Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 281: 88

Answer

The given statement is does not make sense.

Work Step by Step

The standard form of a circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$, where the center is $\left( h,k \right)$ and the radius is $r$. The given equation can be written as: ${{\left( x-\left( -1 \right) \right)}^{2}}+{{\left( y-5 \right)}^{2}}={{\left( 2\sqrt{-1} \right)}^{2}}$ Now, comparing the equation to the standard form, we get $\left( h,k \right)=\left( -1,5 \right)$ and $r=2\sqrt{-1}$. The equation does not represent a circle. It represents a set of imaginary points. Thus, the equation ${{\left( x+1 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=-4$ cannot be used to identify the circle center and radius.
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