Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 855: 75

Answer

The equation of a circle with center $\left( 3,-5 \right)$ and radius $3$ is${{\left( x-3 \right)}^{2}}+{{\left( y+5 \right)}^{2}}=9$

Work Step by Step

From the figure we see that the circle touches the y-axis at $\left( 0,-5 \right)$, Since the radius is the distance between $\left( 0,-5 \right)$ and $\left( 3,-5 \right)$, $\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( -5-\left( -5 \right) \right)}^{2}}}$ $r=3$ Put the value of radius $3$ and center coordinate $\left( 3,-5 \right)$ in the standard equation of the circle, $\begin{align} & {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} \\ & {{\left( x-\left( 3 \right) \right)}^{2}}+{{\left( y-\left( -5 \right) \right)}^{2}}={{\left( 3 \right)}^{2}} \\ & {{\left( x-3 \right)}^{2}}+{{\left( y+5 \right)}^{2}}=9 \end{align}$ Therefore, the equation of the circle with center $\left( 3,-5 \right)$ and radius $3$ is ${{\left( x-3 \right)}^{2}}+{{\left( y+5 \right)}^{2}}=9$.
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