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 center 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 = 2 r

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 $π$ (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 ( $π$ equals $c$ divided by $d$ ). As a fraction the number $π$ is equal to about $\frac{22}{7}$ or $\frac{355}{113}$ (which is closer) and as a number it is about $3.1415926536$ .

	$π = \frac{c}{d}$
$∴ (therefore)$	$c = 2 π r$
	$c = π d$

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

A = π r^{2}

Calculating π

$π$ 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 $π$ times its diameter.

π = \frac{c}{d}

$π$ can also be calculated by only using mathematical methods. Most methods used for calculating the value of $π$ 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:

π = \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \frac{4}{11} + \dots

While that series is easy to write and calculate, it is not easy to see why it equals $π$ . 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 $π$ times the radius squared, $π$ can be approximated by using the following formula:

π = \frac{A}{r^{2}}

Calculating measures of a circle

Area

Using the radius: $A = π r^{2} = \frac{τ r^{2}}{2}$

Using the diameter: $A = \frac{π d^{2}}{4} = \frac{τ d^{2}}{8}$

Using the circumference: $A = \frac{c^{2}}{2 τ} = \frac{c^{2}}{4 π}$

Circumference

Using the radius: $c = τ r = 2 π r$

Using the diameter: $c = π d = \frac{τ d}{2}$

Using the area: $c = \sqrt{2 τ A} = 2 \sqrt{π A}$

Diameter

Using the radius: $d = 2 r$

Using the circumference: $d = \frac{c}{π} = \frac{2 c}{τ}$

Using the area: $d = 2 \sqrt{\frac{A}{π}} = 2 \sqrt{\frac{2 A}{τ}}$

Radius

Using the diameter: $r = \frac{d}{2}$

Using the circumference: $r = \frac{c}{τ} = \frac{c}{2 π}$

Using the area: $r = \sqrt{\frac{A}{π}} = \sqrt{\frac{2 A}{τ}}$

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, $r (θ) = R$

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: $\begin{matrix} x = & r \cos θ \\ y = & r \sin θ \end{matrix}$ Substituting this into the equation for a circle in rectangular coordinates gives $\begin{matrix} (r \cos θ - h)^{2} + (r \sin θ - k)^{2} = & R^{2} \\ r^{2} \cos^{2} θ - 2 r h \cos θ + h^{2} + r^{2} \sin^{2} θ - 2 r k \sin θ + k^{2} = & R^{2} \\ r^{2} - 2 (h \cos θ + k \sin θ) r + h^{2} + k^{2} - R^{2} = & 0 \end{matrix}$ This is a quadratic equation in $r$ . The quadratic formula can be used: $\begin{matrix} a = & 1 \\ b = & - 2 (h \cos θ + k \sin θ) \\ c = & h^{2} + k^{2} - R^{2} \\ r = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{2 (h \cos θ + k \sin θ) \pm \sqrt{4 (h \cos θ + k \sin θ)^{2} - 4 (h^{2} + k^{2} - R^{2})}}{2} \\ = & h \cos θ + k \sin θ \pm \sqrt{h^{2} \cos^{2} θ + 2 h k \cos θ \sin θ + k^{2} \sin^{2} θ - h^{2} - k^{2} + R^{2}} \\ = & h \cos θ + k \sin θ \pm \sqrt{R^{2} - h^{2} \sin^{2} θ + 2 h k \cos θ \sin θ - k^{2} \cos^{2} θ} \\ = & h \cos θ + k \sin θ \pm \sqrt{R^{2} - (h \sin θ - k \cos θ)^{2}} \end{matrix}$ Whether the square root is added or subtracted does not matter as long as $θ$ is allowed to make a full rotation: $\begin{matrix} r (θ + π) = & h \cos (θ + π) + k \sin (θ + π) + \sqrt{R^{2} - (h \sin (θ + π) - k \cos (θ + π))^{2}} \\ = & - h \cos θ - k \sin θ + \sqrt{R^{2} - (- h \sin θ + k \cos θ)^{2}} \\ = & - (h \cos θ + k \sin θ - \sqrt{R^{2} - (h \sin θ - k \cos θ)^{2}}) \end{matrix}$ In polar coordinates, the points $(θ, r)$ and $(θ + π, - r)$ are equal, so this means that both the positive and negative conjugates are included by other half of the domain.

Related pages

References

1 2 Weisstein, Eric W. "Circle". mathworld.wolfram.com. Retrieved 2020-09-24.
↑ "Basic information about circles (Geometry, Circles)". Mathplanet. Retrieved 2020-09-24.

Other websites

Calculate the measures of a circle online

[:1-1] 1 2 Weisstein, Eric W. "Circle". mathworld.wolfram.com. Retrieved 2020-09-24.

[2] "Basic information about circles (Geometry, Circles)". Mathplanet. Retrieved 2020-09-24.