SECTION 1.6 Circles 49 Figure 72 Unit circle + = x y 1 2 2 1 21 1 21 y x (0,0) THEOREM The standard form of an equation of a circle of radius r with center at the origin ( ) 0, 0 is + = x y r 2 2 2 DEFINITION Unit Circle If the radius = r 1, the circle whose center is at the origin is called the unit circle and has the equation + = x y 1 2 2 See Figure 72. Notice that the graph of the unit circle is symmetric with respect to the x -axis, the y -axis, and the origin. Writing the Standard Form of the Equation of a Circle Write the standard form of the equation of the circle with radius 5 and center ( ) −3, 6 . Solution EXAMPLE 1 Substitute the values = = − r h 5, 3, and = k 6 into equation (1). ( ) ( ) ( ) ( ) ( ) [ ] ( ) − + − = − − + − = + + − = x h y k r x y x y 3 6 5 3 6 25 2 2 2 2 2 2 2 2 Equation (1) Now Work PROBLEM 11 2 Graph a Circle by Hand and by Using a Graphing Utility In Words The symbol ± is read “plus or minus.” It means to add and subtract the quantity following the ± symbol. For example, ±5 2 means “ − = 5 2 3 or + = 5 2 7. ” Figure 73 ( ) ( ) + + − = x y 3 2 16 2 2 (–3, 6) (–7, 2) (–3, 2) y 6 2 –5 –10 4 (–3, –2) (1, 2) x Graphing a Circle by Hand and by Using a Graphing Utility Graph the equation: ( ) ( ) + + − = x y 3 2 16 2 2 EXAMPLE 2 Solution The given equation is in the standard form of an equation of a circle. To graph the circle by hand, compare the equation to equation (1). The comparison gives information about the circle. x y x y x h y k r 3 2 16 3 2 4 2 2 2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) + + − = − − + − = ↑ ↑ ↑ − + − = We see that = − = h k 3, 2, and = r 4. The circle has center ( ) −3, 2 and radius 4. To graph this circle, first plot the center ( ) −3, 2 . Since the radius is 4, locate four points on the circle by plotting points 4 units to the left, to the right, up, and down from the center.These four points can then be used as guides to obtain the graph. See Figure 73. To graph a circle on some graphing utilities, we must write the equation in the form y x expression involving . { } = We must solve for y in the equation x y y x y x y x 3 2 16 2 16 3 2 16 3 2 16 3 2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) + + − = − = − + − = ± − + = ± − + Subtract x 3 2 ( ) + from both sides. Use the Square Root Method. Add 2 to both sides. (continued)
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