Equations of a Circumference: A Comprehensive Guide

Circumference as Geometric Locus

The geometric locus is a group of points that satisfy certain geometric conditions. The geometric locus of the circumference is the group of points that are at the same distance r from a point called the center.

Example

Determine the geometric locus that describes a point that moves a distance of 4 units around a fixed point with coordinates (0,0).

Steps

  1. Identify the fixed point on the plane.
  2. Draw a point at a distance of 4 units to each of the four cardinal points.
    e5-2.jpg
  3. Join the drawn points as a circle at 4 units from the fixed point (0, 0).

Answer

The resulting figure is a circumference with center at (0, 0) and radius equal to 4.

e5-4.jpg

Types of Equations of the Circumference

There are three ways to represent the equation of the circumference:

  1. Canonical form
  2. Ordinary form
  3. General equation

For the canonical form, the center of the circumference is considered to be at the origin C (0, 0), the distance is from the center to any of the points of the circumference P(x,y), and its radius is a length equal to r:

e5-5.jpg

Given points C(0,0) and P(x,y), the distance between them will be the length of the radius. Applying the formula of the distance between two points, we have:

e5-1.gif

By raising both sides of the equation to the square, we have:

e5-2.gif

This expression is known as the equation of the circumference with its center at the origin, also known as the canonical form.

Example

Determine the equation of the circumference with its center at the origin and a radius equal to 7.

Steps

  1. If the center of the circumference is at the origin, then the following equation is used:
    e5-3.gif
  2. Substitute the value of the radius in the previous equation:
    e5-4.gif

Answer

Raise the radius to the square to obtain the following expression:

e5-5.gif

Example

Determine the equation of the circumference that contains point (2,5) and its center is at the origin C(0,0).

Steps

  1. In this exercise, the length of the radius is not known, it should be determined with the given data and the following equation:
    e5-6.gife5-8.gif
  2. Once the radius has been established, substitute its value in the canonical form of the equation of the circumference.
    e5-9.gif

Answer

The square is canceled with the square root and the following equation is obtained:

e5-10.gif

The equation of the circumference in its ordinary form has its center outside the origin at C(h,k) and its radius is the distance from the center to any point within the circumference P(x,y):

e5-6.jpg

Calculating the distance from the center to the point for obtaining the radius, we have:

e5-11.gif

Both sides of the equality are raised to the square:

e5-12.gif

This equation is known as the equation of the circumference with its center outside the origin, also known as the ordinary form of the equation of the circumference.

Example

Determine the equation of the circumference with center at (-1,3) and with radius length equal to 6.

Steps

  1. To find the equation, use the following expression because its center is outside the origin.
    e5-14.gif
  2. Substitute the center C(-1,3) and the radius in the previous equation:
    e5-15.gif
  3. Apply the rules of signs and raise to the square:
    e5-16.gif

Answer

e5-17.gif

Example

Determine the equation of the circumference with center at (-2,-1) that passes through point P(2,4).

Steps

  1. Obtain the value of the radius with the following equation:
    e5-7.gife5-18.gif
  2. Apply the rules of signs and obtain the square root:
    e5-19.gif
  3. Knowing the value of the radius e5-20.gif and the center C(-2,-1), substitute the values in the following equation:
    e5-21.gif

Answer

The resulting equation is:

e5-22.gif

To obtain the general form of the circumference equation, we start from its ordinary form.

e5-13.gif

We work with the binomials to the square and the resulting equation is as follows:

e5-23.gif

Making the equation equal zero, we have:

e5-24.gif

Substitute e5-25.gif to obtain the following:

e5-26.gif

The previous expression is known as the general equation of the circumference.

From the value of the radius, we can determine its graphic representation:

  1. If r < 0, then the equation does not represent a circumference, so it cannot be graphically represented because there are no negative radiuses.
  2. If r = 0, then it represents a point which is represented by the coordinates of the center.
  3. If r > 0, the geometric locus is from a circumference and it is possible to represent it in a plane.

Example

Given the equation e5-29.gif of the circumference on ordinary form, obtain the equation in its general form.

Steps

  1. Expand binomials to the square, as follows:
    e5-30.gif
  2. Arrange the quadratic terms first, then the linear ones and finally, the equation must equal to zero:
    e5-31.gif

Answer

Simplify the expression by adding all independent terms. The resulting equation is:

e5-32.gif

Equation of a Circumference that Satisfies Three Conditions

Consider the case in which you have a circumference that passes through three points A(x1,y1), B(x2,y2) and C(x3,y3), these three points are substituted in the equation of the circumference in its general form e5-27.gif where the missing values are D, E and F. In this way, you will obtain a system of three equations with three variables to solve.

You can see this case illustrated in the following example.

Example

Determine the equation of the circumference that passes through points O(2,-2), P(3,1) and Q(-3,-2).

Steps

  1. Substitute the points in the general formula:
    e5-28.gif
  2. Evaluate the values of each of the equations and simplify as follows:
    e5-33.gif
  3. You have 3 equations with three variables. You can use any method you know to solve the system of equations: substitution, addition or elimination.
    e5-34.gif
  4. Applying the elimination method, you add equations P and Q:
    e5-35.gif
  5. Equation O is multiplied by 3 and equation Q is multiplied by 2, then both are added:
    e5-36.gif
  6. Equation R is multiplied by -10 and is added to equation S for then solving for variable F:
    e5-37.gif
  7. Knowing the value for variable F, substitute in equation S and solve for variable E:
    e5-38.gif
  8. Knowing the values F = -12 and E = -1, substitute in any of the three original equations, O, P or Q to obtain the value for D.
    e5-39.gif

Answer

Substitute the values for D, E and F in the general equation of the circumference to obtain the following expression:

e5-40.gif

Polar Form of the Circumference Equation

To obtain the polar form of the equation of the circumference, you should express it in terms of r and e5-41.gif, and substitute e5-42.gif and e5-43.gi. The following example can help you understand this concept.

Example

Given the equation of the circumference e5-45.gif, obtain its polar form.

Steps

  1. Expand binomials to the square, as follows:
    e5-46.gif
  2. Substitute e5-47.gif and e5-44.gif in the resulting expression:
    e5-48.gif
  3. From the first two terms, take out r2 as a common factor:
    e5-49.gif
  4. Considering the identity e5-50.gif=1
    substitute it by 1 in the equation and factorize:
    e5-51.gif

Answer

The equation in its polar form is:

e5-52.gif