Difference between revisions of "Circle"

Revision as of 14:52, 1 December 2015

A circle is a geometric figure commonly used in Euclidean geometry.

$[asy]unitsize(2cm);draw(unitcircle,blue);[/asy]$

A basic circle.

Traditional Definition

A circle is defined as the set (or locus) of points in a plane with an equal distance from a fixed point. The fixed point is called the center and the distance from the center to a point on the circle is called the radius.

The radius and center of a circle.

Coordinate Definition

Using the traditional definition of a circle, we can find the general form of the equation of a circle on the coordinate plane given its radius, $r$ , and center $(h,k)$ . We know that each point, $(x,y)$ , on the circle which we want to identify is a distance $r$ from $(h,k)$ . Using the distance formula, this gives $\sqrt{(x - h)^2 + (y - k)^2} = r$ which is more commonly written as $(x - h)^2 + (y - k)^2 = r^2.$

Example: The equation $(x - 3)^2 + (y + 6)^2 = 25$ represents the circle with center $(3,-6)$ and radius 5 units.

Area of a Circle

The area of a circle is $\pi r^2$ where $\pi$ is the mathematical constant pi and $r$ is the radius.

Archimedes' Proof

We shall explore two of the Greek mathematician Archimedes demonstrations of the area of a circle. The first is much more intuitive.

Archimedes envisioned cutting a circle up into many little wedges (think of slices of pizza). Then these wedges were placed side by side as shown below:

As these slices are made infinitely thin, the little green arcs in the diagram will become the blue line and the figure will approach the shape of a rectangle with length $r$ and width $\pi r$ thus making its area $\pi r^2$ .

Archimedes also came up with a brilliant proof of the area of a circle by using the proof technique of reductio ad absurdum.

Archimedes' actual claim was that a circle with radius $r$ and circumference $C$ had an area equivalent to the area of a right triangle with base $C$ and height $r$ . First let the area of the circle be $A$ and the area of the triangle be $T$ . We have three cases then.

Case 1: The circle's area is greater than the triangle's area.

Case 2: The triangle's area is greater than the circle's area.

Case 3: The circle's area is equal to the triangle's area.

Assume that $A>T$ . Let $P$ be the area of a regular polygon that is closest to the circle's area. Therefore we have $A-P<A-T$ so $P>T$ . Let the apothem be $a$ and the perimeter be $p$ so the area of a regular polygon is one half of the product of the perimeter and apothem. The perimeter is less than the circumference so $p<2\pi r$ and the apothem is less than the radius so $a<r$ . Therefore $P=\frac{1}{2}ap<\frac{1}{2}r\cdot 2\pi r=T$ . However it cannot be both $P>T$ and $P<T$ . So $A\not >T$ .

This article is a stub. Help us out by expanding it.

Proof Using Calculus

Let the circle in question be $x^2 + y^2 = r^2$ , where r is the circle's radius. By symmetry, the circle's area is four times the area in the first quadrant. The area in the first quadrant can be computed using a definite integral from 0 to r of the function $f(x) = \sqrt(r^2 - x^2)$ . Using the substitution $x = r \sin u, dx = r \cos u$ gives the indefinite integral as $\frac{r^2}{2} (u - \frac{\sin 2u}{2}) + C$ , so the definite integral equals $\frac{r^2}{2} * \frac{\pi}{2}$ . Multiplying by four gives the area of the circle as $\pi r^2$ .

Related Formulae

The area of a circle with radius $r$ is $\pi r^2$ .
The circumference of a circle with radius $r$ is $2\pi r$ .

Other Properties and Definitions

$[asy]draw(unitcircle);draw((-0.8,1)--(1,1),Arrow);draw((1,1)--(-0.8,1),Arrow);draw((0,1)--(1,0));[/asy]$

A circle with a tangent and a chord marked.

A line that touches a circle at only one point is called the tangent of that circle. Note that any point on a circle can have only one tangent.
A line segment that has endpoints on the circle is called the chord of the circle. If the chord is extended to a line, that line is called a secant of the circle.
The measure of an inscribed angle is always half the measure of the central angle with the same endpoints.
Chords, secants, and tangents have the following properties:
- The perpendicular bisector of a chord is always a diameter of the circle.
- The perpendicular line through the tangent where it touches the circle is a diameter of the circle.
- The Power of a point theorem.

Other interesting properties are:

A right triangle inscribed in a circle has a hypotenuse that is a diameter of the circle.
Any angle formed by the two endpoints of a diameter of the circle and a third distinct point on the circle as the vertex is a right angle.

Problems

Introductory

Under what constraints is the circumference (in inches) of a circle greater than its area (in square inches)?

Intermediate

Circles with centers $A$ and $B$ have radii 3 and 8, respectively. A common internal tangent intersects the circles at $C$ and $D$ , respectively. Lines $AB$ and $CD$ intersect at $E$ , and $AE=5$ . What is $CD$ ?

$\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\mathrm{(E) \ } \frac{55}{3}\qquad$

(Source)

Let

$S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}$

and

$S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}$ . What is the ratio of the area of $S_2$ to the area of $S_1$ ?

$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$

(Source)

Olympiad

Consider a circle $S$ , and a point $P$ outside it. The tangent lines from $P$ meet $S$ at $A$ and $B$ , respectively. Let $M$ be the midpoint of $AB$ . The perpendicular bisector of $AM$ meets $S$ in a point $C$ lying inside the triangle $ABP$ . $AC$ intersects $PM$ at $G$ , and $PM$ meets $S$ in a point $D$ lying outside the triangle $ABP$ . If $BD$ is parallel to $AC$ , show that $G$ is the centroid of the triangle $ABP$ .

(<url>viewtopic.php?=217167 Source</url>)

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 Assume that <math>A>T</math>. Let <math> P </math> be the area of a regular polygon that is closest to the circle's area. Therefore we have <math>A-P<A-T</math> so <math>P>T</math>. Let the apothem be <math>a</math> and the perimeter be <math>p</math> so the area of a regular polygon is one half of the product of the perimeter and apothem. The perimeter is less than the circumference so <math>p<2\pi r</math> and the apothem is less than the radius so <math>a<r</math>. Therefore <math> P=\frac{1}{2}ap<\frac{1}{2}r\cdot 2\pi r=T</math>. However it cannot be both <math>P>T</math> and <math>P<T</math>. So <math>A\not >T</math>.
-{{incomplete|proof}}
+{{stub}}
 ===Proof Using Calculus===
 Let the circle in question be <math>x^2 + y^2 = r^2</math>, where r is the circle's radius. By symmetry, the circle's area is four times the area in the first quadrant. The area in the first quadrant can be computed using a definite integral from 0 to r of the function <math>f(x) = \sqrt(r^2 - x^2)</math>. Using the substitution <math>x = r \sin u, dx = r \cos u</math> gives the indefinite integral as <math>\frac{r^2}{2} (u - \frac{\sin 2u}{2}) + C</math>, so the definite integral equals <math>\frac{r^2}{2} * \frac{\pi}{2}</math>. Multiplying by four gives the area of the circle as <math>\pi r^2</math>.

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