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properties of circle

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Research, Knowledge and Information :

Circle - Wikipedia

Properties. The circle is the shape with the largest area for a given length of perimeter. (See Isoperimetric inequality.) The circle is a highly symmetric shape: ...
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Circle - Math word definition - Math Open Reference

Properties of a circle. Center: A point inside the circle. All points on the circle are equidistant (same distance) from the center point. Radius:
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Circles | Geometry (all content) | Math | Khan Academy

Explore, prove, and apply important properties of circles that have to do with things like arc length, radians, inscribed angles, and tangents.
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Properties of Circles.

PROPERTIES OF CIRCLES Introduction A circle is a simple, beautiful and symmetrical shape. When a circle is rotated through any angle about its centre, its orientation ...
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Properties of the Circle | eMathZone

In geometry, a large number of facts about circles and their relations to straight lines, angles and polygons can be proved. These facts are called the properties of ...
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Lesson Basic properties of a circle - Algebra.Com

In this lesson, we will look at the properties of the Circle. Before we start looking that the properties of the circle. we need to know basic
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Properties of Circle | Circle Properties | TutorCircle

Read on Properties of Circle and learn basic concepts of Properties of Circle and its applications
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Properties of Shapes: Circles -

Circles are fundamental to everything we do. But, did you know they're much more than just round shapes? In this lesson, we'll look at the various parts of circles ...
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Properties of Circles - Home

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Andrews University

10.1 Use Properties of Tangents. Diameter (d) – chord that goes through the center of the circle (longest chord = 2 radii) d = 2r. What is the radius of a circle if ...
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Properties of Circles - Mrs. Luthi's geometry

Properties of Circles In previous chapters, you learned the following skills, which you’ll use in Chapter 10: classifying triangles, finding angle measures, and solving
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Suggested Questions And Answer :

what property do the circled polygons have in common?

all hav a serkel round it if regular polygon, reel simpel tu draw serkel that go thru all points ("vertexes") nonregular...mae hav a problem
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proving the elips equation

The general equation of an ellipse is x^2/a^2+y^2/b^2=1, where a and b are the lengths of the semimajor and semiminor axes. The following proof uses the distance property: the sum of the lengths of the lines between any point on the ellipse and its two foci is constant. Both foci lie on the major axis (x) inside the ellipse. We start with a line AB of length p. Now we imagine a length of string s>p. The string is attached to the line by fixing one end to A and the other to B. Now we attach a pointer or marker to the string using a very small loop so that the pointer can move along the length of the string, keeping it taut. The pointer starts at a point that is an extension of AB, and the pointer is moved out of the line AB to describe the ellipse on both sides of AB, until the pointer returns to its starting position. The geometry is that, if the pointer's position at any time is P, AP+BP=s, where s is a constant. The midpoint of AB is O and represents the origin of the rectangular coordinates x and y. The point A is (-p/2,0) and B is (p/2,0). P is the general point (x,y). By Pythagoras, AP^2=(x+p/2)^2+y^2; BP^2=(p/2-x)^2+y^2; s=sqrt((x+p/2)^2+y^2)+sqrt((x-p/2)^2+y^2). ((p/2-x)^2=(x-p/2)^2, so these may be used interchangeably. Note also that (x+p/2)^2+(x-p/2)^2=x^2+p^2/4 and (x+p/2)(x-p/2)=x^2-p^2/4.) s-sqrt((x+p/2)^2+y^2)=sqrt((x-p/2)^2+y^2) Squaring both sides: s^2+(x+p/2)^2+y^2-2ssqrt((x+p/2)^2+y^2)=(x-p/2)^2+y^2; but (x+p/2)^2=x^2+xp+p^2/4 and (x-p/2)^2=x^2-xp+p^2/4; so, s^2+2xp=2ssqrt((x+p/2)^2+y^2); Squaring again: s^4+4xps^2+4x^2p^2=4s^2(x+p/2)^2+y^2)=4s^2(x^2+xp+p^2/4+y^2); s^4+4x^2p^2=4s^2x^2+s^2p^2+4s^2y^2; 4x^2(s^2-p^2)+4s^2y^2=s^2(s^2-p^2); 4x^2+4s^2y^2/(s^2-p^2)=s^2; 4x^2/s^2+4y^2/(s^2-p^2)=1; Let a^2=s^2/4 and b^2=(s^2-p^2)/4: x^2/a^2+y^2/b^2=1 is the standard equation of an ellipse. Things to note: When p=0 the ellipse should be a circle: 4x^2/s^2+4y^2/s^2=1 is the equation of a circle radius s/2. When x=0, y=+sqrt(s^2-p^2)/2=+b and when y=0, x=+s/2=+a. The difference between the extreme points for x, (-a,0) and (a,0), is 2a=s, so s represents the major axis of the ellipse, and a the semimajor axis. Similarly the difference between extreme points for y, (0,-b) and (0,b), is 2b=sqrt(s^2-p^2), and b represents the semiminor axis. The points A and B of line AB are the foci of the ellipse. Since s=2a and b^2=(s^2-p^2)/4, we can find p. 4b^2=4a^2-p^2; p=2sqrt(a^2-b^2) and A(-p/2,0) and B(p/2,0)=A(-sqrt(a^2-b^2),0) and B(sqrt(a^2-b^2),0). For a circle, a=b and the foci converge to the centre of the circle.
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Please help me find the answers and solve the problems, I am having a terrible time!!

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prove that C,D,H,E are concyclic

The picture shows the secants from point P. The quadrilateral CDHE is required to be proved to be a cyclic quadrilateral. That means that CDH+CEH=180=DCE+EHD. Join CB and AD, and join AH and BE. From this construction we get two pairs of similar triangles: APD and BPC, and APH and BPE, because of the common angle at P, and equal angles PCB=DAP, PAH=BEP (angles in the same segment). We can therefore write: PD/PB=PA/PC=DA/BC (triangles PCB and DAP) and PB/PH=PE/PA=BE/HA (triangles PAH and BEP). From this we get: PB.PA=PC.PD=PE.PH. So PC.PD=PE.PH and therefore PC/PE=PH/PD. Therefore, triangles PCE and PHD are also similar because P is the included common angle. This means PDH=PEC. But CDH=180-PDH (supplementary angles on a straight line), so CDH=180-PEC (PEC is the same angle as CEH). These are opposite angles of the quadrilateral CDHE, and this is a definitive property of cyclic quadrilaterals, so CDHE is cyclic. The other two angles must also be supplementary because the angles of a quadrilateral add up to 360 degrees.
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3(9x + 4) > 35x - 4 27x + 12 > 35x - 4 {used distributive property} 12 > 8x - 4 {subtracted 27x from both sides} 16 > 8x {added 4 to both sides} 2 > x {divided both sides by 8} x < 2 {flipped around so that x is first} To graph: - put an open circle (not filled in) on 2, with a line and arrow pointing to the left Ask
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what is geometric proof

It means you use geometry to prove a theorem, congruency, similar figures, the size of an angle or length of a side, etc. So you refer to geometrical theorems such as Pythagoras', Appollonius', and others, and properties of geometrical figures like triangles, circles, ellipses, rectangles, parallelograms, etc.
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What is the radius of this vicious circle?

A vicious circle The ends of the diameter of a unit circle are joined to make a smaller circle. The ends of the diameter of this smaller circle are joined to make an even smaller circle. And so on indefinitely. If the radii of all the circles (including the initial unit circle) are laid end to end in a line, and the ends of that line are joined to make a circle, what is the radius of the circle?   The diameter of each circle becomes the circumference of the next smaller circle down. Let C1 be the circumference of the unit circle. i.e. C1 = 2πR1, where R1 is the radius of the unit circle, so R1 = 1. The diameter, D1 = 2R1 And this is the circumference of the 2nd circle, C2. i.e. C2 = D1. C2 = 2πR2 = D1 = 2R1. i.e. R2 = R1/π Continuing ... C3 = 2πR3 = D2 = 2R2. i.e. R3 = R2/π C4 = 2πR4 = D3 = 2R3. i.e. R4 = R3/π C5 = 2πR5 = D4 = 2R4. i.e. R5 = R4/π Cn = 2πRn = D_(n-1) = 2R_(n-1). i.e. Rn = R_(n-1)/π If the radii of all these circles were laid end to end, the total length would be, L = R1 + R2 + R3 + ... + Rn, n = 1 to ∞, L = R1 + R1/π + R2/π + R3/π + ... + Rn/π, n = 1 to ∞, L = R1 + (1/π){R1 + R2 + R3 + ... + Rn}, n = 1 to ∞, L = R1 + (1/π)*L L – L/π = R1 L(π – 1) = πR1 L = πR1/(π-1) This length, of all the radii joined together, now becomes the circumference of a final circle. The radius of this final circle now is, Rf = L/(2π) Rf = (R1/2)/(π – 1) Since R1 is the radius of a unit circle, then R1 = 1 and Rf = 0.5/(π – 1) = 0.23347
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Find the point(s) where the line through the origin with slope 6 intersects the unit circle.

The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).  If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis. Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6),  to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2. Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0. If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k. Recap There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36. When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)). When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37. We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6). The values of h and k are not restricted to integers.
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. A car dealership has 556 new cars on its lot.

There seems to be one number missing: the number of black cars with automatic transmission and a sun roof, so we'll call this number X. It's helpful to use a diagram (Venn diagram). Draw a large circle (ellipse or other enclosure) containing three interlocking circles. I'll use the word "circle" to mean any completely enclosed area. The three circles represent black cars (B), automatics (A), and cars fitted with a sunroof (S) respectively. The large circle represents all the cars (C), so those that are not black, not automatic and have no sunroof are represented by the interior of the large circle outside the other three circles. We know there are 133 black cars, and 116 of these are automatic, so the remaining 17 must be manual transmission (non-automatic). We also know that 10 black cars have a sunroof, so 123 don't have a sunroof. And we know that X black automatics have a sunroof.  Of the 401 automatics, 42 have sunroofs, so 359 automatics have no sunroof; and of the 57 cars fitted with a sunroof, 42 are automatics, so 15 are manual. In the Venn diagram the intersection of the three sets are represented by the areas enclosed by two or three interlocking circles. I need a symbol to express intersection, normally represented by an inverted U symbol, which I don't have on my tablet, so I'll use ^, which is normally used to represent "to the power of". A^B means black automatics because it's the set of all black cars with automatic transmission, intersection of the black set B with the automatic set A. On the diagram it's the area enclosed by circles A and B. A^B^S is the area enclosed by the three interlocking circles and represents all black automatics with a sunroof. In all, the overlapping circles produce four enclosed areas: A^B, A^S, B^S and A^B^S. We can give values to these: 116, 42, 10 and X, respectively. Remember, we weren't given a value for X. We also have A=401, B=133, S=57. We can represent "not" by underlining a set, so, for example, A represents non-automatic and B^A would mean black cars without automatic transmission (AT), so B^A=17 because 17 cars are black without AT. Similarly, B^S=123. So B^A^S=A^B^S=116-X. That is, a) there are 116-X black automatics without a sunroof. There are 556-133=423 cars that are not black; 556-401=155 cars that are not automatic; and 556-57=499 cars without a sunroof. b) The final part of the question is easier to work out by looking at the diagram. We need to work out the total number of cars within the interlocking circles. We can't simply add the numbers of cars in the three circles and subtract the sum from 556, because the circles interlock. Take A and B for example. They interlock enclosing 116 cars (black automatics), so we deduct 116 from A and add the result to B; or we deduct 116 from B and add the result to A: 285+133=418=17+401.  If you look at the Venn diagram you'll see there are seven areas produced by the interlocking circles. Let's call the areas a, b, c, d, e, f and g, such that all automatics are given by a+d+g+e=401; black cars by b+d+g+f=133; and sunroofs by c+e+f+g=57. Diagrammatically, a is the set of automatics that are not black and have no sunroof; b the set of black cars that are neither automatic nor have a sunroof; and c the set of sunroofed cars that are neither black or automatic. There is an eighth area, h, outside of all the interlocking circles representing the residual cars that are not black, not automatic, and have no sunroof. Area g contains X cars (black automatics with a sunroof), so g=X; d+g=116; f+g=10; e+g=42; a+b+c+d+e+f+g+h=556. So d=116-X; f=10-X; e=42-X; a=401-(116+42-X)=43+X; b=133-(116+10-X)=7+X; c=57-(42-X+10)=5+X; (43+X)+(7+X)+(5+X)+(116-X)+(42-X)+(10-X)+X+h=556; so h=556-(223+X)=333-X. This is the number of cars that are not black, not automatic and have no sunroof. Although we don't know the value of X, we know it must be less than 10 because area f contains a number of cars that can't, obviously, be negative. The wording of the question suggests that X is greater than 1, so the possible answers range from 324 to 331 (1 Read More: ...

Prove that any chord [AB] of the larger circle is bisected by the smaller circle.

The small circle passes through O and touches the large circle at A, so AO is a radius of the large circle and a diameter of the small circle, since point A is a tangent point to both circles, and N is the intersection of the small circle with the large circle. Angle ANO is a right angle, because the angle in a semicircle is a right angle and the circle is the small circle.  OB is a radius of the large circle. ON is perpendicular to AB. Angle ANO=ONB=90 degrees. ABO is an isosceles triangle because OA=OB=radius. ON is a common side, so triangles ANO and ONB are congruent (RHS-Right angle, Hypotenuse, Side). Therefore AN=NB so the small circle bisects the chord AB.
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