Archive for circle

It’s Maths!!

Posted in General with tags , , on January 20, 2010 by Admin

Maths Forum

Not only talks about Conic section(Circle) but also

other math topics.

KLIK to view;)

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Here’s another link which I think helps a lot for those who had troubles with Conic Section

Click HERE or HERE

or maybe HERE.

The Tab Tutor program sure is useful;)

 

moderator,

Nor Hidayah Binti Kamin

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Picture of circle applications

Posted in Gallery with tags , , , on January 20, 2010 by ctsuraidah

Post by,

ctsuraidah

Circle Theorems

Posted in The Angle of Circle with tags , , , , , , , , , , , , on January 10, 2010 by merrusmsk

Circle Theorems

Circles

A circle is a set of points which are all a certain distance from a fixed point known as the centre.
A line joining the centre of a circle to any of the points on the circle is known as a radius.

The circumference of a circle is the length of the circle. The circumference of a circle = 2 × π × the radius.

a sector, an arc and chord

The red line in the second diagram is called a chord. It divides the circle into a major segment and a minor segment.

Theorems

Angles Subtended on the Same Arc

Angles subtended on the same arc

Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points.

Angle in a Semi-Circle

angle in a semi-circle

Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So c is a right angle.

This proof is higher tier Proof

We can split the triangle in two by drawing a line from the centre of the circle to the point on the circumference our triangle touches.

Divide the triangle in two

We know that each of the lines which is a radius of the circle (the green lines) are the same length. Therefore each of the two triangles is isosceles and has a pair of equal angles.

Two isosceles triangles

But all of these angles together must add up to 180°, since they are the angles of the original big triangle.

Therefore x + y + x + y = 180, in other words 2(x + y) = 180.
and so x + y = 90. But x + y is the size of the angle we wanted to find.

Tangents

A tangent to a circle is a straight line which touches the circle at only one point (so it does not cross the circle- it just touches it).

A tangent to a circle forms a right angle with the circle’s radius, at the point of contact of the tangent.

angle with a tangent

Also, if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same.

Tangents from an external point are equal in length

Angle at the Centre

Angle at the centre

The angle formed at the centre of the circle by lines originating from two points on the circle’s circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.

This proof is higher tier Proof

You might have to be able to prove this fact:

proof diagram 1

OA = OX since both of these are equal to the radius of the circle. The triangle AOX is therefore isosceles and so ∠OXA = a
Similarly, ∠OXB = b

proof diagram 2
Since the angles in a triangle add up to 180, we know that ∠XOA = 180 – 2a
Similarly, ∠BOX = 180 – 2b
Since the angles around a point add up to 360, we have that ∠AOB = 360 – ∠XOA – ∠BOX
= 360 – (180 – 2a) – (180 – 2b)
= 2a + 2b = 2(a + b) = 2 ∠AXB

This section is higher tier Alternate Segment Theorem

Alternate segment theorem

This diagram shows the alternate segment theorem. In short, the red angles are equal to each other and the green angles are equal to each other.

Proof

You may have to be able to prove the alternate segment theorem:

proof of alternate segment theorem

We use facts about related angles:

A tangent makes an angle of 90 degrees with the radius of a circle, so we know that ∠OAC + x = 90.
The angle in a semi-circle is 90, so ∠BCA = 90.
The angles in a triangle add up to 180, so ∠BCA + ∠OAC + y = 180
Therefore 90 + ∠OAC + y = 180 and so ∠OAC + y = 90
But OAC + x = 90, so ∠OAC + x = ∠OAC + y
Hence x = y

Cyclic Quadrilaterals

cyclic quadrilateral is a four-sided figure in a circle, with each vertex (corner) of the quadrilateral touching the circumference of the circle. The opposite angles of such a quadrilateral add up to 180 degrees.

This proof is higher tier Area of Sector and Arc Length

A sector

If the radius of the circle is r,
Area of sector = πr2 × A/360
Arc length = 2πr × A/360

In other words, area of sector = area of circle × A/360
arc length = circumference of circle × A/360

Fof more detail just type circle theorem in ur search list..

by

Melson Manggis

Equation Of Circle

Posted in Equation Of Circle with tags on January 10, 2010 by msrica

This article is taken from the notes for my Math 107 web page. It’s purpose is to explain how to find equations for circles and also how to use this knowledge to provide a shortcut for graphing such equations. In order to do this you need to know two things.

  1. The distance formula
  2. The definition of a circle

For the distance formula see my notes.

What is a circle?

ãA circle is a set of all points equidistant from a point.

That is the standard geometry definition for a circle, and it is this kind of definition that allows us to write down equations for things like circles. Since we have a definition in terms of distances, we can use the distance formula to come up with an equation. Remember that the equation for a geometrical figure is an equation that all of the points in it have to satisfy, a kind of entrance test to get on the graph.

The point we want to make our points equidistant from is called the center and the distance is called the radius. (Think about how you draw a circle with a compass.) So let’s find the equation of the circle with radius r and center (h,k). I am using (h,k) for the coordinates of the circle rather than (x,y) because I have to reserve (x,y) for the points on the circle.

By the distance formula, the distance between (x,y) and (h,k) is given by this expression. By our definition of a circle, in order for (x,y) to pass the entrance test, this must be r, so we get this. This looks nicer if we square both sides, so we get this. for the standard equation of a circle with center (h,k) and radius r.

This means that if you were asked to write down the equation for a circle with a given center and radius you should be able to do this by just putting the numbers into this formula. You would substitute the x coordinate of the center for h and the y coordinate of the center for k and the radius for r.

Example: Find the equation of the circle with center (2,3) and radius 5.

Solution: Just put the 2 where the h is and the 3 where the k is and the 5 where the r is to get the equation.

(x-2)2+(y-3)2=25

This means that if someone were to plot all of the points that satisfy this equation, they would get exactly the circle that we want them to get, namely the one that has a center at (2,3) and a radius of 5.

Knowing about this standard equation for a circle can also be useful for making it easier to graph equations, because now whenever you see an equation like this, you don’t need to plot any points, you just need to recognize it is of this form and draw the circle. Before when you graphed equations you could only do it be just substituting x’s and y’s and finding points.  If you have an equation of this kind of form and are asked to graph it, you need to find the center and the radius of the circle, so you need to recognize what the h, k, and r are. The right side of the equation will be r2, so to find r, you have to take its square root. You need to put the left side into the form (x-h)2+(y-k)2 in order to determine h and k. One problem you might have is that you might have addition instead of subtraction, so to deal with this you would have to see addition as subtracting a negative, which is kind of backwards of the way we normally think about these things. My suggestion for an approach so that you don’t have to remember too many things like that is just always find x and y so that what is in the parentheses is 0, and that will be your h and k, the coordinates of the center. Then once you have found the center and the radius of the circle, you can graph it by plotting the center and then drawing a circle of radius r around that center. A good way to do this is to plot four points, r to the right, r to the left, r up, and r down from the center and then just sketch a circle through those points. It doesn’t have to be perfect, just try your best. The main thing is that you show you teacher that you know what the center and radius is and what that means.

But since math instructors are all are mean and wicked, there is one more problem I need to tell how to deal with. The problem is, somebody just might have taken a nice equation for a circle like

(x+1)2+(y-2)2=9,

and expanded everything out and rearranged things like this

x2+2x+1+y2-4y+4=9
x2+y2+2x-4y-5=0,

and then presented you with the equation

x2+y2+2x-4y-5=0

to graph. The problem is that it is a lot easier to expand that out than to put it back to where it came from. But there is a way to do it and it is to use the method of completing the squares. So here is what you do.

Get all the x’s together and all the y’s together and add something to both sides to get the constant on the right side of the equation. Leave space so that there is room to complete the square.

x2+2x         +y2-4y        =5

Then complete the square to find something to add to the x’s part and something to add to the y’s part so that they are both perfect squares. Remember to complete the square you take half of the coefficient and the square it, so for the x’s you get 2/2=1, 12=1, so you add 1, and for the y’s 4/2=2, 22=4, so you add 4. (You don’t have to worry about the minus on the 4 because you are going to square, so it will go away anyway.)

x2+2x +1+y2-4y +4=5+1+4

Then write the left side as squares and add up the right side and you get

(x+1)2+(y-2)2=9,

and now you can find the center and radius and graph it. Here the center would be (-1,2) and the radius would be 3. So to graph it, all you need to do is find the point (-1,2) and then plot the points 3 up, 3 down, 3 to the right, and 3 to the left of it, and draw the circle through them.

by,

Erica Olivia Henry

ABOUT CIRCLE

Posted in Introduction with tags , on December 28, 2009 by ayen89

Circle In Euclidean geometry, a circle is the set of all points in a plane at a fixed distance, called the radius, from a fixed point, called the centre. Circles are simple closed curves, dividing the plane into an interior and exterior. Sometimes the word circle is used to mean the interior, with the circle itself called the circumference. More usually, the circumference means the length of the circle, and the interior of the circle is called a disk.

In an x-y coordinate system, the circle with centre (x0, y0) and radius r is the set of all points (x, y) such that

(x − x0)2 + (y − y0)2 = r2.
If the circle is centered at the origin (0, 0), then this formula can be simplified to

x2 + y2 = r2.
The circle centered at the origin with radius 1 is called the unit circle.
All circles are similar; as a consequence, a circle’s circumference and radius are proportional, as are its area and the square of its radius. The constants of proportionality are 2π and π, respectively. In other words:

Length of a circle’s circumference = 2 × π × radius
Area of a circle = π × (radius)2
The formula for the area of a circle can be derived from the formula for the circumference and the formula for the area of a triangle, as follows. Imagine a regular hexagon (six-sided figure) divided into equal triangles, with their apices at the center of the hexagon. The area of the hexagon may be found by the formula for triangle area by adding up the lengths of all the triangle bases (on the exterior of the hexagon), multiplying by the height of the triangles (distance from the middle of the base to the center) and dividing by two. This is an approximation of the area of a circle. Then imagine the same exercise with an octagon (eight-sided figure), and the approximation is a little closer to the area of a circle. As a regular polygon with more and more sides is divided into triangles and the area calculated from this, the area becomes closer and closer to the area of a circle. In the limit, the sum of the bases approaches the circumference 2πr, and the triangles’ height approaches the radius r. Multiplying the two and dividing by 2, we get the area π r².

A line cutting a circle in two places is called a secant, and a line touching the circle in one place is called a tangent. The tangent lines are necessarily perpendicular to the radii, segments connecting the centre to a point on the circle, whose length matches the definition given above. The segment of a secant bound by the circle is called a chord, and the longest chords are those that pass through the centre, called diameters and divided into two radii. The part of a circle cut off by a chord is called a circle segment.

It is possible (Circle points segments proof) to find the maximum number of unique segments generated by running chords between a number of points on the perimeter of a circle.

If only (part of) a circle is known, then the circle’s center can be constructed as follows: take two non-parallel chords, construct perpendicular lines on their midpoints, and find the intersection point of those lines.

A part of a circle bound by two radii is called an arc, and the ratio between the length of an arc and the radius defines the angle between the two radii in radians.

Every triangle gives rise to several circles: its circumcircle containing all three vertices, its incircle lying inside the circle and touching all three sides, the three excircles lying outside the triangle and touching one side and the extensions of the other two, and its nine point circle which contains various important points of the triangle. Thales’ theorem states that if the three vertices of a triangle lie on a given circle with one side of the triangle being a diameter of the circle, then the angle opposite to that side is a right angle.

Given any three points which do not lie on a line, there exists precisely one circle containing those points (namely the circumcircle of the triangle defined by the points).

A circle is a kind of conic section, with eccentricity zero. In affine geometry all circles and ellipses become (affinely) isomorphic, and in projective geometry the other conic sections join them. In topology all simple closed curves are homeomorphic to circles, and the word circle is often applied to them as a result. The 3-dimensional analog of the circle is the sphere.

Squaring the circle refers to the (impossible) task of constructing, for a given circle, a square of equal area with ruler and compass alone. Tarski’s circle-squaring problem, by contrast, is the task of dividing a given circle into finitely many pieces and reassembling those pieces to obtain a square of equal area. Assuming the axiom of choice, this is indeed possible.

Three-dimensional shapes whose cross-sections in some planes are circles include spheres, spheroids, cylinders, and cones.

by,

Gloria Haren Saging

How To Construct A Tangent To A Circle

Posted in Tangent of circle with tags , , , on December 27, 2009 by gloria terrence firaon
Step-by-step Instructions  
After doing this Your work should look like this
We start with a point P somewhere on a given circle, with center point O.If the center is not given, you can use: “Finding the center of a circle with compass and straightedge or ruler“,
or
“Finding the center of a circle with any right-angled object”.
Geometry construction with compass and straightedge or ruler or ruler
1. Draw a straight line through the center O of the circle and the point P right across the circle. This is a diameter of the circle. Geometry construction with compass and straightedge or ruler or ruler
2. Mark a point Q anywhere. For best accuracy, avoid putting it too close to the diameter line. Geometry construction with compass and straightedge or ruler or ruler
3. Place the compass on the point Q just drawn, and set it’s width to the point P. Geometry construction with compass and straightedge or ruler or ruler
4.  Without changing the compass width, draw an arc across the diameter line, creating point R. Geometry construction with compass and straightedge or ruler or ruler
5.  Again, without changing the compass width, draw another arc on the opposite side of Q. Geometry construction with compass and straightedge or ruler or ruler
6. Using the straightedge, draw a line through R and Q, extending it onwards so it crosses the arc just drawn. Mark this point S. Geometry construction with compass and straightedge or ruler or ruler
7. Using the straightedge, draw a line through P and S, extending it in both directions. Geometry construction with compass and straightedge or ruler or ruler
8. Done. The line just drawn is the tangent to the circle O through point P. Geometry construction with compass and straightedge or ruler or ruler

Posted by,

Gloria Terrence Firaon

tangent of circle

Posted in Tangent of circle with tags , , , on December 27, 2009 by shuyen22

Tangent

  • The line drawn perpendicular to a radius through the end point of the radius is a tangent to the circle.
  • A line drawn perpendicular to a tangent through the point of contact with a circle passes through the center of the circle.
  • Two tangents can always be drawn to a circle from any point outside the circle, and these tangents are equal in length.

Tangent lines

Main article: Tangent lines to circles

The tangent line through a point P on the circle is perpendicular to the diameter passing through P. If P = (x1, y1) and the circle has center (a, b) and radius r, then the tangent line is perpendicular to the line from (a, b) to (x1, y1), so it has the form (x1a)x+(y1b)y = c. Evaluating at (x1, y1) determines the value of c and the result is that the equation of the tangent is

(x1a)x + (y1b)y = (x1a)x1 + (y1b)y1

or

(x1a)(xa) + (y1b)(yb) = r2.

This application is controlled by the application menu on top of the screen (figure 1.24) and the selection menu on bottom of the screen (figure 1.18). It is used to build and manipulate 2D objects made up of points, segments, arcs and splines. Circles and straight lines are only used as construction aids.

  • The application uses elementary geometry to define points, straight lines, circles, segments and arcs. For example, by using the tangency property, it is possible to define a circle tangent to three straight lines or a straight line tangent to two circles.
  • The application uses splines—a series of points—to define open or closed curves.
  • The application is able to cut one object with another object, except the circles and straight lines, which are only construction aids. It is for example possible to cut an object by a straight line. However, the straight line will not be affected by this operation.
  • The application is able to round off angles.
  • The application is able to duplicate an object n times by using the classical affine transformations: symmetry, rotation, homothety.
  • The application is able to “outline” figures.
  • The application is able to invert arcs and segments.
  • etc.

 Page II.2 of the exhibition René Descartes
Géométrie

Let CE be the curve, and assume that through point C we need to draw a straight line that forms straight angles with it. I imagine everything already done and assume CP as the sought line, a line that I prolong until P where it meets straight line GA, which I imagine being that to which all the points of line CE must refer. Thus position MA or CB = y, CM or BA = x, I will obtain a certain equation that expresses the relation between x and y. […]

With a familiar language and formalism, Descartes continues by explaining that if one poses for the unknown circle PC=s and PA=v, observing that the triangle PMC has a straight angle, one has $s^2=x^2+v^2-2vy+y^2$, from which one can recover x (or equivalently, y) and substitute it in the equation of the given curve. Then,

“after having found [that equation] instead of using it to know the quantities x or y […] that are already given because point C [in which we must determine the normal to the curve] is given, we must use it to find v or s that determine the requested point P [centre of the sought circle]. For this reason, we must consider that if this point P is the way we want it, the circle of which this is the centre and that will pass through C, will touch the curve CE there without intersecting it. On the other hand, if P is a little closer or a little further from A of that which it should be, the circle will intersect the curve not only in point C but necessarily also in some other [E]. […] but the more these two points, C and E, will be near, the less the difference between the roots [of the equation]. Finally, if these two points are one (that is if the circle that goes through C touches the curve there without intersecting it), the roots will be exactly the same […].”

Therefore it will be sufficient to impose that the polynomial has two double roots. If the equation of the curve was of degree $m$, the resulting polynomial will have degree $2m$ and will be of the form $(y-y_0)^2Q(y)$ where $Q(y)$is a generic polynomial of degree $2(m-1)$. Equalising the coefficients of homologous powers we obtain $2m+1$ equations from which we can get the $2m-1$ coefficients of $Q$ as well as the parameters v and s.

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The Géométrie was diffused among mathematicians mostly through the two Latin editions edited by Franz van Schooten. The first was published in 1649 and, together with the translation of the Géométrie , it contains De Beaune’s Notae Breves and Schooten’s own Comentarii. In the second edition, in two volumes, many other pamphlets are added, among which two letters by John Hudde containing a theorem on double roots that brings to a simplification of the previous method. Schooten’s commentary cover all three books of the Géométrie with precise notes, observations, integrations and applications. Regarding the problem of tangents, Schooten illustrates how to apply Descartes’ method in various examples, among which the determination of the normal to the conchoid.

* Page II.3 in the exhibition

Franz van Schooten
In geometriam Renati Des Cartes Commentarii

Let CE be the first conchoid of the ancients, with a pole G and a centre line AB, such that all segments whose prolongations intersect in G and are comprised between the curve CE and the straight line AB (like AE, LC) are equal. We require to draw a straight line (like CP), that intersects the conchoid at a straight angle in a given point C. […] It is then enough to take on the straight line CG the segment CD equal to CB, which is perpendicular to AB, and then from point D draw DF parallel to AG and equal to GL; thus we will have point F, through which we will draw the required line CP.

[follows the construction with Descartes’ method]

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The kinematic construction of tangents

In 1644 Mersenne divulged a method to draw the tangents to a curve, communicated to him by Gilles Personne de Roberval, professor at Paris’s College Royal. In the same year, Torricelli published a similar method in his Opera Geometrica. Both methods presuppose the knowledge of the kinematic decomposition of the curve of which the tangent must be drawn: in the parabola, for example, a point distances itself from the focus with the same speed with which it distances itself from the centre line, in the ellipse it approaches a focus with the same speed with which it distances itself from the other, in the spiral a point rotates around the origin at the same speed with which it distances itself from it, a fact already known and used by Archimedes. Roberval’s manuscript, written down by a pupil of his, was presented in 1668 to the Académie des Sciences and published in a collection of writings only in 1693. The “axiom or principle of invention” at the basis of the method is that “the direction of movement of a point describing a curved line is the tangent to the curved line in any position of that point”, a principle which is “sufficiently intelligible” that “one will easily accept if one considers it with some attention.” Hence descends the “general rule” to follow for the tangents:

For the specific properties of the curved line (which will be given to you), examine the various movements that the point describing the line has where you want to draw the tangent: compose all these movements in one, trace the line of the direction of the composite movement, and you will have the tangent to the curved line.

By applying the rule “word by word”, one can study several curves:

tangents to conical sections, tangents to the other main curves known to the ancients and to other recently described curves, like Mr. Pascal’s snail, Mr. Roberval’s cycloid, Mr. Descartes’s parabola of the second type, etc.

The eleventh example of the pamphlet deals actually with the cycloid that Robertval calls “roulette” o “trochoïde”. The curve is described by a point B that is on a circumference as this rolls on a straight line BC. Another way of generating the curve is by saying that the circumference translates with an uniform motion so that the centre a describes the base segment, and at the same time point B uniformly traces the circumference. If the length of the basis is the same than the circumference, we have a roulette of the first type, but in general we can consider the cases in which the base is longer or shorter than the circumference. After describing the construction by points of the curve, Roberval describes the construction of the tangent in any point E on the basis of the decomposition into two simultaneous motions.

* Page II.4 in the exhibition

Gilles Personne de Roberval
Observations sur la composition des mouvements et sur le moyen de trouver les touchantes des lignes courbes.

be given; we require its tangent in point E. Describe the circle BDF of the roulette […]; from point E draw the straight line EF parallel to AC which intersects in F the circumference of the roulette’s semicircle […]; draw FG tangent to the circle, then take H on the tangent to the circle so that AC is to the circumference of the circle like EF is to FH; from point H draw HE, and this will be the tangent to the roulette. 

Roberval’s construction is then compared to Fermat’s:

“one draws EF as above, draws the straight line FB and through the point E draws AH parallel to FB. EH will be the tangent”.

It is proven that the two construction agree, but, it is said, Fermat’s method is not as general because it only works for the cycloid of the first type.

by,

Seah Shu Yen