Circle

Revision as of 15:53, 9 March 2014 by Greenpepper9999 (talk | contribs) (Proof Using Calculus)

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

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

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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.

Circlecoordinate1.PNG

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:

Pizzawedges2.PNG

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 $A\not >T$.

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Proof Using Calculus

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Related Formulae

Other Properties and Definitions

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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

  • What is the area of a circle with radius $3?$
  • Under what constraints is the circumference of a circle greater than its area? Assume they are both expressed in the same units.

Intermediate

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

(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

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

See Also

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