Difference between revisions of "Circle"

Revision as of 12:31, 30 March 2010

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

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

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.

Template:Incomplete

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.
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

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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 ===Olympiad===
+*Consider a [[circle]] <math>S</math>, and a [[point]] <math>P</math> outside it. The [[tangent line]]s from <math>P</math> meet <math>S</math> at <math>A</math> and <math>B</math>, respectively. Let <math>M</math> be the [[midpoint]] of <math>AB</math>. The [[perpendicular bisector]] of <math>AM</math> meets <math>S</math> in a point <math>C</math> lying inside the [[triangle]] <math>ABP</math>. <math>AC</math> intersects <math>PM</math> at <math>G</math>, and <math>PM</math> meets <math>S</math> in a point <math>D</math> lying outside the triangle <math>ABP</math>. If <math>BD</math> is [[parallel]] to <math>AC</math>, show that <math>G</math> is the [[centroid]] of the triangle <math>ABP</math>.
+(<url>viewtopic.php?=217167 Source</url>)
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