<p> Worksheet on Equations of Circles The equation of a circle is based upon its definition and the Pythagorean theorem. Since a locus for the circle is “the set of points equidistant from a single point called the center” and that distance between the center and the point on the circle is a constant radius, then we can consider the radius to be the hypotenuse of all right triangles whose sides are the differences between the ( x , y ) coordinates of the center and the ( x , y ) coordinates of any point on the circle as illustrated in the graph to the right.</p><p>Since the radius is constant and the center is equi- distant from all points on the circle, then we can apply our distance formula to develop an equation for the circle.</p><p>If a circle has a radius of r and center coordinates of ( h , k ), and any point on the circle has coordinates of ( x , y ), then the standard form of the equation that relates these elements is as follows:  x  h  2   y  k  2  r 2 In the diagram above, the center of the circle occurs at ( 6 , 4 ) with a radius of 5 units. Hence, the standard form of the equation of this circle would be</p><p>2 2 2  x  6   y  4  5     If we were to expand this equation and put it into general form, then it would be x 2 + y 2 – 12 x – 8 y + 27 = 0 Your Turn: Sample problems: A. Identify the center and radius of each equation below. Eq. # Standard Graphing Equation Radius Center 1 ( x – 2 ) 2 + ( y – 5 ) 2 = 144 2 ( x – 0 ) 2 + ( y – 7 ) 2 = 625 3 ( x – 15 ) 2 + ( y + 8 ) 2 = 289 4 ( x + 11 ) 2 + ( y + 6 ) 2 = 324 5 ( x + 4 ) 2 + ( y – 9 ) 2 = 196 B. Given each center and radius below, write the equation of the circle in standard graphing form. Eq. # Center Radius Standard Graphing Equation 1 ( -5 , 12 ) 13 2 ( 9 , -12 ) 15 3 ( -7 , 0 ) 10 4 ( 11 , 14 ) 9 5 ( -6 , -8 ) 5 Now, let’s reverse the process. Suppose that we had an equation in general form. How could we convert it into standard form suitable for graphing? </p><p>That process would involve completing the square of the binominals for ( x – h ) and ( y – k ) .</p><p>Consider the general equation of the circle defined by x2 + y2 – 14x – 18y + 105 = 0.</p><p>First, group the terms by variable in brackets: [ x2– 14x ] + [ y2 – 18y ] + 105 = 0.</p><p>Then, take half of the linear term and square it for each variable: = 49 =81.</p><p>Add AND Subtract the appropriate quantity inside each pair of brackets so that we are adding ZERO to the equation thereby changing only its form but preserving its value: [ x2– 14x + 49 – 49 ] + [ y2 – 18y + 81 – 81 ] + 105 = 0. Group appropriate terms into expanded binomials within the brackets. [ ( x2– 14x + 49 ) – 49 ] + [ ( y2 – 18y + 81 ) – 81 ] + 105 = 0. Factor the trinomials into binomials squared and collect the remaining terms: ( x – 7 ) 2 + ( y – 9 ) 2 – 49 – 81 + 105 = 0. Combine the constants: ( x – 7 ) 2 + ( y – 9 ) 2 – 25 = 0 Add the total to both sides of the equation to produce the graphing form of the equation of a circle. ( x – 7 ) 2 + ( y – 9 ) 2 = 25 = 5 2 </p><p>Your turn: transform the following equation into standard graphing form, then identify the center and radius of the circle, and graph the result.</p><p> x2 + y2 – 6x – 8y – 75 = 0</p>