# 3. The Circle

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## a. Circle Formulas

### Center at the Origin

The circle with centre (0, 0) and radius *r* has the equation:

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

This means any point (*x*, *y*) on the circle will give the radius squared when substituted into the circle equation.

### Center not at the Origin

The circle with centre (*h*, *k*) and radius *r* has the equation:

(

x−h)^{2}+ (y−k)^{2}=r^{2}

These formulas are a direct result of Pythagoras' Formula for the length of the hypotenuse of a right triangle.

### Need Graph Paper?

### Example 1

Sketch the circle *x*^{2} + *y*^{2} = 4.

Find the center and radius first.

Answer

The equation is in the form *x*^{2} + *y*^{2} = *r*^{2}, so we have a circle, center (0, 0) with radius `r=sqrt4=2`.

### Example 2

Sketch the circle (*x* − 2)^{2} + (*y* − 3)^{2} = 16

Find the center and radius first.

Answer

The equation is in the form (*x* − *h*)^{2} + (*y* − *k*)^{2} = *r*^{2}, so we have a circle with centre at (2, 3) and the radius is `r=sqrt(16)=4`.

### Example 3

Sketch the circle (*x* + 4)^{2} + (*y* − 5)^{2} = 36

Answer

The equation is in the form (*x* − *h*)^{2} + (*y* − *k*)^{2} = *r*^{2}.

We have a circle with centre at (−4, 5) and the radius is `r=sqrt(36)=6`.

Be careful with **positive **and **negative** *x*- and *y*-values in this work!

## b. The General Form of the Circle

An equation which can be written in the following form (with constants *D*,* E*,* F*) represents a **circle**:

x^{2}+y^{2}+Dx + Ey + F= 0

This is called the **general form of the circle**.

### Example 4

Find the centre and radius of the circle

x^{2}+y^{2}+ 8x+ 6y= 0

Sketch the circle.

Answer

Please revise **Completing
the Square** first...

Our aim is to get the equation into the form: (*x* − *h*)^{2} + (*y* − *k*)^{2} = *r*^{2}

We complete the square on the *x*-related portion and on the *y*-related portion, at the same time.

`x^2+y^2+8x+6y=0`

Group the *x* parts together and the *y* parts togther:

`(x^2+8x)+(y^2+6y)=0`

Complete the square on each of the *x* and *y* parts.

`(x^2+8x+16)+(y^2+6y+9)` `=16+9` `=25`

`(x+4)^2+(y+3)^2=5^2`

This is now in the format we require and we can determine the center and radius of the circle.

So the centre of the circle is (−4, −3) and the radius is 5 units.

Note that the circle passes through (0, 0). This is logical, since:

- The circle has radius 5
- Considering the right triangle formed by the points (−4, −3), (0, −3), and (0, 0), we can apply Pythagoras' Theorem and obtain: (−4)
^{2}+ (−3)^{2}= (5)^{2} - The formula for a circle follows from Pythagoras' Theorem.

### Exercises

1. Find the equation of the circle with centre `(3/2, -2)` and radius `5/2`.

Answer

Centre `(3/2, -2)` radius `5/2`.

General form of the equation of a circle:

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

The required equation for this case:

`(x-3/2)^2+(y-(-2))^2=(5/2)^2`

`(x-3/2)^2+(y+2)^2=(5/2)^2`

There is no need to expand this out, since this is the most useful form of the equation.

2. Determine the centre and radius and then sketch the circle:

3

x^{2}+ 3y^{2}− 12x+ 4 = 0

Answer

We **complete the square** as we did in an earlier example above.

First, we collect the *x* parts together and the *y* parts together, then divide throughout by 3.

`3x^2+3y^2-12x+4=0`

`3x^2-12x+3y^2+4=0`

`x^2-4x+y^2+4/3=0`

Then we complete the square on the *x* part. We do not need to do so for the *y* part because there is no single *y* term (only a *y*^{2} term).

`(x^2-4x+4)+y^2+4/3=4`

`(x-2)^2+y^2=4-4/3`

`(x-2)^2+y^2=8/3`

So the circle has centre `(2,0)` and has radius `sqrt(8/3)~~1.63`.

3. Find the points of intersection of the circle

x^{2}+y^{2}−x− 3y= 0

with the line

y=x− 1.

Answer

We solve the 2 equations simultaneously by substituting the expression

`y = x -1`

into the expression

`x^2+y^2-x-3y=0`

[See some background to this at: Algebraic Solution of Systems of Equations.]

We have:

`x^2+(x-1)^2-x-3(x-1)=0`

`x^2+x^2-2x+1-x-3x+3=0`

`2x^2-6x+4=0`

`x^2-3x+2=0`

`(x-1)(x-2)=0`

So we see that the solutions for *x* are `x = 1` or `x = 2`. This gives the corresponding y-values of `y = 0` and `y = 1`.

So the points of intersection are at: `(1, 0)` and `(2, 1)`.

We can see that our answer is correct in a sketch of the situation:

### Exercise for You

Where did (0.5, 1.5) for the center of the circle come from?

### Conic section: Circle

How can we obtain a circle from slicing a cone?

Each of the lines and curves in this chapter are **conic sections**, which means the curves are formed when we slice a cone at a certain angle.

If we slice a cone with a plane at right angles to the axis of the cone, the shape formed is a circle.

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