Circle

A **circle** is a round, two-dimensional shape. All points on the edge of the circle are at the same distance from the center.

The point on the side. Mathematicians use the letter r for the length of a circle's radius. The  of a circle is the point in the very middle. It is often written as O.

The diameter (meaning "all the way across") of a circle is a straight line that goes from one side to the opposite and right through the center of the circle. Mathematicians use the letter d for the length of this line. The diameter of a circle is equal to twice its radius (d equals 2 times r):

:d=2r

The circumference (meaning "all the way around") of a circle is the line that goes around the center of the circle. Mathematicians use the letter c for the length of this line.

The number \pi (written as the Greek letter *pi*) is a very useful number. It is the length of the circumference divided by the length of the diameter (\pi equals c divided by d). As a fraction the number \pi is equal to about \frac{22}{7} or \frac{355}{113} (which is closer) and as a number it is about 3.1415926536.

The area, A, inside a circle is equal to the radius multiplied by itself, then multiplied by \pi (A equals \pi times r times r).

:A=\pi r^2

## Calculating π
\pi can be measured by drawing a circle, then measuring its diameter (d) and circumference (c). This is because the circumference of a circle is always equal to \pi times its diameter.

:\pi=\frac cd

\pi can also be calculated by only using mathematical methods. Most methods used for calculating the value of \pi have desirable mathematical properties. However, they are hard to understand without knowing trigonometry and calculus. However, some methods are quite simple, such as this form of the Gregory-Leibniz series:

:\pi=\frac41-\frac43+\frac45-\frac47+\frac49-\frac{4}{11}+\cdots

While that series is easy to write and calculate, it is not easy to see why it equals \pi. A much easier way to approach is to draw an imaginary circle of radius r centered at the origin. Then any point (x,y) whose distance d from the origin is less than r, calculated by the Pythagorean theorem, will be inside the circle:

:d=\sqrt{x^2+y^2}

Finding a set of points inside the circle allows the circle's area A to be estimated, for example, by using integer coordinates for a big r. Since the area A of a circle is \pi times the radius squared, \pi can be approximated by using the following formula:

:\pi=\frac{A}{r^2}

## Calculating measures of a circle
### Area
Using the radius: A=\pi r^2=\frac{\tau r^2}{2}

Using the diameter: A=\frac{\pi d^2}{4}=\frac{\tau d^2}{8}

Using the circumference: A=\frac{c^2}{2\tau}=\frac{c^2}{4\pi}

### Circumference
Using the radius: c=\tau r=2\pi r

Using the diameter: c=\pi d=\frac{\tau d}{2}

Using the area: c=\sqrt{2\tau A}=2\sqrt{\pi A}

### Diameter
Using the radius: d=2r

Using the circumference: d=\frac c\pi=\frac{2c}{\tau}

Using the area: d=2\sqrt\frac A\pi=2\sqrt\frac{2A}{\tau}

### Radius
Using the diameter: r=\frac d2

Using the circumference: r=\frac c\tau=\frac{c}{2\pi}

Using the area: r=\sqrt\frac A\pi=\sqrt\frac{2A}{\tau}

## Equations 
### Cartesian coordinates 
The equation of a circle in Cartesian coordinates, with the centre at (h,k) and radius r, is given by
(x-h)^2 + (y-k)^2 = r^2

### Polar coordinates 
A circle centred at the origin with radius R is the equivalent of a constant function,
r(\theta)=R

A circle with a centre at a different location (h,k) and radius R can be found by using the conversion between polar and Cartesian coordinates:
\begin{align}
x = {} & r\cos\theta\\
y = {} & r\sin\theta
\end{align}
Substituting this into the equation for a circle in rectangular coordinates gives
\begin{align}
(r\cos \theta - h)^2 + (r\sin\theta - k)^2 = {} & R^2\\
r^2 \cos^2 \theta - 2rh\cos\theta + h^2 + r^2\sin^2\theta - 2rk\sin\theta + k^2 = {} & R^2\\
r^2 -2(h\cos\theta + k\sin\theta)r + h^2 + k^2 - R^2 = {} & 0
\end{align}
This is a quadratic equation in r. The quadratic formula can be used:
\begin{align}
a = {} & 1\\
b = {} & -2(h\cos\theta + k\sin\theta)\\
c = {} & h^2 + k^2 - R^2\\
r = {} & \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
= {} & \frac{2(h\cos\theta + k\sin\theta) \pm \sqrt{4(h\cos\theta + k\sin\theta)^2 - 4(h^2+k^2-R^2)}}{2}\\
= {} & h\cos\theta + k\sin\theta \pm \sqrt{h^2\cos^2\theta + 2hk\cos\theta\sin\theta + k^2\sin^2\theta - h^2 - k^2 + R^2}
\\
= {} & h\cos\theta + k\sin\theta \pm \sqrt{R^2-h^2\sin^2\theta + 2hk\cos\theta\sin\theta - k^2\cos^2\theta}\\
= {} & h\cos\theta + k\sin\theta \pm \sqrt{R^2-(h\sin\theta - k\cos\theta)^2}
\end{align}
Whether the square root is added or subtracted does not matter as long as \theta is allowed to make a full rotation:
\begin{align}
r(\theta+\pi) = {} & h\cos(\theta + \pi) + k\sin(\theta+\pi) + \sqrt{R^2-(h\sin(\theta+\pi)-k\cos(\theta+\pi))^2}\\
= {} & -h\cos\theta - k\sin\theta + \sqrt{R^2 -(-h\sin\theta+k\cos\theta)^2}\\
= {} & -(h\cos\theta + k\sin\theta - \sqrt{R^2 -(h\sin\theta - k\cos\theta)^2})
\end{align}
In polar coordinates, the points (\theta,r) and (\theta+\pi,-r) are equal, so this means that both the positive and negative conjugates are included by other half of the domain.

## Related pages
* Concentric
* Semicircle
* Sphere
* Squaring the circle
* Pi
* Tau

## References
## Other websites
* Calculate the measures of a circle online