## Equation of a Circle Video Lessons

### How to Find the Equation of a Circle

### How to Graph a Circle

## What is the Equation of a Circle?

**The standard form equation of a circle is (π₯ – a)^{2} + (y – b)^{2} = r^{2}, where (a,b) is the centre and r is the radius. For example, a circle with centre (5, -3) and radius 3 has the equation (π₯ – 5)^{2} + (y + 3)^{2} = 9.**

(π₯ – a)^{2} + (y – b)^{2} = r^{2} is known as the standard form of the equation of a circle.

This is because the centre and radius of the circle can be easily read from the equation and they can be easily substituted in to make the equation.

The equation of a circle is not a function. This is because circle equations do not pass the ‘vertical line test’ required for functions. Instead we say that the equation of a circle is a relation.

To be a function, any given input (π₯ value) must have exactly one output (y value). However for most π₯ values, circles produce two outputs. Therefore a circle is not a function.

We can see in this example of (π₯ – 5)^{2} + (y – 5)^{2} = 16 that the single input of π₯ = 5 gives two outputs: y = 1 and y = 9.

It is possible to rearrange the equation of a circle in terms of π₯ or y. In this case, to ensure that it remains a function, we must only consider the positive values of the square root.

In terms of y, the equation of a circle is .

In terms of π₯, the equation of a circle is

## How to Find the Equation of a Circle

**To find the equation of a circle:**

- Find the coordinates of the centre of the circle
- Find the length of the radius: r
- Substitute these values into the equation of a circle: (π₯ – a)
^{2}+ (y – b)^{2}= r^{2}

### For example, find the equation of the circle with centre (5,7) and radius 3.

The standard equation of a circle is (π₯ – a)^{2} + (y – b)^{2} = r^{2}, where:

*‘a’*is the π₯-coordinate of the centre of the circle*‘b’*is the y-coordinate of the centre of the circle*‘r’*is the radius of the circle

If the centre is (5,7) then *‘a’* = 5 and *‘b’* = 7.

If the radius is 3 then r = 3 and r^{2} = 9.

The equation of the circle, (π₯ – a)^{2} + (y – b)^{2} = r^{2}, becomes (π₯ – 5)^{2} + (y – 7)^{2} = 9.

### For example, find the equation of the circle with centre (-6, -3) and radius 2.

In this example, *‘a’* = -6, *‘b’* = -3 and r = 2

The values of *‘a’* and *‘b’* are both negative. This means that when they are subtracted, they become positive. Subtracting a negative number is the same as adding it.

Therefore (π₯ – a)^{2} + (y – b)^{2} = r^{2}, becomes (π₯ + 6)^{2} + (y + 3)^{2} = 4.

## How to Graph a Circle

**To graph a circle, read the coordinates of the centre, (a,b) and the radius, r from the circle equation (π₯ – a) ^{2} + (y – b)^{2} = r^{2}. First plot the centre coordinates and from here, use the radius length to find the outer points on the circle. **

### For example, graph the circle with equation (π₯ – 7)^{2} + (y – 4)^{2} = 4

In this example, the centre is (7, 4).

We first plot this centre coordinate by finding 7 on the π₯-axis and 4 on the y-axis.

To draw the circle, it helps to form a cross using vertical and horizontal lines coming out from the centre that are each as long as the radius.

We draw lines that are 2 long because the radius of this circle is 2.

We can then graph the circle using a compass so that the circle passes through these outer points.

## The Equation of a Circle with a Centre at the Origin

**The equation of a circle with its centre at the origin and a radius of r has the equation π₯ ^{2} + y^{2} = r^{2}. For example, a circle with centre (0,0) and radius 6 has the equation π₯^{2} + y^{2} = 36. **

The origin is where the π₯ and y axes meet and it has the coordinates (0,0).

If a circle has a centre at the origin, then centre of the circle is (0,0). Therefore *‘a’* = 0 and *‘b’* = 0 and the standard equation of a circle becomes (π₯ – a)^{2} + (y – b)^{2} = r^{2} to (π₯ – 0)^{2} + (y – 0)^{2} = r^{2}.

This reduces to π₯^{2} + y^{2} = r^{2}.

To find the equation of a circle with a centre at the origin, find its radius and substitute it into π₯^{2} + y^{2} = r^{2} .

## How to Write the Equation of a Circle in General Form

**The general form of the circle equation is π₯^{2} + aπ₯ + y^{2} + by + c= 0. To write the equation of a circle in general form, simply expand the two brackets in its standard form (π₯ – a)^{2} + (y – b)^{2} = r^{2}. Then collect like terms. **

### For example write the equation of a circle with centre (2, 0) and radius 1 in general form

**Step 1. Write the equation of a circle in standard form using the centre and radius.**

The standard form of the circle equation can be written as (π₯ – a)^{2} + (y – b)^{2} = r^{2} , where the centre is (a, b) and the radius is r.

With a centre of (2, 0) and radius of 1, the circle equation becomes (π₯ – 2)^{2} + y^{2} = 1.

**Step 2. Expand the brackets in the standard form of the equation.**

(π₯ – 2)^{2} + y^{2} = 1 means (π₯ – 2)(π₯ – 2) + y^{2} = 1.

Expanding the brackets (π₯ – 2)(π₯ – 2), we have π₯^{2} – 4π₯ + 4.

Therefore (π₯ – 2)(π₯ – 2) + y^{2} = 1 becomes π₯^{2} – 4π₯ + 4 + y^{2} = 1.

**Step 3. Collect like terms**

We now collect like terms, which means to add up all of the π₯^{2} terms, all of the y^{2} terms and all of the constant terms.

We can do this by subtracting 1 from both sides of the equation.

π₯^{2} – 4π₯ + 4 + y^{2} = 1 becomes π₯^{2} – 4π₯ + y^{2} + 3= 0.

The expanded form of the equation of the circle is π₯^{2} – 4π₯ + y^{2} + 3= 0.

We tend to write the π₯ terms first, then the y terms and then the constant terms. We also set the equation equal to zero.

**The advantage of the general form of the circle equation is that it is easier to substitute values of π₯ and y into. This makes it an easier choice for finding particular coordinates on the circle or axis intercepts.**

**The advantage of the standard form of the circle equation is that we can easily read off the coordinates of the centre of the circle and its radius. When written as (π₯ – a)^{2} + (y – b)^{2} = r^{2}, (a,b) is the centre and r is the radius. **

## How to Write the Equation of a Circle in Standard Form

**The equation of a circle in standard form is (π₯ – a) ^{2 }+ (y – b)^{2} = r^{2}. To write the equation of a circle in standard form, collect the π₯^{2} and π₯ terms together, the y^{2} and y terms together and complete the square with them separately. Then move any constant terms to the other side of the equals sign.**

### For example, find the centre and radius of the circle with equation π₯^{2} + y^{2} + 4π₯ – 8y + 3 = 0

**Step 1. Collect the π₯ terms and y terms together**

We move the π₯^{2} and π₯ terms next to each other and the y^{2} and y terms next to each other so that:

π₯^{2} + y^{2} + 4π₯ – 8y + 3 = 0 becomes π₯^{2} + 4π₯ + y^{2} – 8y + 3 = 0

**Step 2. Complete the square for the π₯ terms and the y terms**

We will complete the square for π₯^{2} + 4π₯ and then complete the square for y^{2} – 8y.

To complete the square for π₯^{2} + 4π₯, we start with (π₯ + )^{2}.

To find the number that goes in the brackets with the π₯, we halve the coefficient of 4π₯. Half of 4 is 2 and so we get (π₯ + 2)^{2}.

Finally we subtract this number squared. 2 squared is 4 and so we get: (π₯ + 2)^{2} – 4.

Now we will complete the square for y^{2} – 8y.

To complete the square for y^{2} – 8y, we start with (y – )^{2}.

To find the number that goes in the brackets with the y, we halve the coefficient of 8y. Half of 8 is 4 and so we get (y – 4)^{2}.

Finally we subtract this number squared. 4 squared is 16 and so we get: (y – 4)^{2} – 16.

**Step 3. Substitute these expressions into the original equation**

Completing the square for π₯^{2} + 4π₯ gave us (π₯ + 2)^{2} – 4.

Completing the square for y^{2} – 8y gave us (y – 4)^{2} – 16.

We substitute these expressions into the original equation from step 1: π₯^{2} + 4π₯ + y^{2} – 8y + 3 = 0.

This becomes (π₯ + 2)^{2} – 4 + (y – 4)^{2} – 16 + 3 = 0.

**Step 4. Move all constant terms to the right hand side of the equals sign**

We can collect the constant terms in (π₯ + 2)^{2} – 4 + (y – 4)^{2} – 16 + 3 = 0 so this becomes

(π₯ + 2)^{2} + (y – 4)^{2} – 17 = 0

We then add 17 to both sides of the equation to get:

** (π₯ + 2) ^{2} + (y – 4)^{2} = 17 **

The equation of the circle has now been converted to standard form. When written in standard form we can easily read the centre and radius of the circle.

The centre is (-2, 4) and the radius is β17.

## Equation of a Circle Derivation

**The equation of a circle (π₯ – a) ^{2} + (y – b)^{2} = r^{2} can be derived from the equation for the distance between two points. Simply substitute the coordinates of (a,b) as the centre of the circle and r as the radius of the circle and then square both sides.**

Here is the algebraic proof of the equation of a circle.

Using Pythagoras’ theorem, the distance between two points with coordinates (π₯_{1},y_{1}) and (π₯_{2},y_{2}) is:

For a circle, the two points in question are the centre and a point on the circle radius.

This means that (π₯_{1},y_{1}) becomes the centre: (a, b)

(π₯_{2},y_{2}) is any point on the radius of the circle, which is just (π₯,y).

The distance between the points, d, is therefore just the radius, r.

We can then square both sides of the equation to find the standard equation of a circle:

**(π₯ – a) ^{2} + (y – b)^{2} = r^{2}**

## How to Find the **π₯** and y Intercepts for a Circle

**To find the π₯ intercepts of a circle, substitute y = 0 into the circle equation. Then solve the resulting quadratic equation for π₯. To find the y intercepts of a circle, substitute**

**π₯ = 0 into the circle equation and solve the resulting quadratic equation for y.**

There may be 0, 1 or 2 axis intercepts depending on the equation of the circle.

### For example, find the π₯-axis intercepts of (π₯ – 5)^{2} + (y + 3)^{2} = 16

**Step 1. Substitute y = 0 into the circle equation.**

(π₯ – 5)^{2} + (y + 3)^{2} = 16 becomes (π₯ – 5)^{2} + 3^{2} = 16.

**Step 2. Rearrange this quadratic equation so that it is equal to zero.**

(π₯ – 5)^{2} + 3^{2} = 16 can be written as (π₯ – 5)(π₯ – 5) + 3^{2} = 16 and expanding the brackets we get π₯^{2} – 10π₯ + 25 + 3^{2} = 16.

Collecting the constant terms, we get: π₯^{2} – 10π₯ + 18 = 0.

**Step 3. Solve the quadratic equation**

Using the quadratic formula, , where a = 1, b = -10 and c = 18.

The quadratic formula becomes .

and π₯ = 2.35 and π₯ = 7.65.

Therefore the π₯-axis intercepts of the circle are found at π₯ = 2.35 and π₯ = 7.65.

An alternative method to find the π₯ or y intercepts is to simply graph the circle and observe where it crosses the axes. However this will only result in an approximation which will need verifying using the algebraic method above.

The quadratic equation had two solutions and so, the circle had 2 axis intercepts.

If the quadratic equation has no solutions, this means that the circle does not have axis intercepts.

If the quadratic equation has just one solution, this means that the circle has one axis intercept. To have one axis intercept, the circle will be touching the axis.

### For example, find the y-axis intercepts of (π₯ + 3)^{2} + (y – 5)^{2} = 9

**Step 1. Substitute** **π₯ = 0 into the circle equation**

(π₯ + 3)^{2} + (y – 5)^{2} = 9 becomes 3^{2} + (y – 5)^{2} = 9.

**Step 2. Rearrange this quadratic equation so that it is equal to zero.**

3^{2} + (y – 5)^{2} = 9 can be written as 3^{2} + (y – 5)(y – 5) = 9 and expanding the brackets we get 9 + y^{2} – 10y + 25 = 9.

Subtracting 9 from both sides, we get: y^{2} – 10y + 25 = 0.

**Step 3. Solve the quadratic equation**

y^{2} – 10y + 25 = 0 can factorise to (y – 5)(y – 5) = 0.

There is only one solution to this quadratic equation. Setting the factor ( y – 5) = 0, we get y = 5.

We can see that there is only one y-axis intercept for this circle equation.

If a circle is just touching the axis, there is only one axis intercept. This means that there is only one solution to the quadratic equation created when setting **π₯** or y to 0.

## The Parametric Form of a Circle

**The parametric form of the circle equation, π₯**^{2}** + y**^{2}** = r**^{2}** , is π₯ = r cosΞΈ, y = r sinΞΈ. If the circle is translated a units right and b units up, the circle equation (π₯ – a)^{2} + (y – b)^{2} = r^{2} can be written parametrically as π₯ = a + r cosΞΈ, y = b+ r sinΞΈ**.

To derive the parametric equations of a circle, start with basic trigonometry for a right-angled triangle.

For any given right-angled triangle, the opposite side is equal to the hypotenuse multiplied by sine of the angle.

The adjacent side is equal to the hypotenuse multiplied by the cosine of the angle.

A right-angled triangle can be drawn from the centre of a circle to its edge, so that the radius of the circle is the hypotenuse of the triangle.

Therefore the π₯ coordinate of any point on the outside of the circle is the same as the adjacent side of the triangle and the y coordinate of the same point is the same as the opposite side of the triangle.

We can substitute the hypotenuse equal to r, the opposite equal to y and the adjacent equal to π₯.

Therefore, we get two parametric equations for the coordinates of any point on the circle:

**π₯ = r cosΞΈ, y = r sinΞΈ**

As ΞΈ changes, we rotate to a new coordinate on the circle. Increasing ΞΈ from 0 to 360Β° (2*Ο*), we obtain coordinates that form the complete circumference of the circle.

We say that ΞΈ is the parameter in these parametric equations.

The parametric equations for a circle with centre (a, b) and radius r are:

**π₯ = a + r cosΞΈ, y = b+ r sinΞΈ**

We simply add *‘a’* to the π₯ coordinate to move it *‘a’* units right.

We add *‘b’* to the y coordinate to move it *‘b’* units up.

## The Equation of a Circle in Complex Form

**On the complex plane, |z| represents the distance of the complex number z from the origin. Therefore the equation of any circle radius r, with the origin as its centre can be written as |z| = r. For a circle with centre z _{0}, where a is also a complex number, the equation is |z – z_{0}| = r.**

For example, a circle with centre (4, 3) and radius 2 can be represented with complex numbers.

On the complex plane (4, 3) means 4 in real numbers and 3 in imaginary numbers. (4, 3) is equivalent to 4 + 3i.

The centre, z_{0} = 4 + 3i.

The radius, r is 2.

Therefore |z – z_{0}| = r becomes |z – (4 + 3i)| = 2.