What is Equation of a Circle

The equation of a circle represents all points that are equidistant from a fixed center point. In a coordinate plane, a circle’s equation is generally written in two forms: the standard form and the general form. Each form has its specific uses, with the standard form being especially useful for quickly identifying the center and radius of the circle.

Standard Form of the Equation of a Circle

The standard form of a circle’s equation is given by:

\( (x – h)^2 + (y – k)^2 = r^2 \)

In this formula:

  • \( (h, k) \) represents the center of the circle
  • \( r \) is the radius of the circle
  • All points \( (x, y) \) that satisfy this equation are at a distance \( r \) from the center \( (h, k) \)

General Form of the Equation of a Circle

The general form of a circle’s equation is expanded from the standard form and appears as:

\( x^2 + y^2 + Dx + Ey + F = 0 \)

In this form:

  • \( D \), \( E \), and \( F \) are constants
  • The center and radius can be determined by completing the square to convert this form back to the standard form

Deriving the Standard Form of a Circle’s Equation

To derive the standard form, we start with the definition of a circle. A circle is defined as the set of points \( (x, y) \) that are at a fixed distance \( r \) (the radius) from a fixed point \( (h, k) \) (the center). Using the distance formula, the distance between \( (x, y) \) and \( (h, k) \) is:

\( \sqrt{(x – h)^2 + (y – k)^2} = r \)

Squaring both sides gives us the standard form:

\( (x – h)^2 + (y – k)^2 = r^2 \)

This is the standard form of the equation of a circle.

Examples of the Equation of a Circle

Example 1: Writing the Equation of a Circle in Standard Form

Problem: Write the equation of a circle with center \( (3, -4) \) and radius \( 5 \).

Solution:

  1. Use the standard form \( (x – h)^2 + (y – k)^2 = r^2 \).
  2. Substitute \( h = 3 \), \( k = -4 \), and \( r = 5 \): \( (x – 3)^2 + (y + 4)^2 = 5^2 \).
  3. Simplify: \( (x – 3)^2 + (y + 4)^2 = 25 \).

The equation of the circle is \( (x – 3)^2 + (y + 4)^2 = 25 \).

Example 2: Converting from General Form to Standard Form

Problem: Convert the general form \( x^2 + y^2 – 6x + 8y – 11 = 0 \) to standard form and identify the center and radius.

Solution:

Rewrite the equation: \( x^2 – 6x + y^2 + 8y = 11 \).

Complete the square for the \( x \)-terms and \( y \)-terms:

For \( x^2 – 6x \): add and subtract \( 9 \) (since \( \left(\dfrac{-6}{2}\right)^2 = 9 \)) to get \( (x – 3)^2 \).

For \( y^2 + 8y \): add and subtract \( 16 \) (since \( \left(\dfrac{8}{2}\right)^2 = 16 \)) to get \( (y + 4)^2 \).

Rewrite the equation with these perfect squares: \( (x – 3)^2 + (y + 4)^2 = 36 \).

The equation of the circle in standard form is \( (x – 3)^2 + (y + 4)^2 = 36 \), with center \( (3, -4) \) and radius \( r = 6 \) (since \( r^2 = 36 \)).

The equation of a circle can be effectively represented in either standard or general form, making it flexible for various calculations and interpretations on the coordinate plane.