Equation of Circles

3 Key excerpts on "Equation of Circles"

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  eBook - ePub

  Plain Plane Geometry

  • Amol Sasane(Author)
  • 2015(Publication Date)
  • WSPC

    (Publisher)

  concyclic .

  Definition 5.2 (Cyclic quadrilateral) . A quadrilateral is said to be cyclic if all of its vertices lie on a circle.

  Exercise 5.1. Show that if two circles intersect in two points, then the line through their centers is the perpendicular bisector of their common chord. Exercise 5.2. (Coming a full circle). Given an arc of a circle, provide a method for completing the circle.

  5.1Area and circumference of a circle

  If we imagine a circle made of string, then we can cut the circular string at any point, stretch out the string along a line, and measure the length of the string. This length is called the circumference of the circle.

  Theorem 5.4 . The ratio of the circumference of any circle to its diameter is a constant, denoted by π .

  Proof . We will give an argument for this fact based on the following diagram showing concentric circles, of diameters d, d ′, each having an inscribed polygon with n sides.

  By the SAS Similarity Rule, the two triangles OAB and OA ′B ′ are similar, and thus

  But

  and as n becomes larger and larger, we expect the error in the above approximation to tend to 0, so that

  So the ratio of the circumference of a circle to its diameter is constant.

  Theorem 5.5 . The area of a circle of radius r is πr 2 .

  Proof . We inscribe a regular polygon with n sides inside the circle of radius r , and triangulate it by joining the center of the circle to the vertices of the polygon. By looking at the following picture, we will justify the expression πr 2 for the area of the circle.

  The area of the polygon is the area of the shaded parallelogram, and this is approximately the height (which differs from radius d /2 by a small amount) times the length of the base (which differs from half the circumference by a tiny amount). As n becomes larger and larger, we expect the errors above to go to 0, and so the area of the circle should be

  Remark 5.1 . (π /∈ℚ). Using tools from the subject of “calculus”, it can be shown that π

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  Geometry Revisited

  • H. S. M. Coxeter, S. L. Greitzer(Authors)
  • 1967(Publication Date)
  • American Mathematical Society

    (Publisher)

  32 PROPERTIES OF CIRCLES In terms of rectangular Cartesian coordinates, the square of the distance d between any two points (x, y ) and (a, b) is ( x - a)* + ( y - b)2. Therefore the power of ( x , y ) with respect to the circle with center (a, b) and radius r is 8 -r ' = ( x - a)' + ( y - b)2 - f. In particular, the circle itself, being the locus of points (x, y ) of power zero, has the equation (2.22) ( x - a)' + ( y - b)2 - f = 0. The same equation, in the form ( x - a)* + ( y - b)2 = f , expresses the circle as the locus of points whose distances from (a, b) have the constant value r. When this circle is expressed in the form (2.23) x 2 + y 2 -2 ~ ~ -2 b y + c = 0 (where c = a2 + bz - r 2 ) , the power of an arbitrary point ( x , y ) is again expressed by the left side of the equation, namely x2 + y' - ~ U X - 2by + C. Another circle having the same center (a, b) but a different radius has an equation of the same form with a different c, and any circle having a different center has an equation of the form (2.24) X ' + r* - ~ U ' X - 2b'y + C' = 0, where either a' # a or b' # b or both. We are thus free to use the equations (2.23) and (2.24) for the two non-concentric circles mentioned in Theorem 2.21. The locus of all points (x, y) whose powers with respect to these two circles are equal is x2 + y' - ~ U X - 2by + c = x2 + 7 - ~ U ' X - 2b'y + c'. Since x2 + y2 cancels, this locus is the line (a' - U ) X +. (b' - b ) y = $(c' - c). By choosing our frame of reference so that the x-axis joins the two centers, we may express the two circles in the simpler form (2.25) x2 + 7 - ~ U X + c = 0, x2 + y2 - 2a'x + C' = 0, where a' # a. Then the locus becomes c' - t x = 2(a' - a) - This line, being parallel to the y-axis, is perpendicular to the x-axis, RADICAL AXIS 33 which is the join of centers. Since the line can be defined geometrically in terms of the circles (as containing all points of equal power), we could have taken it to be the y-axis itself, as in Figure 2.2A.

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  Technical Mathematics with Calculus

  • Paul A. Calter, Michael A. Calter(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  We observe that, by the definition of a circle, the distance OP must be constant and equal to r. But, by the distance formula (Eq. 204), Squaring, we get a standard equation of a circle (also called the standard form of the equation). 218 Note that both have the same coefficient. Otherwise, the graph is not a circle. ◆◆◆ Example 24: Write, in standard form, the equation of a circle of radius 3, whose center is at the origin. Solution: ◆◆◆ ◆◆◆ Example 25: Find the center and radius of the circle with the equation Solution: We recognize this as the equation of a circle whose center is at the origin. Finding the radius, ◆◆◆ ◆◆◆ Example 26: An Application. Circular Arches. Portions of a circle are commonly used for arches, as shown in Fig. 22–38. (a) Write the equation for the circle in the r  164  8 x 2  y 2  64 x 2  y 2  3 2  9 x 2 and y 2 x 2  y 2  r 2 Standard Equation, Circle of Radius r : Center at Origin OP  4 x 2  y 2  r O y x r r r P(x, y) FIGURE 22–37 Circle with center at origin. FIGURE 22–38 Arches based on the circle. 0 Horseshoe or Moorish r 0 Circular r 0 Semicircular r Pointed or gothic r r 0 0 696 Chapter 22 ◆ Analytic Geometry horseshoe arch ( also called a Moorish arch), Fig. 22–39. Take the origin at the cen- ter of the circle. (b) Use your equation to find h. Solution: (a) The circle has a radius of 2.76 m, so substituting into the equation for a circle with center at the origin, Eq. 218, gives (b) The x coordinate of point P is half the width of the doorway, or 2.20 m. Substituting into our equation gives Solving for y, We take the negative value, since P is below the center of the circle. Finally, ◆◆◆ Standard Equation of a Circle: Center Not at the Origin Figure 22–40 shows a circle whose center has the coordinates (h, k). We can think of the difference between this circle and the one in Fig. 22–37 as having its center moved or translated h units to the right and k units upward.

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