The Geometry and Algebraic Representation of Circles

Introduction to the Geometry and Algebra of Circles

The study of geometry often begins with the visual representation of coordinates and distances, as demonstrated by the Department of Mathematics at Lehman College. A map featuring the Mayon Volcano at the origin of a coordinate system serves as a primary example. On this map, the cardinal directions are designated as North, South, East, and West, with scale markings extending from $-14$ to $14$ on the horizontal axis and $-10$ to $12$ on the vertical axis. A specific point $(S)$ is identified at the coordinates $(6\,km\,\text{East}, 7\,km\,\text{South})$. This spatial orientation provides the foundation for defining circular boundaries, such as a distance of $8\,km$ from a fixed landmark. As noted by the mathematician George P3lya, "No idea is really bad, unless we are uncritical. What is really bad is to have no idea at all." This philosophy encourages students to explore the mathematical properties of a circle through critical analysis of its algebraic equations.

Fundamental Parts and Relationships of a Circle

A circle is formally defined as the set of all points in a plane that are equidistant from a fixed point known as the center. This fixed distance between the center and any point located on the circle itself is called the radius. Another critical component is the diameter, which is a line segment that passes through the center with its endpoints resting on the circle. The relationship between these components is linear and reflexive. The diameter is exactly twice the length of the radius, expressed by the formula $d = 2r$. Conversely, the radius is half the length of the diameter, represented by the equation $r = \frac{d}{2}$.

Derivation of the Standard Form Equation

The equation of a circle is derived using the distance formula in a two-dimensional coordinate plane. If we let $C(h, k)$ represent the coordinates of the center and $(x, y)$ represent any arbitrary point on the circle, the radius $r$ is the distance between these two points. Starting with the distance formula $r = \sqrt{(x - h)^2 + (y - k)^2}$, we can square both sides to remove the radical. This results in the standard form equation of a circle: $(x - h)^2 + (y - k)^2 = r^2$. This equation allows for the identification of the circle's properties simply by looking at the specific values of $h$, $k$, and $r$. In the specific case where the center is located at the origin $(0, 0)$ and the radius is $1$, the equation simplifies to $x^2 + y^2 = 1$, which is known as the unit circle.

The General Form of a Circle Equation

Beyond the standard form, a circle's equation can be expressed in general form by expanding the binomials in the standard form expression. Expanding $(x - h)^2 + (y - k)^2 = r^2$ yields $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = 0$. By rearranging the terms to group the quadratic, linear, and constant components, we arrive at $x^2 + y^2 + (-2h)x + (-2k)y + h^2 + k^2 - r^2 = 0$. This is generalized as $Ax^2 + By^2 + Cx + Dy + E = 0$, where $A = B$ and neither is zero. Specifically, the transcript notes the form $Ax^2 + By^2 + Cx + Dy + C = 0$. Transitioning between the standard and general forms often requires algebraic techniques such as foil expansion or completing the square.

Computational Examples using Standard Form

Several examples illustrate how to apply the standard form equation $(x - h)^2 + (y - k)^2 = r^2$. In Example 1, a circle with a radius of $3$ centered at $(2, 1)$ results in the equation $(x - 2)^2 + (y - 1)^2 = 3^2$, which simplifies to $(x - 2)^2 + (y - 1)^2 = 9$. In Example 2, finding the equation for a circle with radius $5$ centered at $(-3, 2)$ involves substituting the negative coordinate: $(x - (-3))^2 + (y - 2)^2 = 5^2$, resulting in $(x + 3)^2 + (y - 2)^2 = 25$. Example 3 demonstrates the reverse process: given $(x - 5)^2 + (y - 3)^2 = 4$, one identifies $h = 5$, $k = 3$, and $r^2 = 4$. Consequently, the center is $(5, 3)$ and the radius is $2$. Similarly, Example 4 provides the equation $(x + 3)^2 + (y - 5)^2 = 9$, wherein the center is identified as $(-3, 5)$ and the radius is $\sqrt{9} = 3$.

Conversion and Completion of the Square

To convert a circle's equation from standard form to general form, one must expand the binomials. For example, Example 5 takes $(x + 3)^2 + (y - 5)^2 = 9$ and expands it to $x^2 + 6x + 9 + y^2 - 10y + 25 = 9$. Grouping terms leads to $x^2 + y^2 + 6x - 10y + 34 = 9$, which simplifies to the general form $x^2 + y^2 + 6x - 10y + 25 = 0$. To move from general form back to standard form, the process of completing the square is utilized. In Example 6, the equation $x^2 + y^2 - 2x - 8y + 8 = 0$ is solved by grouping $(x^2 - 2x)$ and $(y^2 - 8y)$. Adding and subtracting the necessary constants gives $(x^2 - 2x + 1 - 1) + (y^2 - 8y + 16 - 16) + 8 = 0$, which becomes $(x - 1)^2 + (y - 4)^2 - 17 + 8 = 0$, or $(x - 1)^2 + (y - 4)^2 = 9$. This identifies the center as $(1, 4)$ and the radius as $3$. Example 7 follows a similar logic with $x^2 + y^2 - 6x + 2y = -1$, leading to $(x - 3)^2 + (y + 1)^2 = 9$, noting a center of $(3, -1)$ and a radius of $3$.

Finding Circle Equations from Specific Geometric Data

Equations can also be constructed when the circle is defined by a center and a point it passes through, or by the endpoints of a diameter. Example 8 describes a circle centered at $(1, 2)$ that passes through the point $(3, 6)$. Substituting the center into the equation gives $(x - 1)^2 + (y - 2)^2 = r^2$. To find $r^2$, the point $(3, 6)$ is substituted: $(3 - 1)^2 + (6 - 2)^2 = r^2$, resulting in $2^2 + 4^2 = 20 = r^2$. The final equation is $(x - 1)^2 + (y - 2)^2 = 20$. Example 9 uses center $(-2, 0)$ and point $(-4, -1)$. Solving $(-4 + 2)^2 + (-1 - 0)^2 = r^2$ gives $(-2)^2 + (-1)^2 = 5 = r^2$, yielding $(x + 2)^2 + y^2 = 5$.

When given diameter endpoints, the center must first be found using the midpoint formula: $(h, k) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$. In Example 10, endpoints $(2, 4)$ and $(4, -2)$ result in a center of $(\frac{2+4}{2}, \frac{4-2}{2}) = (3, 1)$. Squaring the distance from this center to the point $(2, 4)$ gives $(2 - 3)^2 + (4 - 1)^2 = 1 + 9 = 10 = r^2$, resulting in $(x - 3)^2 + (y - 1)^2 = 10$. Example 11 follows with endpoints $(-4, 1)$ and $(2, 5)$, finding a center of $(-1, 3)$. Substituting point $(2, 5)$ into $(x + 1)^2 + (y - 3)^2 = r^2$ gives $(2 + 1)^2 + (5 - 3)^2 = 9 + 4 = 13$, leading to the final equation $(x + 1)^2 + (y - 3)^2 = 13$.