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: 64

Answer

See below:
1569143429

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: $\begin{align} & {{x}^{2}}+{{y}^{2}}+3x+5y+\frac{9}{4}=0 \\ & \left( {{x}^{2}}+3x \right)+\left( {{y}^{2}}+5y \right)=-\frac{9}{4} \\ & \left( {{x}^{2}}+3x+\frac{9}{4} \right)+\left( {{y}^{2}}+5y+\frac{25}{4} \right)=-\frac{9}{4}+\frac{9}{4}+\frac{25}{4} \\ & {{\left( x+\frac{3}{2} \right)}^{2}}+{{\left( y+\frac{5}{2} \right)}^{2}}=\frac{25}{4} \end{align}$ Thus, the equation of the circle in the standard form is given as: ${{\left( x-\left( -\frac{3}{2} \right) \right)}^{2}}+{{\left( y-\left( \frac{5}{2} \right) \right)}^{2}}={{\left( \frac{5}{2} \right)}^{2}}$ Now compare this equation to the standard form, to get the value of $h=-\frac{3}{2},k=-\frac{5}{2}\,,\,\text{ and }\,\,r=\frac{5}{2}$. Therefore, the center is $\left( -\frac{3}{2},-\frac{5}{2} \right)$ and the radius is $\frac{5}{2}$ units.
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