Difference between revisions of "Circle"
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− | + | A '''circle''' is a geometric figure commonly used in Euclidean [[geometry]]. | |
− | A '''circle''' is | + | {{asy image|<asy>unitsize(2cm);draw(unitcircle,blue);</asy>|right|A basic circle.}} |
− | < | ||
− | |||
− | |||
− | + | ==Definition== | |
+ | === Traditional Definition === | ||
+ | A circle is defined as the [[set]] (or [[locus]]) of [[point]]s 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]]. | ||
+ | [[Image:circle1.PNG|thumb|right|The radius and center of a circle.]] | ||
− | '''Example:''' The equation <math> (x-3)^2 + (y+6)^2 = 25 </math> represents the circle with center <math> (3,-6) </math> and radius 5 units. | + | === 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, <math>r</math>, and center <math>(h,k)</math>. We know that each point, <math>(x,y)</math>, on the circle which we want to identify is a distance <math>r</math> from <math>(h,k)</math>. Using the [[distance formula]], this gives <math>\sqrt{(x - h)^2 + (y - k)^2} = r</math> which is more commonly written as | ||
+ | <cmath>(x - h)^2 + (y - k)^2 = r^2.</cmath> | ||
+ | |||
+ | '''Example:''' The equation <math>(x - 3)^2 + (y + 6)^2 = 25</math> represents the circle with center <math>(3,-6)</math> and radius 5 units. | ||
<center>[[Image:Circlecoordinate1.PNG]]</center> | <center>[[Image:Circlecoordinate1.PNG]]</center> | ||
− | == Area | + | ==Circumference and Area== |
− | + | ||
+ | Given a circle of radius <math>r</math>, the [[circumference]] (distance around a circle) is <math>2 \pi r</math> and the area is <math>\pi r^2</math>. Both formulas involve the mathematical constant [[pi]] (<math>\pi</math>). | ||
− | === Archimedes' Proof === | + | === Archimedes' Proof of Area=== |
We shall explore two of the Greek [[mathematician]] [[Archimedes]] demonstrations of the area of a circle. The first is much more intuitive. | We shall explore two of the Greek [[mathematician]] [[Archimedes]] demonstrations of the area of a circle. The first is much more intuitive. | ||
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<center>[[Image:Pizzawedges2.PNG]]</center> | <center>[[Image:Pizzawedges2.PNG]]</center> | ||
− | 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 <math> r </math> and width <math> \pi r </math> thus making its area <math> | + | 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 <math> r </math> and width <math> \pi r </math> thus making its area <math> \pi r^2 </math>. |
Archimedes also came up with a brilliant proof of the area of a circle by using the [[proof]] technique of [[reductio ad absurdum]]. | Archimedes also came up with a brilliant proof of the area of a circle by using the [[proof]] technique of [[reductio ad absurdum]]. | ||
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'''Case 1:''' The circle's area is greater than the triangle's area. | '''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 <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>. | ||
+ | |||
+ | {{stub}} | ||
+ | ===Area 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>. | ||
+ | |||
+ | ==Lines in Circles== | ||
+ | <asy>draw(unitcircle);draw((-0.8,1)--(1,1),Arrow);draw((1,1)--(-0.8,1),Arrow);draw((0,1)--(1,0));</asy> | ||
+ | |||
+ | A line that touches a circle at only one point is called the [[Tangent (Geometry)|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 longest chord of the circle is the diameter; it passes through the center of the circle. | ||
+ | |||
+ | When two secants intersect on the circle, they form an [[inscribed angle]]. | ||
+ | |||
+ | ===Properties=== | ||
+ | *The measure of an [[inscribed angle]] is always half the measure of the [[central angle]] with the same endpoints. | ||
+ | **Since the diameter divides the circle into two equal parts, any angle formed by the two endpoints of a diameter and a third distinct point on the circle as the vertex is a right angle. | ||
+ | **Also, a right triangle inscribed in a circle has a hypotenuse that is a diameter of the circle. | ||
+ | *Similarly, if a tangent line and a secant line intersects at the point of tangency, the measure of the angle formed is always half the measure of the [[central angle]] with the same endpoints. | ||
+ | **From that property, the angle formed by the diameter and a tangent line with the point of tangency on the diameter is a right angle. | ||
+ | **The perpendicular line through the tangent where it touches the circle is a diameter of the circle. | ||
+ | *The perpendicular bisector of a chord is always a diameter of the circle. | ||
+ | *When two chords <math>AB</math> and <math>CD</math> intersect at point <math>P</math> inside the circle, <math>\angle APC = \frac{m\widehat{AC} + m\widehat{BD}}{2}</math>. | ||
+ | *When two chords <math>AB</math> and <math>CD</math> intersect at point <math>P</math> outside the circle, <math>\angle APC = \frac{m\widehat{AC} - m\widehat{BD}}{2}</math>. | ||
+ | *Lengths of chords can be calculated by using the [[Power of a point]] theorem. | ||
+ | |||
+ | ==Problems== | ||
+ | ===Introductory=== | ||
+ | *Under what constraints is the circumference (in inches) of a circle greater than its area (in square inches)? | ||
− | == | + | ===Intermediate=== |
− | * | + | *[[Circle]]s with [[center]]s <math>A</math> and <math>B</math> have [[radius |radii]] 3 and 8, respectively. A [[common internal tangent line | common internal tangent]] [[intersect]]s the circles at <math>C</math> and <math>D</math>, respectively. [[Line]]s <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>? |
− | |||
− | |||
− | + | <cmath>\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</cmath> | |
− | + | ([[2006 AMC 12A Problems/Problem 16|Source]]) | |
+ | *Let | ||
+ | <cmath>S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}</cmath> | ||
+ | :and | ||
+ | <cmath>S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}</cmath>. | ||
+ | What is the ratio of the area of <math>S_2</math> to the area of <math>S_1</math>? | ||
− | *[ | + | <cmath> \mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102</cmath> |
− | + | ||
+ | ([[2006 AMC 12A Problems/Problem 21|Source]]) | ||
+ | |||
+ | ===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>) | ||
== See Also == | == See Also == | ||
− | |||
* [[Geometry]] | * [[Geometry]] | ||
* [[Pi]] | * [[Pi]] | ||
* [[Power of a point]] | * [[Power of a point]] | ||
− | * [[ | + | * [[Homothety]] |
− | + | ||
+ | |||
+ | [[Category:Definition]] | ||
+ | [[Category:Geometry]] |
Revision as of 16:24, 1 October 2020
A circle is a geometric figure commonly used in Euclidean geometry.
|
A basic circle. |
Contents
Definition
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.
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, , and center . We know that each point, , on the circle which we want to identify is a distance from . Using the distance formula, this gives which is more commonly written as
Example: The equation represents the circle with center and radius 5 units.
Circumference and Area
Given a circle of radius , the circumference (distance around a circle) is and the area is . Both formulas involve the mathematical constant pi ().
Archimedes' Proof of Area
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 and width thus making its area .
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 and circumference had an area equivalent to the area of a right triangle with base and height . First let the area of the circle be and the area of the triangle be . 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 . Let be the area of a regular polygon that is closest to the circle's area. Therefore we have so . Let the apothem be and the perimeter be 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 and the apothem is less than the radius so . Therefore . However it cannot be both and . So .
This article is a stub. Help us out by expanding it.
Area Proof Using Calculus
Let the circle in question be , 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 . Using the substitution gives the indefinite integral as , so the definite integral equals . Multiplying by four gives the area of the circle as .
Lines in Circles
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 longest chord of the circle is the diameter; it passes through the center of the circle.
When two secants intersect on the circle, they form an inscribed angle.
Properties
- The measure of an inscribed angle is always half the measure of the central angle with the same endpoints.
- Since the diameter divides the circle into two equal parts, any angle formed by the two endpoints of a diameter and a third distinct point on the circle as the vertex is a right angle.
- Also, a right triangle inscribed in a circle has a hypotenuse that is a diameter of the circle.
- Similarly, if a tangent line and a secant line intersects at the point of tangency, the measure of the angle formed is always half the measure of the central angle with the same endpoints.
- From that property, the angle formed by the diameter and a tangent line with the point of tangency on the diameter is a right angle.
- The perpendicular line through the tangent where it touches the circle is a diameter of the circle.
- The perpendicular bisector of a chord is always a diameter of the circle.
- When two chords and intersect at point inside the circle, .
- When two chords and intersect at point outside the circle, .
- Lengths of chords can be calculated by using the Power of a point theorem.
Problems
Introductory
- Under what constraints is the circumference (in inches) of a circle greater than its area (in square inches)?
Intermediate
- Circles with centers and have radii 3 and 8, respectively. A common internal tangent intersects the circles at and , respectively. Lines and intersect at , and . What is ?
(Source)
- Let
- and
. What is the ratio of the area of to the area of ?
(Source)
Olympiad
- Consider a circle , and a point outside it. The tangent lines from meet at and , respectively. Let be the midpoint of . The perpendicular bisector of meets in a point lying inside the triangle . intersects at , and meets in a point lying outside the triangle . If is parallel to , show that is the centroid of the triangle .
(<url>viewtopic.php?=217167 Source<url>)