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 280: 60

Answer

See below:
1569143320

Work Step by Step

We know that for a circle, the standard form 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 $r$. So, the equation can be written as: $\begin{align} & {{x}^{2}}+{{y}^{2}}-6y-7=0 \\ & {{x}^{2}}+\left( {{y}^{2}}-6y \right)=7 \\ & {{x}^{2}}+\left( {{y}^{2}}-6y+9 \right)=7+9 \\ & {{\left( x-0 \right)}^{2}}+{{\left( y-3 \right)}^{2}}=16 \end{align}$ So, the equation of the circle in the standard form is: ${{\left( x-0 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{4}^{2}}$ Now, compare this equation to the standard form, to get the value of $h=0,k=1\,,\,\text{ and }\,\,r=4$. Therefore, the center is $\left( 0,3 \right)$ and the radius is $4$ units.
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