Parametric Equations for a Circle
Parametric Equations for a Circle

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Equation of Circle

The equation of circle provides an algebraic way to describe a circle, given the center and the length of the radius of a circle. The equation of a circle is different from the formulas that are used to calculate the area or the circumference of a circle. This equation is used across many problems of circles in coordinate geometry.

To represent a circle on the Cartesian plane, we require the equation of the circle. A circle can be drawn on a piece of paper if we know its center and the length of its radius. Similarly, on a Cartesian plane, we can draw a circle if we know the coordinates of the center and its radius. A circle can be represented in many forms:

• General form
• Standard form
• Parametric form
• Polar form

In this article, let’s learn about the equation of the circle, its various forms with graphs and solved examples.

## What is the Equation of Circle?

An equation of a circle represents the position of a circle in a Cartesian plane. If we know the coordinates of the center of the circle and the length of its radius, we can write the equation of a circle. The equation of circle represents all the points that lie on the circumference of the circle.

A circle represents the locus of points whose distance from a fixed point is a constant value. This fixed point is called the center of the circle and the constant value is the radius r of the circle. The standard equation of a circle with center at $$(x_1, y_1)$$ and radius r is $$(x – x_1)^2 + (y – y_1)^2 = r^2$$.

## Different Forms of Equation of Circle

An equation of circle represents the position of a circle on a cartesian plane. A circle can be drawn on a piece of paper given its center and the length of its radius. Using the equation of circle, once we find the coordinates of the center of the circle and its radius, we will be able to draw the circle on the cartesian plane. There are different forms to represent the equation of a circle,

• General form
• Standard form
• Parametric form
• Polar form

Let’s look at the two common forms of the equation of circle-general form and standard form of the equation of circle here along with the polar and parametric forms in detail.

### General Equation of a Circle

The general form of the equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0. This general form is used to find the coordinates of the center of the circle and the radius, where g, f, c are constants. Unlike the standard form which is easier to understand, the general form of the equation of a circle makes it difficult to find any meaningful properties about any given circle. So, we will be using the completing the square formula to make a quick conversion from the general form to the standard form.

### Standard Equation of a Circle

The standard equation of a circle gives precise information about the center of the circle and its radius and therefore, it is much easier to read the center and the radius of the circle at a glance. The standard equation of a circle with center at $$(x_1, y_1)$$ and radius r is $$(x – x_1)^2 + (y – y_1)^2 = r^2$$, where (x, y) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – x_1)^2 + (y – y_1)^2} = r$$

Squaring both sides, we get the standard form of the equation of the circle as:

$$(x – x_1)^2 + (y – y_1)^2 = r^2$$

Consider this example of an equation of circle (x – 4)2 + (y – 2)2 = 36 is a circle centered at (4,2) with a radius of 6.

### Parametric Equation of a Circle

We know that the general form of the equation of a circle is x2 + y2 + 2hx + 2ky + C = 0. We take a general point on the boundary of the circle, say (x, y). The line joining this general point and the center of the circle (-h, -k) makes an angle of $$\theta$$. The parametric equation of circle can be written as x2 + y2 + 2hx + 2ky + C = 0 where x = -h + rcosθ and y = -k + rsinθ.

### Polar Equation of a Circle

The polar form of the equation of the circle is almost similar to the parametric form of the equation of circle. We usually write the polar form of the equation of circle for the circle centered at the origin. Let’s take a point P(rcosθ, rsinθ) on the boundary of the circle, where r is the distance of the point from the origin. We know that the equation of circle centered at the origin and having radius ‘p’ is x2 + y2 = p2.

Substitute the value of x = rcosθ and y = rsinθ in the equation of circle.

(rcosθ)2 + (rsinθ)2 = p2
r2cos2θ + r2sin2θ = p2
r2(cos2θ + sin2θ) = p2
r2(1) = p2
r = p
where p is the radius of the circle.

Example: Find the equation of the circle in the polar form provided that the equation of the circle in standard form is: x2 + y2 = 9.

Solution:

To find the equation of the circle in polar form, substitute the values of $$x$$ and $$y$$ with:

x = rcosθ
y = rsinθ

x = rcosθ
y = rsinθ
x2 + y2 = 9
(rcosθ)2 + (rsinθ)2 = 9
r2cos2θ + r2sin2θ = 9
r2(cos2θ + sin2θ) = 9
r2(1) = 9
r = 3

## Equation of a Circle Formula

The equation of a circle formula is used for calculating the equation of a circle. We can find the equation of any circle, given the coordinates of the center and the radius of the circle by applying the equation of circle formula. The equation of circle formula is given as, $$(x – x_1)^2 + (y – y_1)^2 = r^2$$.

where,$$(x_1, y_1)$$ is the center of the circle with radius r and (x, y) is an arbitrary point on the circumference of the circle.

## Derivation of Circle Equation

Given that $$(x_1, y_1)$$ is the center of the circle with radius r and (x, y) is an arbitrary point on the circumference of the circle. The distance between this point and the center is equal to the radius of the circle. So, let’s apply the distance formula between these points.

$$\sqrt{(x – x_1)^2 + (y – y_1)^2} = r$$.

Squaring both sides, we get: $$(x – x_1)^2 + (y – y_1)^2 = r^2$$. So, the equation of a circle is given by:

$$(x – x_1)^2 + (y – y_1)^2 = r^2$$

Example: Using the equation of circle formula, find the center and radius of the circle whose equation is (x – 1)2 + (y + 2)2 = 9.

Solution:

We will use the circle equation to determine the center and radius of the circle.
Comparing $$(x – 1)^2 + (y + 2)^2 = 9$$ with $$(x – x_1)^2 + (y – y_1)^2 = r^2$$, we get

$$x_1$$ = 1, $$y_1$$ = -2 and r = 3

So, the center and radius are (1, -2) and 3 respectively.

Answer: The center of the circle is (1, -2) and its radius is 3.

## Graphing the Equation of Circle

• For this, we only need to change the constant 9 to match with r2 as (x -3)2 + (y – 2)2 = 32.
• Here, we need to note that one of the common mistakes to commit is to consider $$x_{1}$$ as -3 and $$y_{1}$$ as -2.
• In the equation of circle, if the sign preceding $$x_{1}$$ and $$y_{1}$$ are negative, then $$x_{1}$$ and $$y_{1}$$ are positive values and vice versa.
• Here, $$x_{1}$$ = 3, $$y_{1}$$ = 2 and r = 3

Thus, the circle represented by the equation (x -3)2 + (y – 2)2 = 32, has its center at (3, 2) and has a radius of 3. The below-given image shows the graph obtained from this equation of the circle.

## How to Find Equation of Circle?

There are so many different ways of representing the equation of circle depending on the position of the circle on the cartesian plane. We have studied the forms to represent the equation of circle for given coordinates of center of a circle. There are certain special cases based on the position of the circle in the coordinate plane. Let’s learn about the method to find the equation of circle for the general and these special cases.

### Equation of Circle With Center at (x$$_1$$, y$$_1$$)

To write the equation of circle with center at (x$$_1$$, y$$_1$$), we will use the following steps,

• Step 1: Note down the coordinates of the center of the circle(x$$_1$$, y$$_1$$) and the radius ‘r’.
• Step 2: Apply the equation of circle formula, $$\sqrt{(x – x_1)^2 + (y – y_1)^2} = r$$.
• Step 3: Express the answer in the required circle equation form.

### Equation of Circle With Center at the Origin

The simplest case is where the circle’s center is at the origin (0, 0), whose radius is r. (x, y) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – 0)^2 + (y – 0)^2} = r$$

Squaring both sides, we get:

$$(x – 0)^2 + (y – 0)^2 = r^2$$

$$x^2 + y^2 = r^2$$

Example: What will be the equation of a circle if its center is at the origin?

Solution:

The equation of a circle is given by $$(x – x_1)^2 + (y – y_1)^2 = r^2$$.

If center is at origin, then $$x_1$$= 0 and $$y_1$$= 0.

Answer: The equation of the circle if its center is at origin is x2+ y2= r2.

### Equation of Circle With Centre on x-Axis

Consider the case where the center of the circle is on the x-axis: (a, 0) is the center of the circle with radius r. (x, y) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – a)^2 + (y – 0)^2} = r$$

Squaring both sides, we get:

$$(x – a)^2 + (y – 0)^2 = r^2$$

$$(x – a)^2 + (y)^2 = r^2$$

### Equation of Circle With Centre on Y-Axis

Consider the case where the center of the circle is on the y-axis: (0, b) is the center of the circle with radius r. (x, y) is an arbitrary point on the circumference of the circle.

The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – 0)^2 + (y – b)^2} = r$$

Squaring both sides, we get:

$$(x – 0)^2 + (y – b)^2 = r^2$$

$$(x)^2 + (y – b)^2 = r^2$$

### Equation of Circle Touching x-Axis

Consider the case where the circumference of the circle is touching the x-axis at some point: (a, r) is the center of the circle with radius r. If a circle touches the x-axis, then the y-coordinate of the center of the circle is equal to the radius r.

(x, y) is an arbitrary point on the circumference of the circle. The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – a)^2 + (y – r)^2} = r$$

Squaring both sides, we get:

$$(x – a)^2 + (y – r)^2 = r^2$$

### Equation of Circle Touching y-Axis

Consider the case where the circumference of the circle is touching the y-axis at some point: (r, b) is the center of the circle with radius r. If a circle touches the y-axis, then the x-coordinate of the center of the circle is equal to the radius r.

(x, y) is an arbitrary point on the circumference of the circle. The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – r)^2 + (y – b)^2} = r$$

Squaring both sides

$$(x – r)^2 + (y – b)^2 = r^2$$

### Equation of Circle Which Touches Both the Axes

Consider the case where the circumference of the circle is touching both the axes at some point: (r, r) is the center of the circle with radius r. If a circle touches both the x-axis and y-axis, then both the coordinates of the center of the circle become equal to the radius (r, r).

(x, y) is an arbitrary point on the circumference of the circle. The distance between this point and the center is equal to the radius of the circle. Let’s apply the distance formula between these points.

$$\sqrt{(x – r)^2 + (y – r)^2} = r$$

Squaring both sides

$$(x – r)^2 + (y – r)^2 = r^2$$

If a circle touches both the axes, then consider the center of the circle to be (r,r), where r is the radius of the circle. Here, (r,r) can be positive as well as negative. For example, the radius of the circle is 3 and it is touching both the axes, then the coordinates of the center can be (3,3), (3,−3), (−3,3), or (−3,−3).

Example: If the equation of circle in general form is given as $$x^2 + y^2 + 6x + 8y + 9 = 0$$, find the coordinates of the center and the radius of the circle.

Solution:

Given the equation of the circle $$x^2 + y^2 +6x + 8y + 9 = 0$$

The general form of the equation of the circle with center $$(x_1, y_1)$$ and radius $$r$$ is $$x^2 + y^2 + Ax + By + C = 0$$
where $$A = -2x_1$$
$$B = -2y_1$$
$$C = {x_1}^2 + {y_1}^2 -r^2$$

From the equation of the circle $$x^2 + y^2 +6x + 8y + 9 = 0$$

$$A = 6 \\ -2x_1 = 6 \\ x_1 = -3 \\ B = 8 \\ -2y_1 = 8 \\ y_1 = -4 \\ C = 9 \\ {x_1}^2 + {y_1}^2 -r^2 = 9 \\ {-3}^2 + {-4}^2 -r^2 = 9 \\ 9 + 16 -r^2 = 9 \\ r^2 = 16 \\ r = 4$$

## Converting General Form to Standard Form

This is the standard equation of circle, with radius r and center at (a,b): (x – a)2 + (y – b)2 = r2 and consider the general form as: x2 + y2 + 2gx + 2fy + c = 0. Here are the steps to be followed to convert the general form to the standard form:

Step 1: Combine the like terms and take the constant on the other side as x2 + 2gx + y2 + 2fy = – c -> (1)

Step 2: Use the perfect square identity (x + g)2 = x2 + 2gx + g2 to find the values of the expression x2 + 2gx and y2 + 2fy as:

(x + g)2 = x2 + 2gx + g2 ⇒ x2 + 2gx = (x + g)2 – g2 -> (2)

(y + f)2 = y2 + 2fy + f2 ⇒ y2 + 2fy = (y + f)2 – f2 -> (3)

Substituting (2) and (3) in (1), we get the equation as:

(x+g)2 – g2 + (y+f)2 – f2 = – c

(x+g)2 + (y+f)2 = g2 + f2 – c

Comparing this equation with the standard form: (x – a)2 + (y – b)2 = r2 we get,

Center = (-g,-f) and radius = $$\sqrt{g^2+f^2 – c}$$

We need to make sure that the coefficients of x2 and y2 are 1 before applying the formula.

Consider an example here to find the center and radius of the circle from the general equation of the circle: x2 + y2 – 6x – 8y + 9 = 0.

The coordinates of the center of the circle can be found as: (-g,-f). Here g = -6/2 = -3 and f = -8/2 = -4. So, the center is (3,4).

Radius r = $$\sqrt{g^2+f^2 – c}$$ = $$\sqrt{(-3)^{2}+(-4)^{2} – 9}$$ = $$\sqrt{9 + 16 – 9}$$ = $$\sqrt{16}$$ = 4. So, radius r = 4.

## Converting Standard Form to General Form

We can use the algebraic identity formula of (a – b)2 = a2 + b2 – 2ab to convert the standard form of equation of circle into the general form. Let’s see how to do this conversion. For this, expand the standard form of the equation of the circle as shown below, using the algebraic identities for squares:

$$(x – x_1)^2 + (y – y_1)^2 = r^2$$

$$x^2 +{x_1}^2 -2xx_1 + y^2 +{y_1}^2 -2yy_1 = r^2$$
$$x^2 + y^2 – 2xx_1 – 2yy_1 + {x_1}^2 + {y_1}^2 = r^2$$
$$x^2 + y^2 – 2xx_1 – 2yy_1 + {x_1}^2 + {y_1}^2 -r^2 = 0$$

Replace $$-2x_1$$ with 2g, $$-2y_1$$ with 2f, $${x_1}^2 + {y_1}^2 -r^2$$ with $$c$$, we get:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Now, we get the general form of equation of circle as: $$x^2 + y^2 + 2gx + 2fy + c = 0$$, where g, f, c are constants.

Related Articles on Equation of Circle

Check out the following pages related to the equation of circle

• Equation of a Circle Calculator
• Circumference of a Circle
• All Circle Formulas
• Ratio of Circumference to Diameter

Important Notes on Equation of Circle

Here is a list of a few points that should be remembered while studying the equation of circle

• The general form of the equation of circle always has x2 + y2 in the beginning.
• If a circle crosses both the axes, then there are four points of intersection of the circle and the axes.
• If a circle touches both the axes, then there are only two points of contact.
• If any equation is of the form $$x^2 + y^2 + axy + C = 0$$, then it is not the equation of the circle. There is no $$xy$$ term in the equation of circle.
• In polar form, the equation of circle always represents in the form of $$r$$ and $$\theta$$.
• Radius is the distance from the center to any point on the boundary of the circle. Hence, the value of the radius of the circle is always positive.

## Examples on Circle Equations

1. Example 1: Find the equation of the circle in standard form for a circle with center (2,-3) and radius 3.

Solution:

Equation for a circle in standard form is written as: (x – x$$_1$$)2 + (y – y$$_1$$)2 = r2. Here, (x$$_1$$, y$$_1$$) = (2, -3) is the center of the circle and radius r = 3.

Let’s put these values in the standard form of equation of circle:

(x – 2)2 + (y – (-3))2 = (3)2

(x – 2)2 + (y + 3)2 = 9 is the required standard form of the equation of the given circle.

2. Example 2: Write the equation of circle in standard form for a circle with center (-1, 2) and radius equal to 7.

Solution:

Equation for a circle in standard form is written as: (x – x$$_1$$)2 + (y – y$$_1$$)2 = r2. Here (x$$_1$$, y$$_1$$) = (-1, 2) is the center of the circle and radius r = 7.

Let’s put these values in the standard form of equation of circle:

(x – (-1))2 + (y – 2)2 = 72

(x + 1)2 + (y – 2)2 = 49 is the required standard form of the equation of the given circle.

3. Example 3: Find the equation of the circle in the polar form provided that the equation of the circle in standard form is: x2 + y2 = 16.

Solution:

To find the equation of the circle in polar form, substitute the values of x and y with:

x = rcosθ

y = rsinθ

x2 + y2 = 16

(rcosθ)2 + (rsinθ)2 = 16

r2cos2θ + r2sin2θ = 16

r2(1) = 4

## FAQs on Equation of Circle

### What is the Equation of Circle in Geometry?

The equation of circle represents the locus of point whose distance from a fixed point is a constant value. This fixed point is called the center of the circle and the constant value is the radius of the circle. The standard equation of circle with center at $$(x_1, y_1)$$ and radius r is $$(x – x_1)^2 + (y – y_1)^2 = r^2$$.

What is the Equation of Circle When the Center Is at the Origin?

The simplest case is where the circle’s center is at the origin (0, 0), whose radius is r. (x, y) is an arbitrary point on the circumference of the circle. The equation of circle when the center is at the origin is x2 + y2 = r2.

### What is the General Equation of Circle?

The general form of the equation of circle is: x2 + y2 + 2gx + 2fy + c = 0. This general form of the equation of circle has a center of (-g, -f), and the radius of the circle is r = $$\sqrt{g^2 + f^2 – c}$$.

### What is the Parametric Equation of Circle?

The parametric equation of circle can be written as $$x^2 + y^2 + 2hx + 2ky + C = 0$$ where $$x = -h +rcos \theta$$ and $$y = -k +rsin \theta$$

### What is C in the General Equation of Circle?

The general form of the equation of circle is: x2 + y2 + 2gx + 2fy + c = 0. This general form is used to find the coordinates of the center of the circle and the radius of the circle. Here, c is a constant term, and the equation having c value represents a circle that is not passing through the origin.

### What are the Various Forms of Equations of a Circle?

Let’s look at the two common forms of the equation are:

• General Form x2 + y2 + 2gx + 2fy + C = 0
• Standard Form $$(x – x_1)^2 + (y – y_1)^2 = r^2$$

What is the Equation of Circle When the Center is on x-Axis?

Consider the case where the center of the circle is on the x-axis: (a, 0) is the center of the circle with radius r. (x, y) is an arbitrary point on the circumference of the circle. The equation of circle when the center is on the x-axis is $$(x – a)^2 + (y)^2 = r^2$$

### How do you Graph a Circle Equation?

To graph a circle equation, first find out the coordinates of the center of the circle and the radius of the circle with the help of the equation of the circle.

Then plot the center on a cartesian plane and with the help of a compass measure the radius and draw the circle.

### How do you Find the General Equation of Circle?

If we know the coordinates of the center of a circle and the radius then we can find the general equation of circle. For example, the center of the circle is (1, 1) and the radius is 2 units then the general equation of the circle can be obtained by substituting the values of center and radius.The general equation of the circle is $$x^2 + y^2 + Ax + By + C = 0$$.

$$\text{A} = -2 \times 1 = -2$$
$$\text{B} = -2 \times 1 = -2$$
$$\text{C} = 1^2 + 1^2 – 2^2 = -2$$

Hence the general form of the equation of circle is $$x^2 + y^2 – 2x – 2y – 2 = 0$$.

How do you Write the Standard Form of Equation of a Circle?

The standard form of the equation of a circle is $$(x – x_1)^2 + (y – y_1)^2 = r^2$$, where $$(x_1, y_1)$$ is the coordinate of the center of the circle and $$r$$ is the radius of the circle

How do you Go From Standard Form to a General Form of Equation of a Circle?

Let’s convert the equation of circle: $${(x – 1)}^2 + {(y – 2)}^2 = 4$$ from standard form to gerenal form.

$${(x – 1)}^2 + {(y – 2)}^2 = 4 \\ x^2 + 1 – 2x + y^2 + 4 – 4y = 4 \\ x^2 + y^2 – 2x – 4y + 1 = 0$$

The above form of the equation is the general form of the equation of circle.

How do you Write the Standard Form of a Circle Equation with Endpoints?

Let’s take the two endpoints of the diameter to be (1, 1), and (3, 3). First, calculate the midpoint by using the section formula. The coordinates of the center will be (2, 2). Secondly, calculate the radius by distance formula between (1, 1), and (2, 2). Radius is equal to $$\sqrt{2}$$. Now, the equation of the circle in standard form is $${(x – 2)}^2 + {(y – 2)}^2 = 2$$.

### What is the Polar Equation of a Circle?

The polar equation of the circle with the center as the origin is, r = p, where p is the radius of the circle.

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