216 CHAPTER 2 Graphs and Functions An Application Seismologists can locate the epicenter of an earthquake by determining the intersection of three circles. The radii of these circles represent the distances from the epicenter to each of three receiving stations. The centers of the circles represent the receiving stations. EXAMPLE 4 Finding the Center and Radius by Completing the Square Give the center and radius of the circle with equation 2x2 + 2y2 - 6x + 10y = 1. SOLUTION To complete the square, the coefficient of the x2-term and that of the y2-term must be 1. In this case they are both 2, so begin by dividing each side by 2. 2x2 + 2y2 - 6x + 10y = 1 x2 + y2 - 3x + 5y = 1 2 Divide by 2. 1x2 - 3x 2 + 1y2 + 5y 2 = 1 2 ax2 - 3x + 9 4b + ay2 + 5y + 25 4 b = 1 2 + 9 4 + 25 4 ax - 3 2b 2 + ay + 5 2b 2 = 9 Factor and add. 2 4 + 9 4 + 25 4 = 36 4 = 9 ax - 3 2b 2 + c y - a- 5 2b d 2 = 32 Center-radius form The equation represents a circle with center A3 2 , - 5 2B and radius 3. S Now Try Exercise 31. Complete the square for both x and y; C 1 2 1-32D 2 = 9 4 and C 1 2 152D 2 = 25 4 Rearrange and regroup terms in anticipation of completing the square. EXAMPLE 5 Determining Whether a Graph Is a Point or Nonexistent The graph of the equation x2 + 10x + y2 - 4y + 33 = 0 either is a point or is nonexistent. Which is it? SOLUTION x2 + 10x + y2 - 4y + 33 = 0 x2 + 10x + y2 - 4y = -33 Subtract 33. Think: c 1 2 1102d 2 = 25 and c 1 2 1-42d 2 = 4 1x2 + 10x + 252 + 1y2 - 4y + 42 = -33 + 25 + 4 Complete the square. 1x + 522 + 1y - 222 = -4 Factor and add. Because -4 60, there are no ordered pairs 1x, y2, with x and y both real numbers, satisfying the equation. The graph of the given equation is nonexistent—it contains no points. (If the constant on the right side were 0, the graph would consist of the single point 1-5, 22.) S Now Try Exercise 35. Prepare to complete the square for both x and y.
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