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: 59

Answer

See below:
1569143294

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 of the circle can be written as: $\begin{align} & {{x}^{2}}-2x+{{y}^{2}}-15=0 \\ & \left( {{x}^{2}}-2x \right)+{{y}^{2}}=15 \\ & \left( {{x}^{2}}-2x+1 \right)+{{y}^{2}}=15+1 \\ & {{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=16 \end{align}$ So, the equation of the circle in the standard form is given as: ${{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{4}^{2}}$ Now, compare this equation to the standard form, to get the value of $h=1,k=0,\,\,\text{ and }\,\,r=4$. Therefore, the center is $\left( 1,0 \right)$ and the radius is $4$ units.
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