Stitz-Zeager_College_Algebra_e-book

5 hyperbolas 75 433 hyperbolas in the denition of an

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Unformatted text preview: btain equations like (x − 3)2 + (y + 1)2 = 0 or (x − 3)2 + (y + 1)2 = −1, neither of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if 402 Hooked on Conics any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula, Equation 1.2, in conjunction with the ideas presented so far in this section. Example 7.2.4. Write the standard equation of the circle which has (−1, 3) and (2, 4) as the endpoints of a diameter. Solution. We recall that a diameter of a circle is a line segment containing the center and two points on the circle. Plotting the given data yields y 4 3 r (h, k) 2 1 −2 −1 1 2 3 x Since the given points are endpoints of a diameter, we know their midpoint (h, k ) is the center of the circle. Equation 1.2 gives us (h, k ) = = = x1 + x2 y1 + y2 , 2 2 −1 + 2 3 + 4 , 2 2 17 , 22 The diameter of the circle is the distance between the given points, so we know that half of the distance is the radius. Thus, r= = = = √ Finally, since 10 2 2 = 1 (x2 − x1 )2 + (y2 − y1 )2 2 1 (2 − (−1))2 + (4 − 3)2 2 1√ 2 3 + 12 2...
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