Difference between revisions of "Circle"
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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>. | 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>. | ||
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===Area Proof Using Calculus=== | ===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>. | 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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[[Category:Definition]] | [[Category:Definition]] | ||
[[Category:Geometry]] | [[Category:Geometry]] | ||
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Revision as of 09:58, 9 June 2021
A circle is a geometric figure commonly used in Euclidean geometry.
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A basic circle. |
Contents
[hide]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 .
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>)
See Also
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