College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding: 46

Answer

$\color{blue}{(x+3)^2+(y-3)^2=9}$

Work Step by Step

The given circle has its center at $(-3, 3)$. RECALL: The standard form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$, where $(h, k)$ is the center and $r$ is the radius. Thus, using the standard form above with the center of $(-3, 3)$, the tentative equation of the circle whose graph is given is: $(x-(-3))^2 + (y-3)^2=r^2 \\(x+3)^2+(y-3)^2=r^2$ The point $(-3, 0)$ is a point on the circle. This means that the x and y coordinates of this point satisfy the equation of the circle. Substitute the x and y coordinates of this point into the tentative equation above to obtain: $(x+3)^2+(y-3)^2=r^2 \\(-3+3)^2+(0-3)^2=r^2 \\0^2 + (-3)^2=r^2 \\0+9=r^2 \\9=r^2$ Therefore, the equation of the circle is: $\color{blue}{(x+3)^2+(y-3)^2=9}$
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