## Equation Of A Circle

The standard **equation of a circle** is given by:

**(x-h)**^{2}** + (y-k)**^{2}** = r**^{2}

Where (h,k) is the coordinates of center of the circle and r is the radius.

Before deriving the equation of a circle, let us focus on what is a circle? A circle is a set of all points which are equally spaced from a fixed point in a plane. The fixed point is called the center of the circle. The distance between the center and any point on the circumference is called the radius of the circle. In this article, we are going to discuss what is an equation of a circle formula in standard form, and find the equation of a circle when the center is the origin and the center is not an origin with examples.

## What is the Equation of a Circle?

A circle is a closed curve that is drawn from the fixed point called the center, in which all the points on the curve are having the same distance from the center point of the center. The equation of a circle with (h, k) center and r radius is given by:

(x-h)^{2} + (y-k)^{2} = r^{2}

This is the standard form of the equation. Thus, if we know the coordinates of the center of the circle and its radius as well, we can easily find its equation.

**Example:** Say point (1,2) is the center of the circle and radius is equal to 4 cm. Then the equation of this circle will be:

(x-1)^{2}+(y-2)^{2} = 4^{2}

(x^{2}−2x+1)+(y^{2}−4y+4) =16

X^{2}+y^{2}−2x−4y-11 = 0

### Function or Not

We know that there is a question that arises in case of circle whether being a function or not. It is clear that a circle is not a function. Because, a function is defined by each value in the domain is exactly associated with one point in the codomain, but a line that passes through the circle, intersects the line at two points on the surface.

The mathematical way to describe the circle is an equation. Here, the equation of the circle is provided in all the forms such as general form, standard form along with examples.

## Equation of a Circle When the Centre is Origin

Consider an arbitrary point P(x, y) on the circle. Let ‘a’ be the radius of the circle which is equal to OP.

We know that the distance between the point (x, y) and origin (0,0)can be found using the distance formula which is equal to-

√[x^{2}+ y^{2}]= a

Therefore, the equation of a circle, with the center as the origin is,

**x ^{2}+y^{2}= a^{2}**

Where “a” is the radius of the circle.

### Alternative Method

Let us derive in another way. Suppose (x,y) is a point on a circle, and the center of the circle is at origin (0,0). Now if we draw a perpendicular from point (x,y) to the x-axis, then we get a right triangle, where radius of the circle is the hypotenuse. The base of the triangle is the distance along x-axis and height is the distance along the y-axis. Thus, by applying the Pythagoras theorem here, we get:

**x**^{2}**+y**^{2 }**= radius**^{2}

## Equation of a Circle When the Centre is not an Origin

Let C(h, k) be the centre of the circle and P(x, y) be any point on the circle.

Therefore, the radius of a circle is CP.

By using distance formula,

(x-h)^{2 }+ (y-k)^{2} = CP^{2}

Let radius be ‘a’.

Therefore, the equation of the circle with center (h, k)and the radius ‘a’ is,

**(x-h) ^{2}+(y-k)^{2} = a^{2}**

which is called the **standard form for the equation of a circle.**

## General form of Equation of a Circle

The general equation of any type of circle is represented by:

x^{2} + y^{2} + 2gx + 2fy + c = 0, for all values of g, f and c.

Adding g^{2} + f^{2} on both sides of the equation gives,

x^{2} + 2gx + g^{2}+ y^{2} + 2fy + f^{2}= g^{2} + f^{2} − c ………………(1)

Since, (x+g)^{2 }= x^{2}+ 2gx + g^{2} and (y+f)^{2 }=y^{2 }+ 2fy + f^{2} substituting the values in equation (1), we have

(x+g)^{2}+ (y+f)^{2 }= g^{2 }+ f^{2}−c …………….(2)

Comparing (2) with (x−h)^{2 }+ (y−k)^{2 }= a^{2}, where (h, k) is the center and ‘a’ is the radius of the circle.

h=−g, k=−f

a^{2} = g^{2}+ f^{2}−c

Therefore,

x^{2 }+ y^{2 }+ 2gx + 2fy + c = 0, represents the circle with centre (−g,−f) and radius equal to a^{2} = g^{2 }+ f^{2}− c.

- If g
^{2 }+ f^{2 }> c, then the radius of the circle is real. - If g
^{2 }+ f^{2 }= c, then the radius of the circle is zero which tells us that the circle is a point that coincides with the center. Such a type of circle is called a point circle - g
^{2 }+ f^{2 }<c, then the radius of the circle become imaginary. Therefore, it is a circle having a real center and imaginary radius.

## Polar Equation of a Circle

To find the polar form of equation of a circle, replace the value of x = r cos θ and y = r sin θ, in x^{2} + y^{2} = a^{2}.

Hence, we get;

(r cos θ)^{2} + (r sin θ)^{2} = a^{2}

r^{2} cos^{2} θ + r^{2} sin^{2} θ = a^{2}

r^{2} (cos^{2} θ + sin^{2} θ) = a^{2}

r^{2} (1) = a^{2} [Using trigonometry identity]

r = a

is the polar equation of a circle with radius a and center at the origin (0,0).

## Other Circle Formulas

Here are some formulas are given for circle in terms of radius.

Diameter | 2 x radius |

Circumference | 2π (radius) |

Area | π(radius)^{2} |

## How to Find the Equation of the Circle?

To find the equation of a circle given the radius and center of the circle, we can directly put the values in the standard form of the equation.

**(x-h)**^{2}** + (y-k)**^{2}** = r**^{2}

Here, some solved problems are given to find the equation of a circle in both cases such as when the center of a circle is origin and center is not an origin is given below.

### Solved Examples

**Example 1:**

Consider a circle whose center is at the origin and radius is equal to 8 units.

**Solution:**

Given: Centre is (0, 0), radius is 8 units.

We know that the equation of a circle when the center is origin:

**x ^{2}+ y^{2 }= a^{2}**

For the given condition, the equation of a circle is given as

x^{2 }+ y^{2 }= 8^{2}

x^{2 }+ y^{2}= 64, which is the equation of a circle

**Example 2:**

Find the equation of the circle whose center is (3,5) and the radius is 4 units.

**Solution:**

Here, the center of the circle is not an origin.

Therefore, the general equation of the circle is,

(x-3)^{2 }+ (y-5)^{2} = 4^{2}

x^{2 }– 6x + 9 + y^{2 }-10y +25 = 16

x^{2 }+y^{2 }-6x -10y + 18 =0

**Example 3:**

Equation of a circle is x^{2}+y^{2}−12x−16y+19=0. Find the center and radius of the circle.

**Solution:**

Given equation is of the form x^{2}+ y^{2} + 2gx + 2fy + c = 0,

2g = −12, 2f = −16,c = 19

g = −6,f = −8

Centre of the circle is (6,8)

Radius of the circle = √[(−6)^{2} + (−8)^{2} − 19 ]= √[100 − 19] =

= √81 = 9 units.

Therefore, the radius of the circle is 9 units.

## Frequently Asked Questions on Equation of a Circle

### What is the equation for a circle?

The equation for a circle is given by:

(x-h)^{2}+(y-k)^{2} = a^{2}

Where (h,k) is the center and a is the radius of the circle.

### What are the formulas for circles?

The circumference of a circle is equal to 2 (pi) of radius or pi of diameter.

The area of a circle is equal to the pi of radius-squared.

### What is the equation of a circle when the center is at the origin?

At origin, the value of coordinates is (0,0), therefore, the equation of circle becomes:

(x-0)^{2} + (y-0)^{2} = r^{2}

x^{2} + y^{2} = r^{2}

### If (x-4)^{2}+(y+7)^{2}=9 is the equation of circle, then what is the center of circle?

Given, (x-4)^{2}+(y+7)^{2}=9 is the equation of circle. If we compare this equation with the standard equation we get:

(x-h)^{2}+(y-k)^{2} = a^{2}

h=4 and y = -7

Therefore, (4,-7) is the center of circle

### How do we know if an equation is the equation of circle?

If x and y are squared and the coefficient of x^{2} and y^{2} are same, then it is equation of circle. For example, 3x^{2}+3y^{2} = 12 is a circle’s equation.

## 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 (x1,y1) and radius r is (x−x1)2+(y−y1)2=r2.

## 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: x^{2} + y^{2} + 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 (x1,y1) and radius r is (x−x1)2+(y−y1)2=r2, 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.

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 x^{2} + y^{2} + 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 θ. The parametric equation of circle can be written as x^{2} + y^{2} + 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 x^{2} + y^{2} = p^{2}.

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

(rcosθ)^{2} + (rsinθ)^{2} = p^{2}

r^{2}cos^{2}θ + r^{2}sin^{2}θ = p^{2}

r^{2}(cos^{2}θ + sin^{2}θ) = p^{2}

r^{2}(1) = p^{2}

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: x^{2} + y^{2} = 9.

**Solution:**

To find the equation of the circle in polar form, substitute the values of xx and yy with:

x = rcosθ

y = rsinθ

x = rcosθ

y = rsinθ

x^{2} + y^{2} = 9

(rcosθ)^{2} + (rsinθ)^{2} = 9

r^{2}cos^{2}θ + r^{2}sin^{2}θ = 9

r^{2}(cos^{2}θ + sin^{2}θ) = 9

r^{2}(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−x1)2+(y−y1)2=r2.

where,(x1,y1) 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 (x1,y1) 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.

**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(x−1)2+(y+2)2=9 with (x−x1)2+(y−y1)2=r2(x−x1)2+(y−y1)2=r2, we get

x1x1 = 1, y1y1 = -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

In order to show how the equation of circle works, let’s graph the circle with the equation (x -3)^{2} + (y – 2)^{2} = 9. Now, before graphing this equation, we need to make sure that the given equation matches the standard form (x−x1)^{2}+(y−y1)^{2}=r^{2}.

- For this, we only need to change the constant 9 to match with r
^{2}as (x -3)^{2}+ (y – 2)^{2}= 3^{2}. - Here, we need to note that one of the common mistakes to commit is to consider x1 as -3 and y1 as -2.
- In the equation of circle, if the sign preceding x1 and y1 are negative, then x1 and y1 are positive values and vice versa.
- Here, x1 = 3, y1 = 2 and r = 3

Thus, the circle represented by the equation (x -3)^{2} + (y – 2)^{2} = 3^{2}, 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 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.

**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−x1)^{2}+(y−y1)^{2}=r^{2}.

If center is at origin, then x1= 0 and y1= 0.

**Answer: The equation of the circle if its center is at origin is x ^{2}+ y^{2}= r^{2}.**

### 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.

### 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.

### 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.

### 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.

### 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.

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 x2+y2+6x+8y+9=0, find the coordinates of the center and the radius of the circle.

## 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} = r^{2} and consider the general form as: x^{2} + y^{2} + 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 x^{2} + 2gx + y^{2} + 2fy = – c -> (1)

Step 2: Use the perfect square identity (x + g)^{2} = x^{2} + 2gx + g^{2} to find the values of the expression x^{2} + 2gx and y^{2} + 2fy as:

(x + g)^{2} = x^{2} + 2gx + g^{2 }⇒ x^{2 }+ 2gx = (x + g)^{2 }– g^{2 }-> (2)

(y + f)^{2} = y^{2} + 2fy + f^{2 }⇒ y^{2} + 2fy = (y + f)^{2 }– f^{2} -> (3)

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

(x+g)^{2 }– g^{2} + (y+f)^{2 }– f^{2} = – c

(x+g)^{2} + (y+f)^{2 }= g^{2 }+ f^{2} – c

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

## Converting Standard Form to General Form

We can use the algebraic identity formula of (a – b)^{2} = a^{2} + b^{2} – 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:

**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 x
^{2}+ y^{2}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 x2+y2+axy+C=0x2+y2+axy+C=0, then it is not the equation of the circle. There is no xyxy term in the equation of circle.
- In polar form, the equation of circle always represents in the form of rr and θθ.
- 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

**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 – x1)^{2} + (y – y1)^{2} = r^{2}. Here, (x1, y1) = (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.

**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 – x1)^{2} + (y – y1)^{2} = r^{2}. Here (x1, y1) = (-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} = 7^{2}

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

**Example 3: Find the equation of the circle in the polar form provided that the equation of the circle in standard form is: x ^{2} + y^{2} = 16.**

**Solution:**

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

x = rcosθ

y = rsinθ

x^{2} + y^{2} = 16

(rcosθ)^{2} + (rsinθ)^{2 }= 16

r^{2}cos^{2}θ + r^{2}sin^{2}θ = 16

r^{2}(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 and radius r is .

### 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 x^{2} + y^{2} = r^{2}.

### What is the Parametric Equation of Circle?

The parametric equation of circle can be written as where and

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

The general form of the equation of circle is: x^{2} + y^{2} + 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 x
^{2}+ y^{2}+ 2gx + 2fy + C = 0 - Standard Form

### 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

### 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 .

Hence the general form of the equation of circle is .

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

The standard form of the equation of a circle is , where 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: from standard form to gerenal form.

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 2. Now, the equation of the circle in standard form is .

### 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.

## Circle Equations

In fact **the definition** of a circle is

**Circle:** The set of all points on a plane that are a fixed distance from a center.

## Circle on a Graph

Let us put a circle of radius 5 on a graph:

Now let’s work out **exactly** where all the points are.

We make a right-angled triangle:

And then use Pythagoras:

x^{2} + y^{2} = 5^{2}

There are an infinite number of those points, here are some examples:

In all cases a point on the circle follows the rule x^{2} + y^{2} = radius^{2}

We can use that idea to find a missing value

Example: **x** value of 2, and a **radius** of 5

## More General Case

Now let us put the center at **(a,b)**

So the circle is **all the points (x,y)** that are **“r”** away from the center **(a,b)**.

Now lets work out where the points are (using a right-angled triangle and Pythagoras):

And that is the **“Standard Form”** for the equation of a circle!

It shows all the important information at a glance: the center **(a,b)** and the radius **r**.

## Try it Yourself

## “General Form”

But you may see a circle equation and **not know it**!

Because it may not be in the neat “Standard Form” above.

As an example, let us put some values to a, b and r and then expand it

It is a circle equation, but “in disguise”!

So when you see something like that think *“hmm … that might be a circle!”*

In fact we can write it in **“General Form”** by putting constants instead of the numbers:

x^{2} + y^{2} + Ax + By + C = 0

*Note: General Form always has x ^{2} + y^{2} for the first two terms*.

## Going From General Form to Standard Form

Now imagine we have an equation in **General Form**:

x^{2} + y^{2} + Ax + By + C = 0

How can we get it into **Standard Form** like this?

(x−a)^{2} + (y−b)^{2} = r^{2}

The answer is to Complete the Square (read about that) twice … once for **x** and once for **y**:

Simplify:(x^{2} − 2x + 1) + (y^{2} − 4y + 4) = 9

Finally:(x − 1)^{2} + (y − 2)^{2} = 3^{2}

And we have it in Standard Form!

(Note: this used the a=1, b=2, r=3 example from before, so we got it right!)

## Unit Circle

If we place the circle center at (0,0) and set the radius to 1 we get:

## How to Plot a Circle by Hand

1. Plot the center **(a,b)**

2. Plot 4 points “radius” away from the center in the up, down, left and right direction

3. Sketch it in!

## How to Plot a Circle on the Computer

We need to rearrange the formula so we get “y=”.

We should end up with two equations (top and bottom of circle) that can then be plotted.

Example: Plot (x−4)^{2} + (y−2)^{2} = 25

So the center is at (4,2), and the radius is √25 = 5

Rearrange to get “y=”:

Try plotting those functions on the Function Grapher.

It is also possible to use the Equation Grapher to do it all in one go.

A circle can be defined by a center point and a radius of a certain length. In the equation of a circle

(*x*−*h*)^{2}+(*y*−*k*)^{2}=*r*^{2}

the center is called (h,k) and the radius is *r*. From the equation, you can see that the circle is the collection of all the (x,y) points that are a distance *r* from the center, (h,k).

The collection of (x,y) values are the points on the circle’s circumference.

## Let’s work through four different examples of how to find the equation of a circle

**Example**

What is the equation of the circle?

Now let’s count from the center to a point on the circumference to find the length of the radius.

Sometimes we want to know the center and radius of a circle given the equation of the circle.

**Example**

What is the center and radius of the circle?

*x*^{2}+(*y*−3)^{2}=27

Sometimes we want to know the -intercepts of a circle.

**Example**

What are the x*x*-intercepts of the circle?

(*x*−2)^{2}+(*y*+1)^{2}=16

Sometimes to find out information about a circle you’ll need to know how to complete the square.

**Example**

Find the center and radius of the circle.

*x*^{2}+*y*^{2}+24*x*+10*y*+160=0

✅ Math Formulas ⭐️⭐️⭐️⭐️⭐

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