Circles Maths

Circle equations

Some well-known circle Equations include:

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

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.

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

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.

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.

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:$a^2+b^2=c^2$

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

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

$r=\pm 5$

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 $\sqrt{a^2+b^2-c}$.

Deriving this circle formula

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

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$x^2+y^2-2ax-2by+c=0$

Rearranging back to the basic circle equation

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