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Difference between revisions of "Circle"

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A '''circle''' is a geometric figure commonly used in Euclidean [[geometry]].
 +
{{asy image|<asy>unitsize(2cm);draw(unitcircle,blue);</asy>|right|A basic circle.}}
 +
 +
 
==Definition==
 
==Definition==
−
*The [[locus]] of points distance <math>\displaystyle {r}</math>(radius) from center <math>{O}</math>
+
=== Traditional Definition ===
−
*A cross section of a cone or cylinder parallel to the base. For more on this, see [[Dandelin Spheres]], a commonly used method of deriving all [[Conic sections]].
+
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]].  
−
*<math>(x - a)^{2} + (y - b)^{2} = r^{2} </math> for a circle centered at <math>(a,b)</math> with radius <math>{r}</math>
+
[[Image:circle1.PNG|thumb|right|The radius and center of a circle.]]
  
−
==Formulas==
+
=== Coordinate Definition ===
−
*'''Area''' <math>\displaystyle \pi r^2</math>
+
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
−
*'''circumference''' <math>\displaystyle 2\pi r</math>
+
<cmath>(x - h)^2 + (y - k)^2 = r^2.</cmath>
−
==Other Properties==
 
−
* The perpendicular bisector of any chord in the circle is a diameter
 
−
* awaiting diagrams to add stuff on inscribed angles + tangents.
 
  
−
==Practice Problems==
+
'''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>
  
−
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=349797#p349797 2005 AMC 12A #16]
+
==Circumference and Area==
−
*[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=423848#p423848 2006 AMC 12A #21]
+
 
 +
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 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:
 +
 
 +
<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> \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' actual claim was that a circle with radius <math> r </math> and circumference <math> C </math> had an area equivalent to the area of a [[right triangle]] with base <math> C </math> and height <math> r </math>.  First let the area of the circle be <math> A </math> and the area of the triangle be <math> T </math>.  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 <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>.
 +
 +
===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} \cdot \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.
 +
*Given a segment (a section of a circle bounded by a chord inside the circle and the respective arc on the circumference; not to be confused with [[line segment]]), angles in that segment, i.e. the angle between the [[line segments]] joining the endpoints of the chord and another point on the circumference, will be equal in measure regardless of the point chosen for a given segment (given chord).
 +
 
 +
==Problems==
 +
===Introductory===
 +
*Under what constraints is the circumference (in inches) of a circle greater than its area (in square inches)?
 +
 
 +
*How many circles with radius <math>r</math> can we fit around a circle with radius <math>r</math>?
 +
 
 +
*A square is placed so that it inscribes a circle and is inscribed by a different circle. Find the ratio of the area of the 2 circles.
 +
 
 +
===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]])
 +
 
 +
*Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math>
 +
 
 +
<asy>
 +
draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N);
 +
</asy>
 +
 
 +
([[2020 AMC 8 Problems/Problem 18|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 ==
−
*[[Pi]]
+
* [[Geometry]]
−
*[[Power of a point]]
+
* [[Pi]]
−
*[[Law of sines]]
+
* [[Power of a point]]
−
*[[Inversion]]
+
* [[Homothety]]
−
*[[Homothecy]]
+
 
 +
 
 +
[[Category:Definition]]
 +
[[Category:Geometry]]
 +
 
 +
{{stub}}

Latest revision as of 12:46, 2 September 2024

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

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

Enlarge.png
A basic circle.


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.

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

Circumference and Area

Given a circle of radius $r$, the circumference (distance around a circle) is $2 \pi r$ and the area is $\pi r^2$. Both formulas involve the mathematical constant pi ($\pi$).

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:

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 $A\not >T$.

Area 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 $\frac{r^2}{2} \cdot \frac{\pi}{2}$. Multiplying by four gives the area of the circle as $\pi r^2$.

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 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 $AB$ and $CD$ intersect at point $P$ inside the circle, $\angle APC = \frac{m\widehat{AC} + m\widehat{BD}}{2}$.
  • When two chords $AB$ and $CD$ intersect at point $P$ outside the circle, $\angle APC = \frac{m\widehat{AC} - m\widehat{BD}}{2}$.
  • Lengths of chords can be calculated by using the Power of a point theorem.
  • Given a segment (a section of a circle bounded by a chord inside the circle and the respective arc on the circumference; not to be confused with line segment), angles in that segment, i.e. the angle between the line segments joining the endpoints of the chord and another point on the circumference, will be equal in measure regardless of the point chosen for a given segment (given chord).

Problems

Introductory

  • Under what constraints is the circumference (in inches) of a circle greater than its area (in square inches)?
  • How many circles with radius $r$ can we fit around a circle with radius $r$?
  • A square is placed so that it inscribes a circle and is inscribed by a different circle. Find the ratio of the area of the 2 circles.

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)

  • Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$

[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy]

(Source)

Olympiad

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

See Also

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