Mathematics
Equation of a circle
The equation of a circle in the Cartesian coordinate system is given by (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the center of the circle and r is the radius. This equation allows for the representation and analysis of circles in the coordinate plane, providing a clear relationship between the circle's center, radius, and its equation.
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8 Key excerpts on "Equation of a circle"
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
In 2 2 4 6 x 4 2 2 2 4 6 y The circle centered at (2, 1) with radius 5. other words, the circle centered at (2, 1) with radius 5 is the set of points (x, y) satisfying the equation (x - 2) 2 + (y - 1) 2 = 5. Squaring both sides, we can more conveniently describe this circle as the set of points (x, y) such that (x - 2) 2 + (y - 1) 2 = 25. Using the same reasoning as in the example above, we get the following more general result: For example, the equation (x - 3) 2 + (y + 5) 2 = 7 describes the cir- cle in the xy -plane with radius √ 7 cen- tered at (3, -5). Equation of a circle The circle with center (h, k) and radius r is the set of points (x, y) satisfying the equation (x - h) 2 + (y - k) 2 = r 2 . Sometimes the Equation of a circle may be in a form in which the radius and center are not obvious. You may then need to complete the square to find the radius and center. The following example illustrates this procedure: example 7 Find the radius and center of the circle in the xy -plane described by x 2 + 4x + y 2 - 6y = 12. solution Completing the square, we have Here the completing- the-square technique has been applied separately to the x and y variables. 12 = x 2 + 4x + y 2 - 6y = (x + 2) 2 - 4 + (y - 3) 2 - 9 = (x + 2) 2 + (y - 3) 2 - 13. Adding 13 to the first and last sides of the equation above shows that (x + 2) 2 + (y - 3) 2 = 25. Thus we have a circle with radius 5 centered at (-2, 3). section 2.3 Quadratic Expressions and Conic Sections 81 Ellipses Ellipse Stretching a circle horizontally and/or vertically produces a curve called an ellipse. example 8 Find an equation describing the ellipse in the xy -plane produced by stretching a circle of radius 1 centered at the origin horizontally by a factor of 5 and vertically by a factor of 3.
eBook - PDF- Mark D. Turner, Charles P. McKeague(Authors)
- 2016(Publication Date)
- XYZ Textbooks(Publisher)
14.1 The Distance Formula and Circles 1009 Circles Because of their perfect symmetry, circles have been used for thousands of years in many disciplines, including art, science, and religion. The photograph on the left is of Stonehenge, a 4,500-year-old site in England. The arrangement of the stones is based on a circular plan that is thought to have both religious and astronomical significance. More recently, the design shown in the photo on the right began appearing in agricultural fields in England in the 1990s. Whoever made these designs chose the circle as their basic shape. We can model circles very easily in algebra by using equations that are based on the distance formula. Proof By definition, all points on the circle are a distance r from the center ( h , k ). If we let ( x , y ) represent any point on the circle, then ( x , y ) is r units from ( h , k ). Applying the distance formula, we have r = √ — ( x − h ) 2 + ( y − k ) 2 Squaring both sides of this equation gives the equation of the circle: ( x − h ) 2 + ( y − k ) 2 = r 2 We can use the circle theorem to find the Equation of a circle, given its center and radius, or to find its center and radius, given the equation. Find the equation of the circle with center at ( − 3, 2) having a radius of 5. SOLUTION We have ( h , k ) = ( − 3, 2) and r = 5. Applying our theorem for the Equation of a circle yields [ x − ( − 3)] 2 + ( y − 2) 2 = 5 2 ( x + 3) 2 + ( y − 2) 2 = 25 Give the equation of the circle with radius 3 whose center is at the origin. SOLUTION The coordinates of the center are (0, 0), and the radius is 3. The equation must be ( x − 0) 2 + ( y − 0) 2 = 3 2 x 2 + y 2 = 9 We can see from Example 4 that the equation of any circle with its center at the origin and radius r will be x 2 + y 2 = r 2 The equation of the circle with center at ( h , k ) and radius r is given by ( x − h ) 2 + ( y − k ) 2 = r 2 THEOREM Circle Theorem EXAMPLE 3 EXAMPLE 4 © Masterfile © Masterfile
eBook - PDF- Paul A. Calter, Michael A. Calter(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
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. Our derivation is simi- lar to the preceding one. Squaring, we get the following equation: 219 ◆◆◆ Example 27: Write, in standard form, the Equation of a circle of radius 5 whose center is at Solution: We substitute into Eq. 219 with ◆◆◆ ◆◆◆ Example 28: Find the radius and the coordinates of the center of the circle Solution: We see that so the radius r is 4. Also, since then and since then So the center is at as shown in Fig. 22–41. ◆◆◆ (5, 3) k 3 y k y 3 h 5 x h x 5 r 2 16, (x 5) 2 (y 3) 2 16. (x 3) 2 (y 2) 2 25 (x 3) 2 [y (2)] 2 5 2 r 5, h 3, and k 2. (3, 2). (x h) 2 (y k) 2 r 2 Standard Equation, Circle of Radius r : Center at (h, k ) CP r 4 (x h) 2 (y k) 2 h 10.3 2.76 1.67 5.87 m y 1.67 m y 2 2.78 y 2 (2.76) 2 (2.20) 2 (2.20) 2 y 2 (2.76) 2 x 2 y 2 (2.76) 2 (x + 5) 2 + (y − 3) 2 = 16 0 2 4 6 x y −2 −4 −6 C(−5, 3) r = 4 FIGURE 22–41 FIGURE 22–39 Horseshoe arch over a doorway. y 4.40 m 10.3 m 2.76 m h x P 0 y x P(x, y) r C(h, k) h k FIGURE 22–40 Circle with center at (h, k). Section 3 ◆ The Circle 697 It is easy to get the signs of h and k wrong. In Example 28, do not take Graphing a Circle by Calculator A computer algebra system may be able to graph an equation entered in implicit form. However, to graph a circle with a graphics calculator and most graphics utili- ties, we must put an equation in explicit form by solving for y. When we do that we get two functions, one for the upper half of the circle and another for the lower half. To get the complete circle, we must graph both. ◆◆◆ Example 29: Graph the circle Solution: We first solve for y.
eBook - PDF- Sheldon Axler(Author)
- 2011(Publication Date)
- Wiley(Publisher)
Thus to find the area inside a circle, we must first find the radius of the circle. Finding the radius sometimes requires a preliminary algebraic manipulation such as completing the square, as shown in the following example. example 3 Consider the circle described by the equation x 2 - 8x + y 2 + 6y = 4. (a) Find the center of this circle. (b) Find the radius of this circle. (c) Find the circumference of this circle. (d) Find the area inside this circle. solution To obtain the desired information about the circle, we put its equation in a standard form. This can be done by completing the square: 4 = x 2 - 8x + y 2 + 6y = (x - 4) 2 - 16 + (y + 3) 2 - 9 = (x - 4) 2 + (y + 3) 2 - 25. Adding 25 to the first and last sides above shows that the circle is described by the equation (x - 4) 2 + (y + 3) 2 = 29. (a) The equation above shows that the center of the circle is (4, -3). (b) The equation above shows that the radius of the circle is √ 29. Do not make the mis- take of thinking that this circle has radius 29. (c) Because the circle has radius √ 29, its circumference is 2 √ 29π . (d) Because the circle has radius √ 29, its area is 29π . In Section 2.3 we saw that the ellipse x 2 25 + y 2 9 = 1 is obtained from the circle of radius 1 centered at the origin by stretching horizontally by a factor of 5 and stretching vertically by factor of 3. 104 chapter 2 Combining Algebra and Geometry 5 5 1 1 x 3 3 y 5 5 x 3 3 y Stretching horizontally by a factor of 5 and vertically by a factor of 3 transforms the circle on the left into the ellipse on the right. Because 5 · 3 = 15, the Area Stretch Theorem tells us that the area inside In addition to discovering that the orbits of planets are ellipses, Kepler also discovered that a line joining a planet to the sun sweeps out equal areas in equal times. this ellipse equals 15 times the area inside the circle of radius 1. Because the area inside a circle of radius 1 is π , we conclude that the area inside this ellipse is 15π .
eBook - PDF- 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.
eBook - PDFThe A to Z of Mathematics
A Basic Guide
- Thomas H. Sidebotham(Author)
- 2003(Publication Date)
- Wiley-Interscience(Publisher)
It is obvious that grandmother’s formula only works for children. For example, a person aged 50 would need −8.5 hours of sleep! References: Balancing an Equation, Formula, Perimeter, Rounding, Solving an Equation, Square Root. 74 CIRCLE CHORD A chord is a straight line segment joining two points on a curve. The curve we will study is the circle. In figure a, AB is a chord of the circle. The perpendicular bisector, or mediator, of the chord of a circle always passes through the center of the circle. A A (a) B O (b) The longest possible chord, which passes through the center O of the circle, is called the diameter (see figure b). When one or both ends of a chord are extended the resulting line is called the secant of the circle. Two examples are shown in figure c. (c) References: Center of a Circle, Circle, Mediator, Perpendicular Bisector. CIRCLE This is the set of points in a plane that are equidistant from a fixed point O (see figure a). The fixed point is called the center of the circle. The concept of a circle is explained here. O G (a) Amanda is schooling up her horse George for a show. She stands in one place, at point O , holding the end of a rope. The other end of the rope is tied to George, who is at G. As George trots around her, Amanda turns so that she is always facing him, but CIRCLE GRAPH 75 keeps the rope tight and the same length. George is always the same distance from Amanda, all the positions of George as Amanda does a full turn form a circle. The different parts of the circle are illustrated here in figure b, but more information is available about them under their own separate headings. Tangent Sem m micircle m Sector Segment Arc Radius Diameter Chord Center (b) If the radius of a circle is R, then the area of the circle is given by the formula A = πR 2 . The circumference of the circle is given by the formula C = 2πR, or C = πD, where D is the diameter of the circle. The circumference of a circle is its perimeter. Example.
eBook - PDF- Doug French(Author)
- 2004(Publication Date)
- Continuum(Publisher)
Stretches have the much more dramatic effect of transforming circles into either ellipses or circles of different sizes in the special case of enlargement. The right-hand diagram of Figure 10.10 shows the same initial circle stretched by a factor of 3 in the x direction and factor 2 in the y direction. In each case a pair of parametric equations, given in terms of a parameter B, is shown below the diagram, and this provides a simple route to the Cartesian equation through Pythagoras' theorem in its important trigonometrical form: cos 2 0 + sin 2 0=1. The importance of circles as geometric figures is abundantly obvious. Being able to represent them in an algebraic form with equations provides another tool for dealing with problems while providing a crucial link with Pythagoras' theorem which is also of such fundamental importance in mathematics. The simple action of stretching a circle leads to the ellipse, a curve which has many interesting properties and applications, which provide the focus of the next section. 128 Teaching and Learning Geometry THE ELLIPSE One way of generating an ellipse as a locus, often referred to as the 'slipping ladder', is shown in the left-hand diagram of Figure 10.11. The line AB-the ladder-is of constant length, with the point B moving along the x axis and the point A moving along the y axis. The locus of the point P is an ellipse. This can be proved easily by letting AP = a and BP = b. Then, if the ladder makes an angle 0 with the ground and P has coordinates (x,y), we have x = a cos 9 and y = b sin 0, the parametric equations of an ellipse, where 2a and 2b are the lengths of the major and minor axes respectively. Besides the locus it is also interesting to observe the envelope of lines, shown in the right-hand diagram of Figure 10.11, created by the positions of the ladder as it moves.
eBook - ePub- Dan Pedoe(Author)
- 2013(Publication Date)
- Dover Publications(Publisher)
IICIRCLESThe circle, in various guises, occurs in many branches of mathematics. In this chapter we shall consider those properties which have the habit of appearing in different contexts. Our methods will be mixed. We shall use Euclidean geometry or analytical geometry, according to our needs at the time. We begin with the consideration of certain one-to-one mappings of the plane onto itself, and the product of these mappings. (See 0.11.) Later on we study inversion, which is only one-to-one onto if we cut out a point of the plane, or better still, add a point to the plane.12.1 The nine-point circleWe have already encountered this circle (§7.3 ). It is a circle which passes through nine points intimately connected with any given triangle ABC. (We shall not be using vector methods, and we shall use the customary notation for the points connected with a triangle). The nine-point circle is the first really exciting one to appear in any course on Euclidean geometry which goes far enough.Fig. 12.1To prove the existence of the nine-point circle of a triangle, we make use of a one-to-one mapping of the Euclidean plane onto itself. This mapping is called a central dilatation. We choose a point V, and call it the center of the dilatation (Fig 12.1 ).If P is any point in the plane its image, or map P′ is found by choosing the unique point P′ which lies on VP, is on the same side of V as is P, and is such that where k is a fixed real positive number. If P = V, we define V′ to be V.This mapping is evidently one-to-one. It is also called a central similarity. It will occur again later (§41.5 ) as one of the fundamental mappings of the Euclidean plane onto itself. Here we are only interested in the case when k is positive, and when the point P moves on a circle, center O (Fig. 12.2 ). If O′ is the map of O, then the triangles VOP and VO′P′ are similar, so that , Since O is a fixed point, and | OP | = r, the radius of the circle on which P moves, we deduce that P′ moves on a circle, center O′ of radius kr
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