General and Standard Equation of a Circle

Q: 🔵 What are the ordinary and general equations of a circle and when do I use each one?

The ordinary equation of a circle is used when you know the center and radius of the circle, and it shows the relationship between these two elements clearly. The general equation, on the other hand, is more useful when you're given three points the circle passes through or when you need to analyze the properties of the circle from its algebraic form. The ordinary form makes it easier to visualize the circle's geometric properties, while the general form is better for algebraic manipulation and solving systems of equations involving circles and other curves.

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The Circle and Its Equation

The circle is defined as the locus of points in the plane that are equidistant from a fixed point $\text{[math]}$ that we call the center.

Therefore, each point $\text{[math]}$ of the circle satisfies:

$\text{[math]}$

where the distance $\text{[math]}$ is called the radius. Thus, we have the following:

$\text{[math]}$

Squaring the previous equation, we obtain:

$\text{[math]}$

The previous equation is known as the ordinary equation of the circle. To obtain the general equation, we must expand the squared binomials:

$\text{[math]}$

Then we regroup the terms as follows:

$\text{[math]}$

We consider the following changes:

$\text{[math]}$

Therefore, the equation of the circle can be written as follows:

$\text{[math]}$

which is known as the general equation of the circle. Here, the center is given by:

$\text{[math]}$

and the radius satisfies:

$\text{[math]}$

It is important to note that the equation:

$\text{[math]}$

must satisfy the following for it to describe a circle:

1. The following inequality is satisfied:

$\text{[math]}$

2. There is no $\text{[math]}$ term (that is, $\text{[math]}$ and $\text{[math]}$ are not multiplied).

3. The coefficients of $\text{[math]}$ and $\text{[math]}$ are both 1.

Note: If $\text{[math]}$ and $\text{[math]}$ have a coefficient different from 1, then both must have the same coefficient. In this way, we can divide the equation by this coefficient to obtain the general equation of the circle.

Note: If the center of the circle coincides with the origin of the coordinates, then the equation of the circle (whether ordinary or general) is reduced to:

$\text{[math]}$

which is known as the canonical equation of the circle.

Exercises on Circle Equations

Score:

Write the equation of a circle with a center at $\text{[math]}$ and radius 2.

Solution

The ordinary equation of the circle is:

$\text{[math]}$

while the general equation of the circle is obtained by expanding the squared binomials:

$\text{[math]}$

which, when we group the constants, gives us:

$\text{[math]}$

Given the circle whose equation is $\text{[math]}$ , find its center and radius.

Solution

The center is given by:

$\text{[math]}$

On the other hand, the radius satisfies:

$\text{[math]}$

therefore, the radius is $\text{[math]}$ .

Find the equation of a circle that passes through the points $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

Solution

To find the circle that passes through three points, we should always use the general equation of the circle, since this will make the work easier.

Thus, we substitute the values of $\text{[math]}$ and $\text{[math]}$ into the equation:

$\text{[math]}$

When we substitute $\text{[math]}$ (that is, $\text{[math]}$ and $\text{[math]}$ ), we get $\text{[math]}$ , that is:

$\text{[math]}$

Similarly, when we substitute $\text{[math]}$ we get $\text{[math]}$ ; and when we substitute $\text{[math]}$ we have $\text{[math]}$ . Therefore, we have the following system of equations:

$\text{[math]}$

We solve this system in any way we wish (it is simpler if we start by subtracting the third equation from the second equation); once we solve the system we get:

$\text{[math]}$

Thus, the general equation of the circle is:

$\text{[math]}$

Indicate whether the equation $\text{[math]}$ corresponds to a circle. If so, determine the center and radius.

Solution

1. Note that the coefficients of $\text{[math]}$ and $\text{[math]}$ are equal (although they are not 1); therefore, we divide the entire equation by 4:

$\text{[math]}$

2. Also, note that it has no $\text{[math]}$ term.

3. Finally, let us verify that the inequality is satisfied with the terms $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ :

$\text{[math]}$

Therefore, because all three conditions are satisfied, the equation does describe a circle.

To find the center, we have:

$\text{[math]}$

Similarly, the radius satisfies:

$\text{[math]}$

so $\text{[math]}$ .

Calculate the equation of a circle with its center at and its tangent to the x-axis.

Solution

We are asked to find a circle that is tangent to a line. When we are asked this, the radius will always be the distance between the point and the line (to which we want the circle to be tangent). Therefore, we must find the distance between the line and the center.

First, recall that the x-axis is the line $\text{[math]}$ . Also, the distance between a point $\text{[math]}$ and a line $\text{[math]}$ is given by the formula:

$\text{[math]}$

Therefore, the distance between and the line $\text{[math]}$ is:

$\text{[math]}$

Therefore, the ordinary equation of the circle is:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia con centro en 2 coma menos 3

Calculate the equation of a circle with its center at $\text{[math]}$ and its tangent to the y-axis.

Solution

Similar to the previous exercise, we must find the distance between the point and the y-axis. Recall that the y-axis is given by the equation $\text{[math]}$ . Thus, the distance is:

$\text{[math]}$

Therefore, the equation of the circle is:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia de radio 1

Calculate the equation of a circle with its center at the point of intersection of the lines $\text{[math]}$ and $\text{[math]}$ , and whose radius equals 5.

Solution

To find the equation of the circle, we only need to find the intersection of the two lines (we already have the radius). To do this, we equate the equations:

$\text{[math]}$

from which it follows that $\text{[math]}$ ; that is, $\text{[math]}$ . Substituting into any equation, we get $\text{[math]}$ . Therefore, the center is $\text{[math]}$ and the equation of the circle is:

$\text{[math]}$

or, in general form:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia de radio 5

Find the equation of a circle that passes through the point $\text{[math]}$ and is concentric with the circle $\text{[math]}$ .

Solution

This problem can be solved in two different ways:

The simplest way is to realize that all circles concentric with $\text{[math]}$ will have an equation of the form:

$\text{[math]}$

Therefore, we must substitute the point $\text{[math]}$ in order to find the value of $\text{[math]}$ , which gives us:

$\text{[math]}$

that is, $\text{[math]}$ . Therefore, $\text{[math]}$ . In this way, the equation of the circle would be:

$\text{[math]}$

Note: The other way is to determine the center of $\text{[math]}$ and use the ordinary equation to determine the radius.

The graphical representation of the circle is:

circunferencia 3

The endpoints of the diameter of a circle are the points $\text{[math]}$ and $\text{[math]}$ . What is the equation of this circle?

Solution

To solve this problem, we must find the radius and the center.

The radius is half the diameter, therefore, it will be half the distance between $\text{[math]}$ and $\text{[math]}$ :

$\text{[math]}$

On the other hand, the center is the midpoint between $\text{[math]}$ and $\text{[math]}$ :

$\text{[math]}$

In this way, the ordinary equation of the circle is:

$\text{[math]}$

while the general equation is:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia con centro en menos 1 coma 2

Find the equation of a circle concentric with the circle $\text{[math]}$ and tangent to the line $\text{[math]}$ .

Solution

To solve this equation, we need to find the center of the circle. Therefore, we will work with the ordinary form. The center is given by:

$\text{[math]}$

once we find the center, we must find the distance between the center and the given line; this distance will be the radius:

$\text{[math]}$

Therefore, the equation of the circle is:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia con centro en 2 coma menos 3 y radio 5

Find the equation of a circle that passes through the points $\text{[math]}$ and $\text{[math]}$ , and has its center on the line $\text{[math]}$ .

Solution

Since we need to use the center, we should use the ordinary equation (and not the general) of the circle. Let $\text{[math]}$ be the center of the circle and $\text{[math]}$ be its radius, then we know that the center satisfies:

$\text{[math]}$

On the other hand, the equation of the circle is:

$\text{[math]}$

if we substitute the point $\text{[math]}$ , we have:

$\text{[math]}$

similarly, if we substitute $\text{[math]}$ , then we have:

$\text{[math]}$

In this way, we have the following system of equations (nonlinear):

$\text{[math]}$

To solve it, we equate the first two equations (since both equal $\text{[math]}$ ):

$\text{[math]}$

If we expand the binomials and cancel appropriate terms, we get:

$\text{[math]}$

Then, from the third equation we solve for $\text{[math]}$ to get $\text{[math]}$ . Substituting it into the previous equation gives us:

$\text{[math]}$

From this it follows that $\text{[math]}$ . Finally, substituting $\text{[math]}$ and $\text{[math]}$ into the first equation of the system of equations we get $\text{[math]}$ . Therefore, the ordinary equation of the circle is:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia con centro en menos 2 coma menos 2

Calculate the equation of a circle that passes through the point $\text{[math]}$ , whose radius is $\text{[math]}$ and whose center lies on the bisector of the first and third quadrants.

Solution

First, we should note that the bisector of the first and third quadrants is the line $\text{[math]}$ or $\text{[math]}$ . That is, the center $\text{[math]}$ must satisfy that $\text{[math]}$ ; therefore, we write the center as $\text{[math]}$ .

We know it passes through the point $\text{[math]}$ and that the radius is $\text{[math]}$ , substituting into the ordinary form of the circle we get:

$\text{[math]}$

We only have $\text{[math]}$ as an unknown, so that equation is sufficient. We expand the binomials:

$\text{[math]}$

So we have a quadratic equation. Using the general formula (or any other method), we find that $\text{[math]}$ and $\text{[math]}$ .

Therefore, there are two circles that satisfy the conditions of the problem. The first has its center at $\text{[math]}$ , so its equation is (first in ordinary form and then in general form):

$\text{[math]}$

and the second circle has its center at $\text{[math]}$ , so its equation is:

$\text{[math]}$

The graphical representation of the circle is:

circunferencia con radio raiz de 5

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Agostina Babbo

Agostina Babbo is an English and Italian to Spanish translator and writer, specializing in product localization, legal content for tech, and team sports—particularly handball and e-sports. With a degree in Public Translation from the University of Buenos Aires and a Master's in Translation and New Technologies from ISTRAD/Universidad de Madrid, she brings both linguistic expertise and technical insight to her work.

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Frequently asked questions

🔵 What are the ordinary and general equations of a circle and when do I use each one?

The ordinary equation of a circle is used when you know the center and radius of the circle, and it shows the relationship between these two elements clearly. The general equation, on the other hand, is more useful when you're given three points the circle passes through or when you need to analyze the properties of the circle from its algebraic form. The ordinary form makes it easier to visualize the circle's geometric properties, while the general form is better for algebraic manipulation and solving systems of equations involving circles and other curves.

📍 How do I find the center and radius from a circle's equation or from given geometric conditions?

If you're given the general equation of a circle, you can find the center by taking half of the coefficients of x and y (with opposite signs) and the radius using a special formula involving those coefficients. When given geometric conditions like three points the circle passes through, substitute those points into the general equation to create a system of equations and solve for the unknown coefficients. For circles tangent to a line, the radius equals the distance from the center to that line, and for circles passing through two endpoints of a diameter, the center is simply the midpoint of those two points and the radius is half the distance between them.

Chapters

The Circle and Its Equation
Exercises on Circle Equations

General and Ordinary Equations of the Circle

The Circle and Its Equation

Exercises on Circle Equations

Exercises on the Equation of an Ellipse

Circle Equation Exercises I

Elliptical Exercises: Definition, Practical Questions, & More!

Exercises on Parabolas

Exercises on the Equation of a Parabola