To explore the different forms that an equation of a circle can appear as it is important to understand what a circle and its parts are in mathematics. Basically, a circle is a curve that closes on itself and has vertical and horizontal symmetry.
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A circle can be broken down into two parts: the circumference and center. The circumference is a sequence of points that are equidistant to the center, while the center is a point that marks the central location of the figure. The segment that connects any point of the circumference to the center is called the radius. Paraphrasing one of the postulates developed by Euclid, a circle can be drawn with line segments that have the length equal to the radius of the circle, and that have the center of the circle as an endpoint.
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A simple explanation of the equation of a circle is that it is mathematical equality that generates specific circles. These equations allow one to draw these figures on a coordinate system (x and y-axis).
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To understand how to find the equation of a circle, one needs to be familiar with the Pythagorean Theorem (hypotenuse formula). This theorem is applied to right triangles that, in the case of a circle, have their hypotenuse formed by a segment drawn from the center of a circle to a point in the circumference (radius).
The following sections will go in-depth in using the hypothenuse formula to explore circle equations.
The hypotenuse formula is based on labeling the parts of a right triangle and inserting each of these parts in the equation {eq}a^2 + b^2 = c^2 {/eq}. The variables a and b are lengths of the sides, also known as catheti (plural form of cathetus), of the triangle that are adjacent to the right angle. The variable c is the length of the hypotenuse, which is the side of a right triangle that is opposite to its right angle.
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To apply the formula shown in the previous section to find the equation of a circle, one must follow these steps:
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Figure 4 shows how a segment from the x-axis completes the figure of the right triangle.
Substituting these values in the formula {eq}a^2 + b^2 = c^2 {/eq}, being {eq}r = c {/eq} (the hypotenuse), the result is:
{eq}x^2 + y^2 = r^2 {/eq}
This is the equation of a circle with a center located at the origin of a coordinate system. Any points (x, y) that satisfy the equation of a circle with radius r, are part of the circumference of this circle.
When the center of a circle is displaced from the origin (0, 0) to a different position (A, B), the difference in the position from each axis must be discounted from the terms x and y of the equation given in the previous section. Thus:
{eq}(x-A)^2 + (y-B)^2 = r^2 {/eq}
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The previous sections showed how to find the standard equation of a circle:
The equation of a circle, however, can also be expressed in what is called general form.
The circle equation general form is expressed as:
{eq}x^2 + y^2 + Ax + By - C = 0 {/eq}
Where, considering P (x, y) as a point in the circumference of a circle, some terms of the equation are:
Considering a circle with the center located at (2, 3) and radius equal to 100, the standard form of the equation will be:
{eq}(x-2)^2 + (y-3)^2 = 100^2 {/eq}
Proceeding with the operations:
{eq}(x-2)(x-2) + (y-3)(y-3) = 10,000 {/eq}
{eq}x^2-2x-2x+4+y^2-3y-3y+9 = 10,000 {/eq}
Organizing the terms:
{eq}x^2+y^2-2x-3y+4+9-10,000 = 0 {/eq}
{eq}x^2+y^2-2x-3y-9,987 = 0 {/eq}
In this case, the term C from {eq}x^2 + y^2 + Ax + By - C = 0 {/eq} is equals to 9,987.
The equation {eq}x^2+y^2-2x-3y-9,987 = 0 {/eq} shows that {eq}C = r^2 - A^2 - B^2 {/eq}.
Thus {eq}r^2 = C + A^2 + B^2 {/eq}
Then {eq}r = \sqrt {C + A^2 + B^2} {/eq}
Testing it for {eq}x^2+y^2-2x-3y-9,987 = 0 {/eq}:
{eq}r = \sqrt {9,987 + 2^2 + 3^2} {/eq}
{eq}r = \sqrt {9,987 + 4 + 9} {/eq}
{eq}r = \sqrt {10,000} {/eq}
{eq}r = 100 {/eq}
Examples 1 and 2 include solved problems involving the equation of a circle.
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The circle is not centered at the origin, but at the position (1, 1). Therefore, A=1, and B=1. The radius of the circle extending from the point (1, 1) to the point (4, 1) is equal to 3 units.
Substituting these values in {eq}(x-A)^2 + (y-B)^2 = r^2 {/eq}, results in:
{eq}(x-1)^2 + (y-1)^2 = 9 {/eq}
The standard form for a circle with a center at the origin is given by {eq}x^2 + y^2 = r^2 {/eq}.
The general form, then, will be {eq}x^2 + y^2 - C = 0 {/eq}, with {eq}C = r^2 {/eq}.
The radius is given by the distance between (0, 0) and (2, 0), which results in a length of 2 units.
Thus, the equation will be {eq}x^2 + y^2 - 4 = 0 {/eq}.
This lesson focused on describing a circle. The two main elements that compose this shape are the center and the circumference, being the distance between the center and any point on the circumference called the radius. A circle, then, can be drawn by grouping several segments of equal length emerging from a central point going to different directions.
A circle can be described by an equation that results from the application of the hypothenuse formula for a right triangle drawn in the circle. This triangle will have the hypothenuse equal to the radius of the circle, and the other sides given by segments over the x and y-axis. Substituting these elements in the formula {eq}a^2+b^2=c^2 {/eq}, one can find the circle equation {eq}x^2+y^2=r^2 {/eq}. This is the equation for a circle with a center at the origin. For a circle with a center at a point (A, B) out of the origin, the equation will become {eq}(x-A)^2 + (y-B)^2 = r^2 {/eq}. The aforementioned formulas are called the standard form of a circle equation. A general form of this formula will have a structure given by {eq}x^2 + y^2 + Ax + By - C = 0 {/eq}.
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Considering that one of the points is the center and the other is a point on the circumference, one must find the distance between the points (radius). Then, one needs to substitute r for the value of the radius and A and B for the coordinates of the center.
The standard form of the equation will be given by the sum of two terms equals the square of the radius. The first term will be the square of the subtraction (x-A), being A the x-coordinate of the center. The second term will be the square of the subtraction (y-B). Where B is the y-coordinate of the center.
Choose a point (x, y) on the circumference. Trace a segment connecting the center and the chosen point (radius). Trace a vertical line connecting the point on the circumference and the x-axis. A right triangle will be formed with the hypothenuse being the radius, and the two other sides having the distance x and y. Finally, one simply needs to substitute these variables on the formula of the Pythagorean theorem.
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