Equation of a Circle
Introduction
Equation of a circle is an algebraic way to represent a circle. It is the locus of the point which is at a constant distance from a fixed point, called the center of the circle. The constant distance from the center is called the radius of the circle. The equation of a circle represents all the points that lie on the circumference of a circle.
Numerous circle problems in coordinate geometry employ this equation.
The equation of a circle is very helpful in `2D` and `3D` coordinate geometry for find various factors related to a circle like area, circumference, area of a sector, arc length, etc. Understanding their equations provides insights into their properties, allowing us to analyze and manipulate them with precision.
Defining Equation of a Circle
The equation of a circle represents the relationship between the coordinates of points lying on the circle's circumference and the circle's center. The algebraic equation of a circle is written based on the coordinates of the center of the circle and the length of the radius of the circle. Again, center of a circle is the fixed point from which the radius is calculated.
The general form of the equation of a circle in the cartesian coordinate system is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
where:
`(h, k)` represents the coordinates of the circle's center.
`r` represents the radius of the circle.
This equation demonstrates that for any point ` (x, y) ` lying on the circumference of the circle, the sum of the squares of the differences between the point's coordinates and the coordinates of the center is equal to the square of the radius.
By manipulating this equation, we can explore various properties of circles, such as their position in the coordinate plane, their size, and their intersections with other geometric figures. Additionally, understanding the equation of a circle lays the groundwork for more advanced topics in geometry, calculus, and physics, where circles frequently appear as fundamental shapes in models and calculations.
Various Forms of Equation of a Circle
The equation of a circle can be written in various forms that are listed below:
- General Form
- Standard Form
- Polar Form
- Parametric Form
Equation of a Circle in General Form
The general equation of a circle is expressed in the form:
\( x^2 + y^2 + Dx + Ey + F = 0 \)
where ` D `, ` E `, and ` F ` are constants.
This equation represents all possible circles in a cartesian coordinate system, regardless of their center or radius. It encompasses circles of varying sizes, positions, and orientations within the coordinate plane.
The general form of the equation of a circle is obtained by expanding and simplifying the standard form equation:
\( (x - h)^2 + (y - k)^2 = r^2 \)
Expanding the squares and rearranging terms results in the general form equation.
In the general form, ` D ` and ` E ` represent coefficients of ` x ` and ` y `, respectively, while ` F ` is a constant term. These coefficients determine the position of the circle's center and influence its orientation within the coordinate plane.
While the standard form of the circle's equation is often more intuitive for understanding the center and radius of a circle, the general form provides a broader perspective, allowing for the representation of circles with arbitrary positions and orientations.
Equation of a Circle in Standard Form
The standard equation of a circle represents a circle in the cartesian coordinate system with its center at the point ` (h, k) ` and a radius of ` r `. The standard form of the equation of a circle is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
In this equation
` (h, k) ` represents the coordinates of the center of the circle.
` r ` represents the radius of the circle.
This equation expresses the fundamental relationship between the coordinates of points lying on the circle's circumference and the circle's center. Specifically, it states that the sum of the squares of the differences between the ` x `-coordinates and ` y `-coordinates of any point on the circle and the center of the circle is equal to the square of the radius.
The standard equation of a circle is particularly useful for graphing circles and determining their properties such as center, radius, and position within the coordinate plane. It provides a clear and concise representation of the geometric attributes of a circle, facilitating mathematical analysis and problem-solving involving circles.
Equation of a Circle in Polar Form
Consider a point on the circle `(x, y)` as `a\cos\theta, a\cos\theta` where the line joining the point with the center of the circle makes an angle `\theta` with the `x`-axis. The radius of the circle is ‘`a`’ and the center of the circle is at the origin `(0, 0)`. Having said that , we can write the equation of the circle using
`x = a\cos\theta … (i)`
`y = a\sin\theta … (ii)`
Adding and squaring equations `(i)` and `(ii)` we get
\( x^2 + y^2 = a^2cos^2\theta + a^2\sin^2\theta \)
\( x^2 + y^2 = a^2( \cos^2\theta + \sin^2\theta) \)
As per the trigonometry identity for `\text{sine}` and `\text{cosine}`, `\cos^2\theta + \sin^2\theta = 1`
Hence \( x^2 + y^2 = a^2 \)
Here center `= (0, 0)` and the radius `r = a`.
Equation of a Circle in Parametric Form
In the parametric form of a circle equation, any point on the circle is represented by `(-h + r\cos\theta, -k + r\sin\theta) `. The line joining the center of the circle `(h, k)` to any such point makes an angle `\theta` with the `x`-axis. The parametric equation of a circle is
\( x^2 + y^2 +2hx +2ky + c = 0 \)
Derivation of Equation of a Circle
The equation of a circle is derived from the definition of a circle as the set of all points that are equidistant from a fixed point called the center. Let's go through the derivation step by step:
`1`. Definition of a circle: A circle is defined as the set of all points that are a fixed distance (called the radius) away from a given point (called the center). Mathematically, this means that for any point ` (x, y) ` on the circle, the distance between ` (x, y) ` and the center ` (h, k) ` is equal to the radius ` r `.
`2`. Distance formula: The distance between two points ` (x_1, y_1) ` and ` (x_2, y_2) ` in a cartesian coordinate system is given by the distance formula:
\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
`3`. Distance from a point to the center of the circle: Using the distance formula, the distance between ` (x, y) ` and ` (h, k) `, the center of the circle, is:
\( d = \sqrt{(x - h)^2 + (y - k)^2} \)
`4`. Equidistant Points: For any point ` (x, y) ` on the circle, the distance from ` (x, y) ` to the center ` (h, k) ` is equal to the radius ` r `. Therefore, we can write:
\( r = \sqrt{(x - h)^2 + (y - k)^2} \)
`5`. Squaring Both Sides: To eliminate the square root, we square both sides of the equation:
\( r^2 = (x - h)^2 + (y - k)^2 \)
`6`. Standard Form: Rearranging terms, we obtain the standard form of the equation of a circle:
\( (x - h)^2 + (y - k)^2 = r^2 \)
This equation represents a circle with its center at ` (h, k) ` and a radius of ` r `. It expresses the relationship between the coordinates of any point on the circle and the center of the circle. Therefore, the equation of a circle is derived from the fundamental geometric definition of a circle and the distance formula in a Cartesian coordinate system.
Solved Examples
Example `1`: Given the center ` (h, k) = (2, -3) ` and radius ` r = 5 `, find the equations of the circle in both standard and general forms.
Solution:
Standard Form:
\( (x - h)^2 + (y - k)^2 = r^2 \)
\( (x - 2)^2 + (y - (-3))^2 = 5^2 \)
\( (x - 2)^2 + (y + 3)^2 = 25 \)
General Form:
\( x^2 + y^2 + Dx + Ey + F = 0 \)
We need to expand `(x-2)^2` and `(y-2)^2` to convert from standard form to general form.
Hence ` (x - 2)^2 + (y + 3)^2 = 25 ` can be written as
\( x^2 - 4x + 4 + y^2 + 6y + 9 = 25 \)
\( x^2 + y^2 - 4x + 6y - 9 = 0 \)
The equation of the circle in its standard form is ` (x - 2)^2 + (y + 3)^2 = 25`.
The equation of the circle in its general form is ` x^2 + y^2 - 4x + 6y - 9 = 0`.
Example `2`: Given the circle with the equation ` x^2 + y^2 - 6x + 4y - 12 = 0 `, find the center and radius, and rewrite it in standard form.
Solution:
General Form:
\( x^2 + y^2 - 6x + 4y - 12 = 0 \)
Rearranging terms by completing the square method, we can write ` x^2 + y^2 - 6x + 4y - 12 = 0 ` as
\( x^2 - 6x + 9 + y^2 + 4y + 4 -9 - 4 - 12 = 0 \)
\( (x - 3)^2 + (y + 2)^2 - 25 = 0 \)
\( (x - 3)^2 + (y + 2)^2 = 25 \)
Comparing ` (x - 3)^2 + (y + 2)^2 = 25 ` with the standard form ` (x - h)^2 + (y - k)^2 = r^2 `, we get
Center: ` (3, -2) `, Radius: ` r = \sqrt{25} = 5 `
Example `3`: Find the equation of the circle with the center at ` (-1, 2) ` passing through the point ` (3, -4) ` in standard form.
Solution:
The distance between the center `(-1, 2)` and the point `(3, -4)` is the radius of the circle.
Using the distance formula `d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ` we get
radius `r = sqrt{(3 - (-1))^2 + (-4 - 2)^2}`
`r = sqrt{4^2 + (-6)^2}`
`r = sqrt{16 + 36}`
`r^2 = 52`
Standard Form:
\( (x - (-1))^2 + (y - 2)^2 = 52\)
\( (x + 1)^2 + (y - 2)^2 = 52 \)
The equation of the circle in its standard form is ` (x + 1)^2 + (y - 2)^2 = 52`.
Example `4`: Given the circle with the equation ` x^2 + y^2 + 6x - 4y + 9 = 0 `. How do you find the radius of a circle and the center of the circle? Use these to rewrite the equation in standard form.
Solution:
General Form:
\( x^2 + y^2 + 6x - 4y + 9 = 0 \)
Rearranging terms by completing the square method, we can write ` x^2 + y^2 - 6x - 4y + 9 = 0 ` as
\( x^2 + 6x + 9 + y^2 - 4y + 4 -9 - 4 + 9 = 0 \)
\( (x + 3)^2 + (y - 2)^2 - 4 = 0 \)
Rearranging terms:
\( (x + 3)^2 + (y - 2)^2 = 4 \)
The equation for the circle in its standard form can be written as ` (x + 3)^2 + (y - 2)^2 = 4 ` with center at ` (-3, 2) ` and radius ` r = \sqrt{4} = 2 ` `\text{units}`.
Example `5`: Find the equation of the circle with the endpoints of its diameter at ` (-4, 1) ` and ` (4, -5) ` in standard form.
Solution:
Radius is the midpoint of the diameter.
Using the midpoint formula, we find the coordinates of the center as ` (\frac{-4 + 4}{2}, \frac{1 - 5}{2}) = (0, -2)`.
The distance between the center `(0, -2)` and the endpoint `(-4, 1)` is the radius of the circle.
Using the distance formula `d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ` we get
radius `r = sqrt{(-4 - 0)^2 + (1 - (-2))^2}`
`r = sqrt{4^2 + (3)^2}`
`r = sqrt{16 + 9}`
`r^2 = 25`
As per the circle formula in standard form:
\( (x - 0)^2 + (y - (-2))^2 = 25\)
\( x^2 + (y + 2)^2 = 25 \)
The equation of the circle in its standard form is ` x^2 + (y + 2)^2 = 25`.
Practice Problems
Q`1`. What is the standard form of the equation of a circle?
- ` (x - h)^2 + (y - k)^2 = r^2 `
- ` x^2 + y^2 + Dx + Ey + F = 0 `
- ` (x - h)^2 + (y - k)^2 + Dx + Ey + F = 0 `
- ` x^2 + y^2 = r^2 `
Answer: a
Q`2`. What does the variable ` r ` represent in the standard form of the equation of a circle?
- The `x`-coordinate of the circle's center
- The `y`-coordinate of the circle's center
- The radius of the circle
- The diameter of the circle
Answer: c
Q`3`. Which of the following is the general form of the equation of a circle?
- ` (x - h)^2 + (y - k)^2 = r^2 `
- ` x^2 + y^2 + Dx + Ey + F = 0 `
- ` x^2 + y^2 = r^2 `
- ` (x - h)^2 + (y - k)^2 + Dx + Ey + F = 0 `
Answer: b
Q`4`. Find the center and radius of a circle represented in its general form as `x^2 + y^2 + 6x -8y - 11 = 0`.
- Center: `(-3,4)`, Radius: `6` `\text{units}`
- Center: `(3,4)`, Radius: `4` `\text{units}`
- Center: `(2,4)`, Radius: `6` `\text{units}`
- Center: `(3,-4)`, Radius: `5` `\text{units}`
Answer: a
Q`5`. A circle has a radius of `6` `\text{units}` with its center located at `(-2, 5)`. Write the equation of a circle in standard form.
- `(x - 2)^2 + (y + 5)^2 = 6`
- `(x - 2)^2 + (y + 5)^2 = 36`
- `(x + 2)^2 + (y - 5)^2 = 36`
- `(x - 6)^2 + (y + 5)^2 = 4`
Answer: c
Frequently Asked Questions
Q`1`. What does the equation of a circle represent?
Answer: The equation of a circle represents the relationship between the coordinates of points lying on the circle's circumference and the circle's center. The algebraic equation of a circle is written based on the coordinates of the center of the circle and the length of the radius of the circle.
Q`2`. How to find the radius of a circle and its center from its equation in standard form?
Answer: We compare the given equation in standard from with `(x - h)^2 + (y - k)^2 = r^2` to find the coordinates of the center `(h, k)` and the radius `r`.
Q`3`. Can we determine the center of a circle if the general form of a circle is given?
Answer: Yes, we can . In order to find the center of the circle, we would convert the general form to standard form using completing the square method. Next, we would compare the equation in standard from with `(x - h)^2 + (y - k)^2 = r^2` to find the coordinates of the center `(h, k)`.
Q`4`: Can the equation of a circle have a negative radius?
Answer: No, the equation of a circle can never have a negative radius. Radius measures the distance between the center of the circle and any point on it. Hence, radius can never be a negative value.
Q`5`: What is the equation of a circle with radius `r` and its center at the origin?
Answer: The equation of a circle with radius `r` and its center at the origin will be `x^2 + y^2 = r^2`.