The Equation of a Circle

The standard equation of a circle in the Cartesian plane is: $$ (x - x_0)^2 + (y - y_0)^2 = r^2 $$ where \((x_0, y_0)\) represents the coordinates of the circle's center \(O\) and \(r\) is the radius.

Alternatively, the equation of a circle can be written in the general form:

$$ x^2 + y^2 + ax + by + c = 0 $$

The coefficients are related to the center and radius of the circle by the following equations:

$$ a = -2x_0 $$

$$ b = -2y_0 $$

$$ c = x_0^2 + y_0^2 - r^2 $$

To find the coordinates of the center \(O\) from the general form, use this formula:

$$ (x_0, y_0) = \left(-\frac{a}{2}, -\frac{b}{2}\right) $$

The radius \(r\) can be determined using this formula:

$$ r = \sqrt{\left(-\frac{a}{2}\right)^2 + \left(-\frac{b}{2}\right)^2 - c} $$

It is important to note that for \(x^2 + y^2 + ax + by + c = 0\) to describe a circle, the following condition must be met:

$$ \left(-\frac{a}{2}\right)^2 + \left(-\frac{b}{2}\right)^2 - c \geq 0 $$

This condition ensures that the value under the square root is non-negative, as \(r\), being a length, cannot be negative.

Additionally, the equation of the circle can be expressed explicitly in terms of \(y\):

$$ y = y_0 \pm \sqrt{r^2 - (x - x_0)^2} $$

This shows that for certain values of \(x\), there are two corresponding values of \(y\) (upper and lower), unless the term under the square root is zero, in which case the circle reduces to a single point (degenerate circle, when \(r = 0\)).

A Practical Example

Let's consider constructing a circle with its center at \( (x, y) = (1, 3) \) and radius \( r = 2 \).

The standard equation of a circle with these characteristics is:

$$ (x - x_0)^2 + (y - y_0)^2 = r^2 $$

Substituting the values \(x_0=1\), \(y_0=3\), \(r=2\), we get:

$$ (x - 1)^2 + (y - 3)^2 = 2^2 $$

$$ (x - 1)^2 + (y - 3)^2 = 4 $$

This equation represents all points in the plane that are exactly 2 units away from the center \( (1, 3) \), forming a circle.

example of a circle

The Proof

Consider a circle with radius "r" and center O(x0;y0).

a circle

A generic point P(x;y) on the plane belongs to the circle if and only if it is at a distance from the center O equal to the radius r of the circle.

$$ \overline{OP} = r $$

Squaring both sides of the equation

$$ \overline{OP}^2 = r^2 $$

The length of segment OP is equal to the distance between points O and P, which is measured using the Pythagorean theorem.

$$ \overline{OP} = \sqrt{ (x-x_0)^2 + (y-y_0)^2 } $$

Thus, I can rewrite the equation in this equivalent form

$$ \overline{OP}^2 = r^2 $$

$$ ( \sqrt{ (x-x_0)^2 + (y-y_0)^2 } )^2 = r^2 $$

The final result is the standard form equation of the circle.

$$ (x-x_0)^2 + (y-y_0)^2 = r^2 $$

To derive the alternate formula, I expand the squares

$$ x^2+x_0^2-2xx_0 + y^2+y_0^2-2yy_0 = r^2 $$

$$ x^2+ y^2-2xx_0 -2yy_0+x_0^2 +y_0^2 - r^2 = 0 $$

Letting a=-2x0, b=-2y0, c=x02+y02-r2

$$ x^2+ y^2-2x_0 \cdot x - 2y_0 \cdot y+( x_0^2 + y_0^2 - r^2 ) = 0 $$

$$ x^2+ y^2+ a \cdot x +b \cdot y+ c = 0 $$

The final result is the general equation of the circle.

$$ x^2+ y^2+ ax +by+ c = 0 $$

Knowing that a=-2x0 and b=-2y0 , I find the coordinates x0=-a/2 and y0=-b/2

Thus, the coordinates (x0; y0) of the center of the circle are

$$ (x_0 ; y_0) = ( - \frac{a}{2} ; - \frac{b}{2} ) $$

To derive the radius formula, I compare the two circle equations \( x^2+x_0^2+ \frac{a}{2} x + \frac{b}{2}y + y^2+y_0^2 - r^2 = 0 \) with \( x^2+ y^2+ ax +by+ c = 0 \)

  • Standard circle equation
    $$ x^2+x_0^2+ \frac{a}{2} x + \frac{b}{2}y + y^2+y_0^2 - r^2 = 0 $$
  • General circle equation
    $$ x^2+ y^2+ ax +by+ c = 0 $$

The constant terms in the two equations are r2, x02, and y02 in the first equation and c in the second equation

Note. By "constant term," I refer to those terms that are not multiplied by a variable, and are thus constants because their value does not change regardless of the x and y values in the equation.

Being constants, these terms must be equal in both equations.

In other words, the sum of the constant terms from the first equation x02+y02-r2 equals the "c" term from the second equation of the circumference.

Therefore, I compare and equate the constant terms of the two equations.

$$ x_0^2 + y_0^2 - r^2 = c $$

This way, I can determine the radius r of the circumference

$$ r^2 = x_0^2 + y_0^2 - c $$

$$ \sqrt{ r^2 } = \sqrt{ x_0^2 + y_0^2 - c } $$

$$ r = \sqrt{ x_0^2 + y_0^2 - c } $$

Knowing that a=-2x0, b=-2y0, I derive the coordinates x0=-a/2 and y0=-b/2

$$ r = \sqrt{ ( - \frac{a}{2} )^2 + ( - \frac{b}{2} )^2 - c } $$

This is the formula for the radius that I wanted to derive

Case Studies for the Circle Equation

The general equation of a circle in the Cartesian plane is:

$$ x^2 + y^2 + ax + by + c = 0 $$

Here, the coefficients \( a \), \( b \), and \( c \) are real numbers that determine the position of the circle in the plane. Below are the key special cases:

  • Circle Centered at the Origin
    To form a circle centered at the origin, set the coefficients \( a \) and \( b \) to zero: \[ x^2 + y^2 + c = 0 \] example
  • Circle Centered on the \( x \)-Axis
    A circle centered on the \( x \)-axis has the coefficient \( b \) equal to zero: \[ x^2 + y^2 + ax + c = 0 \] example
  • Circle Centered on the \( y \)-Axis
    A circle centered on the \( y \)-axis has the coefficient \( a \) equal to zero: \[ x^2 + y^2 + by + c = 0 \] example
  • Circle Passing Through the Origin
    A circle passes through the origin when the coefficient \( c \) is zero: \[ x^2 + y^2 + ax + by = 0 \] example

Note: By combining these cases, you can obtain other configurations. For instance, to get a circle that is centered on the \( x \)-axis (\( b = 0 \)) and passes through the origin (\( c = 0 \)), just combine the two conditions: $$ x^2 + y^2 + ax = 0 $$ example of combining two cases

Notes

Here are some observations and additional notes on the equation of a circle:

  • How to derive the equation of a circle given its center and radius
    Given that the distance from any point on the circle to its center is constant and equal to the radius, if we know the center C(x0, y0) and the radius r, we can derive the equation of the circle using the formula: $$ (x - x_0)^2 + (y - y_0)^2 = r^2 $$
  • The Equation of a Circle Given Its Diameter
    To derive the equation of a circle from the endpoints A and B of its diameter, first find the center by calculating the midpoint of segment AB: $$ C \left( \frac{x_A+x_B}{2}, \frac{y_A+y_B}{2} \right) $$ Next, calculate the radius as half the distance between points A and B: $$ r = \frac{1}{2} \cdot \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} $$ With the center and radius known, you can use the standard circle equation: $$ (x - x_0)^2 + (y - y_0)^2 = r^2 $$
  • The Equation of a Circle Passing Through Three Non-Collinear Points
    To find the equation of a circle that passes through three non-collinear points A(x1,y1), B(x2,y2), and C(x3,y3), we solve for the coefficients "a", "b", and "c" in the following system of equations: $$ \begin{cases} x^2_1+y^2_1+ax_1+by_1 + c = 0 \\ x^2_2+y^2_2+ax_2+by_2 + c = 0 \\ x^2_3+y^2_3+ax_3+by_3 + c = 0 \end{cases} $$ Once we have the values of "a", "b", and "c", we substitute them into the general equation of the circle: $$x^2 + y^2 + ax + by + c = 0 $$

    The equation of a circle passing through three non-collinear points

And so on.

 

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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