52 CHAPTER 1 Graphs Now Work PROBLEM 33 Figure 76 ( ) ( ) + + − = x y 2 3 1 2 2 y 4 1 23 1 (22, 3) (22, 2) (23, 3) (21, 3) (22, 4) x (a) 3 5 1 2 (x 1 2)2 Y 2 5 3 2 1 2 (x 1 2)2 Y 1 5 3 1 (b) (c) 25 0 Solution To find the equation of a circle, we need to know its center and its radius. Here, the center is ( ) − 1, 2 . Since the point ( ) − 4, 2 is on the graph, the radius r will equal the distance from ( ) − 4, 2 to the center ( ) − 1, 2 . See Figure 77. Thus, ( ) ( ) [ ] = − + − − − = = r 4 1 2 2 9 3 2 2 The standard form of the equation of the circle is ( ) ( ) − + + = x y 1 2 9 2 2 Eliminate the parentheses and rearrange the terms to get the general equation + − + − = x y x y 2 4 4 0 2 2 Finding the General Equation of a Circle Find the general equation of the circle whose center is ( ) − 1, 2 and whose graph contains the point ( ) − 4, 2. EXAMPLE 5 Now Work PROBLEM 17 Figure 77 ( ) ( ) − + + = x y 1 2 9 2 2 x r y 3 5 25 (1, 22) (4, 22) Overview The discussion in Sections 1.5 and 1.6 about lines and circles deals with two problems that can be generalized as follows: • Given an equation, classify it and graph it. • Given a graph, or information about a graph, find its equation. This text deals with both problems. We shall study various equations, classify them, and graph them. ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1.6 Assess Your Understanding 1. To complete the square of + x x 10 , 2 you would ( add / subtract ) the number . (p. A29) 2. Use the Square Root Method to solve the equation ( ) − = x 2 9. 2 (p. A49) 1. Now Work 1. Modeling 1.ExplainingConcepts Calculus Preview 1.InteractiveFigure Figure 76(b) displays the graph using a TI-84 Plus CE graphing calculator. Figure 76(c) displays the graph using Geogebra.
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