# Circles Maths

Circles can exist within themselves or be mapped onto Graphs where we can give them Equations.

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Circles are a 2-dimensional shape where a single line is rotated around a point at${360}^{\circ }$or$2\mathrm{\pi }$ if in Radians.

## Circle equations

Some well-known circle Equations include:

• $Diameterofthecircle=radius×2$
• $Areaofacircle=\mathrm{\pi }×{\mathrm{radius}}^{2}$
• $CircumferenceoftheCircle=2×\mathrm{\pi }×\mathrm{radius}=\mathrm{\pi }×\mathrm{diameter}$

## Parts of a circle diagram

To understand circle diagrams, you need to be familiar with partitioning circles into parts in maths.

Circle diagram. Jaime Nichols StudySmarter Originals

• Circumference (C): the perimeter of the circle.

• Radius (r): the distance between any point on the circumference and the centre of the circle.

• Diameter (d): the distance from one side of the circumference to the other which goes through the centre of the circle.

• Sector: an area bordered by two radii.

• Chord: the distance from one side of the circumference without going through the centre of the circle.

• Segment: the area between a chord and the circumference.

• Tangent: an exterior line that touches the circumference of the circle at one point.

• Arc: a Proportion of the circumference of the circle

### Circle theorems

There are eight Circle Theorems that describe the angle properties of circles.

• Circle Theorem 1: Angles in the same segment are equal.

• Circle Theorem 2: In a cyclical quadrilateral, the sum of the opposite Angles is equal to 180.

A cyclical quadrilateral is all the points in a quadrilateral that lie on the circumference of the circle.

• Circle Theorem 3: The angle at the centre of a circle is half the angle at the circumference.

• Circle Theorem 4: The perpendicular bisector of a chord travels through the circle centre.

• Circle Theorem 5: The radius and a Tangent of a Circle will meet perpendicular to one another.

• Circle Theorem 6: Two tangent lines that meet at a point have the same length.

• Circle Theorem 7: A triangle within the circle, whereby the hypotenuse of the triangle is the same as the diameter of the circle, has an angle at the circumference of 90.

• Circle Theorem 8: Alternate Angle Theorem.

## The basic circle equation

All circles can be represented by the formula: ${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}$where (a, b) are the coordinates of the centre of this circle and r is the radius of the circle.

A circle with the centre at (5, 9) with a radius of 10 will have the equation of ${\left(x-5\right)}^{2}+{\left(x-9\right)}^{2}={10}^{2}$which is also equal to${\left(x-5\right)}^{2}+{\left(x-9\right)}^{2}=100$

When the circle has a centre at the origin, then it has no constants attached to the x- or y-coordinate: ${x}^{2}+{y}^{2}={r}^{2}$.

To find out if a coordinate lies on the circumference of the circle, you substitute the x- and y- coordinate into the circle equation. If the equation solves to equal the circle radius, the point lies on the circle circumference.

Prove that (4, 12) lies on the circumference of the circle ${x}^{2}+{\left(y-10\right)}^{2}=20$.

${4}^{2}+{\left(12-10\right)}^{2}$

$16+4=20$

The answer to the circle equation when you substitute (4, 12) is the same as the ${r}^{2}$ (20); therefore, the point (4,12) must lie on the circle circumference.

### Forming circle equations from a graph

Forming an equation from a circle requires two pieces of information: the circle centre and the radius of the circle.

To find the radius of the circle, you need to:

1. Draw a line from the centre to the circumference (which will be the radius), then another line from the intersection of the radius down to a horizontal line from the centre.

2. As you are calculating the radius of the circle, you need to find the other two sides of the triangle, and you can do that by counting the squares.

3. As this is a right-angled triangle with no angles shown, you can use Pythagoras theorem to find the radius of the circle.

Pythagoras theorem is a formula to calculate the sides of a right-angled triangle: ${a}^{2}+{b}^{2}={c}^{2}$; where c is the hypotenuse.

For the first step, where possible, try and put the radius on points on the graph. This will help you later because you will be working with whole numbers.

Find the radius of the circle with the centre (4, 1) shown on the graph below.

1. Drawing the radius: The coordinate (7, 5) lies on the circumference. The radius is the line between the centre (4, 1) and the point on the circumference (7, 5).
2. Finding the values of the sides of the triangle: By counting the squares, the horizontal line is 3, and the vertical line is 4.

Pythagoras Triangle derived from Circle 1 - Jaime Nichols - StudySmarter Originals

3. Finding the value for the radius using Pythagoras theorem:${a}^{2}+{b}^{2}={c}^{2}$

${3}^{2}+{4}^{2}={r}^{2}$

$r=\sqrt{9+16}=\sqrt{25}$

$r=±5$

However, the radius of the circle is a distance and, therefore, can only be a positive, so r = 5

## The derived circle equation

Sometimes circle equations can be formed in the expression ${x}^{2}+{y}^{2}+2ax+2bx+c=0$, whereby the circle centre is (-a, -b) and the radius of the circle is $\sqrt{{a}^{2}+{b}^{2}-c}$.

Circle B has an equation of ${x}^{2}+{y}^{2}+16x-10x+8=0$. What is the centre of Circle B? What is the radius of Circle B?

• The centre of the circle is $\left(\begin{array}{cc}\frac{-16}{2},& \frac{--10}{2}\end{array}\right)=\left(-8,5\right)$, where a = -8 and b = 5.
• The radius of the circle is $\sqrt{{\left(-8\right)}^{2}+{5}^{2}-8}=\sqrt{81}=9$

#### Deriving this circle formula

The circle formula ${x}^{2}+{y}^{2}+2ax+2bx+c=0$ can be derived from the basic circle formula ${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}$.

1. Expand out the brackets out of the original circle equation to get ${x}^{2}-2ax+{a}^{2}+{y}^{2}-2by+{b}^{2}={r}^{2}$
2. Rearrange so that the form resembles the derived equation:${x}^{2}+{y}^{2}-2ax+{a}^{2}-2by+{b}^{2}-{r}^{2}=0$
3. Let $c={a}^{2}+{b}^{2}-{r}^{2}$ so the equation becomes${x}^{2}+{y}^{2}-2ax-2by+c=0$

The last stage of deriving the equation can help you remember the formula for the radius. As $c={a}^{2}+{b}^{2}-{r}^{2}$, you can change the subject to ${r}^{2}$: ${r}^{2}={a}^{2}+{b}^{2}-c$. Therefore,$r=\sqrt{{a}^{2}+{b}^{2}-c}$

### Rearranging back to the basic circle equation

You could be asked to rearrange ${x}^{2}+{y}^{2}+2ax+2by+c=0$ into ${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}$ and to do so, you need to complete the square.

It is helpful to group the x-'s and y-'s together to make it easier for you to complete the square.

Rearrange ${x}^{2}+{y}^{2}+12x+4y+20=0$ into the form${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}$

1. Organise the equation so that the x- and y- equation ${x}^{2}+12x+{y}^{2}+4y+20=0$2. Complete the square on each variable For the x-axis: ${\left(x+6\right)}^{2}-36$For the y-axis: ${y}^{2}+4y={\left(y+2\right)}^{2}-4$3. Combining these equations: ${\left(x+6\right)}^{2}-36+{\left(y+2\right)}^{2}-4+20=0$4. Rearrange to get into the form${\left(x-a\right)}^{2}+{\left(y-2\right)}^{2}={r}^{2}$

${\left(x-a\right)}^{2}+{\left(y+2\right)}^{2}-20=0$

${\left(x+6\right)}^{2}+{\left(y+2\right)}^{2}=20$

## Circles - Key takeaways

• Circles can be described and labelled in lots of different ways, which you should be able to recognise and understand.
• Circles can be written as an equation in the form where the centre of the circle is (a, b) and the radius of the circle is r.
• Circles can also be written in the form ${x}^{2}+{y}^{2}+2ax+2bx+c=0$ where the centre of the circle is (-a, -b) and the radius of the circle is$\sqrt{{a}^{2}+{b}^{2}-c}$
• To find the radius of a circle from the graph, you create a right-angled triangle using the radius of the circle and solve using Pythagoras theorem.

#### Flashcards in Circles Maths 65

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How do you find the radius of a circle?

You create a right angle triangle where the hypotenuse is the radius of the circle and you solve for r using Pythagoras theorem.

What are the equations for circles?

All circles can be written as an equation in the form (x-a)^2+(y-b)^2=r^2 as well as in the derived form:  x^2+y^2+2ax+2bx+c=0.

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