Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises - Page 135: 106

Answer

$a)$ This equation represents a point $b)$ There is no need to find the center or radius. There is also no need to sketch a graph

Work Step by Step

$x^{2}+y^{2}-6x-10y+34=0$ Take $34$ to the right: $x^{2}+y^{2}-6x-10y=-34$ On the left side, group the terms with $x$ together and the terms with $y$ together: $(x^{2}-6x)+(y^{2}-10y)=-34$ Complete the square for each group. To do so, add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and do it for both groups. For the first group, $b=-6$ and for the second group, $b=-10$ $\Big[x^{2}-6x+\Big(\dfrac{-6}{2}\Big)^{2}\Big]+\Big[y^{2}-10y+\Big(\dfrac{-10}{2}\Big)^{2}\Big]=-34+9+25$ $(x^{2}-6x+9)+(y^{2}-10y+25)=0$ Now, each group is a perfect square trinomial. Factor them both: $(x-3)^{2}+(y-5)^{2}=0$ Since the part of the equation that represents the radius of the circle is $0$, then this equation represents a point. $b)$ Since the original equation represents a point, there is no need to find the center or radius. There is also no need to sketch a graph
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