term represents the horizontal location in the plane and is modified by subtracting 'h', causing the center of the circle to shift to the right by 'h' units. The y term represents the vertical location and is modified by subtracting 'k', causing the center of the circle to shift up by 'k' units. As a result, the center of the circle is shifted from $\left(0,0\right)$ to $\left(h,k\right)$ , while the rest of the circle remains unchanged relative to its new center.

## Practice problems on solving the equation of a circle

a. Find the equation of a circle where the center is $\left(0,0\right)$ and the radius is 8 cm .

The equation of a circle, when the center is the origin, is:

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

For the given condition, the equation of the circle is given as

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

Therefore, the equation of the circle is ${x}^{2}+{y}^{2}=64$ .

b. Find the equation of a circle with a center of $\left(3,5\right)$ and a radius of 4 cm.

Here, the center of the circle is not the origin.

The general equation of the circle is:

${(x\u20133)}^{2}+{(y\u20135)}^{2}={4}^{2}$

${x}^{2}\u20136x+9+{y}^{2}\u201310y+25=16$

${x}^{2}+{y}^{2}\u20136x\u201310y+18=0$

## Topics related to the Equation of a Circle

Conic Sections and Standard Forms of Equations

## Flashcards covering the Equation of a Circle

Common Core: High School - Geometry Flashcards

## Practice tests covering the Equation of a Circle

Common Core: High School - Geometry Diagnostic Tests

Advanced Geometry Diagnostic Tests

## Get help learning about the equation of a circle

Learning about the equation of a circle can be difficult even for the best students. If your student could use some help figuring out the various formulas to use depending on whether the center of the circle is the origin or not, they should work with a professional tutor. An expert tutor can provide an individualized learning experience that allows your student to work more effectively towards their academic goals using a plan that addresses their academic strengths, areas of opportunity, and individual goals. They can meet together in a location free from the distractions of the classroom so your student can get the questions answered as they arise. Their tutor can show your student how to solve problems step by step so that each part makes sense before they move on to the next part.

Your student's tutor can discover how your child learns best and use that to teach them in the most effective ways. They can also encourage independent learning skills and self-management so that when your student masters the equation of a circle, they will be able to more easily pick up on the next concept in geometry. Contact the Educational Directors at Varsity Tutors today to learn more about how tutoring can help your student with the equation of a circle and other geometry concepts.

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